This error suggests that the expression arm corresponding to the noted pattern will never be reached as for all possible values of the expression being matched, one of the preceding patterns will match.
This means that perhaps some of the preceding patterns are too general, this one is too specific or the ordering is incorrect.
For example, the following match block has too many arms:
match foo { Some(bar) => {/* ... */} None => {/* ... */} _ => {/* ... */} // All possible cases have already been handled }
match blocks have their patterns matched in order, so, for example, putting
a wildcard arm above a more specific arm will make the latter arm irrelevant.
Ensure the ordering of the match arm is correct and remove any superfluous arms.
This error indicates that an empty match expression is invalid because the type it is matching on is non-empty (there exist values of this type). In safe code it is impossible to create an instance of an empty type, so empty match expressions are almost never desired. This error is typically fixed by adding one or more cases to the match expression.
An example of an empty type is enum Empty { }. So, the following will work:
fn foo(x: Empty) { match x { // empty } }
However, this won't:
fn foo(x: Option<String>) { match x { // empty } }
Not-a-Number (NaN) values cannot be compared for equality and hence can never match the input to a match expression. So, the following will not compile:
const NAN: f32 = 0.0 / 0.0; match number { NAN => { /* ... */ }, // ... }
To match against NaN values, you should instead use the is_nan() method in a
guard, like so:
match number { // ... x if x.is_nan() => { /* ... */ } // ... }
This error indicates that the compiler cannot guarantee a matching pattern for one or more possible inputs to a match expression. Guaranteed matches are required in order to assign values to match expressions, or alternatively, determine the flow of execution.
If you encounter this error you must alter your patterns so that every possible
value of the input type is matched. For types with a small number of variants
(like enums) you should probably cover all cases explicitly. Alternatively, the
underscore _ wildcard pattern can be added after all other patterns to match
"anything else".
Patterns used to bind names must be irrefutable, that is, they must guarantee
that a name will be extracted in all cases. If you encounter this error you
probably need to use a match or if let to deal with the possibility of
failure.
This error indicates that the bindings in a match arm would require a value to
be moved into more than one location, thus violating unique ownership. Code like
the following is invalid as it requires the entire Option<String> to be moved
into a variable called op_string while simultaneously requiring the inner
String to be moved into a variable called s.
let x = Some("s".to_string()); match x { op_string @ Some(s) => ... None => ... }
See also Error 303.
Names bound in match arms retain their type in pattern guards. As such, if a name is bound by move in a pattern, it should also be moved to wherever it is referenced in the pattern guard code. Doing so however would prevent the name from being available in the body of the match arm. Consider the following:
match Some("hi".to_string()) { Some(s) if s.len() == 0 => // use s. ... }
The variable s has type String, and its use in the guard is as a variable of
type String. The guard code effectively executes in a separate scope to the
body of the arm, so the value would be moved into this anonymous scope and
therefore become unavailable in the body of the arm. Although this example seems
innocuous, the problem is most clear when considering functions that take their
argument by value.
match Some("hi".to_string()) { Some(s) if { drop(s); false } => (), Some(s) => // use s. ... }
The value would be dropped in the guard then become unavailable not only in the body of that arm but also in all subsequent arms! The solution is to bind by reference when using guards or refactor the entire expression, perhaps by putting the condition inside the body of the arm.
In a pattern, all values that don't implement the Copy trait have to be bound
the same way. The goal here is to avoid binding simultaneously by-move and
by-ref.
This limitation may be removed in a future version of Rust.
Wrong example:
struct X { x: (), } let x = Some((X { x: () }, X { x: () })); match x { Some((y, ref z)) => {}, None => panic!() }
You have two solutions:
Solution #1: Bind the pattern's values the same way.
struct X { x: (), } let x = Some((X { x: () }, X { x: () })); match x { Some((ref y, ref z)) => {}, // or Some((y, z)) => {} None => panic!() }
Solution #2: Implement the Copy trait for the X structure.
However, please keep in mind that the first solution should be preferred.
#[derive(Clone, Copy)] struct X { x: (), } let x = Some((X { x: () }, X { x: () })); match x { Some((y, ref z)) => {}, None => panic!() }
The value of statics and constants must be known at compile time, and they live for the entire lifetime of a program. Creating a boxed value allocates memory on the heap at runtime, and therefore cannot be done at compile time.
Initializers for constants and statics are evaluated at compile time. User-defined operators rely on user-defined functions, which cannot be evaluated at compile time.
Bad example:
use std::ops::Index; struct Foo { a: u8 } impl Index<u8> for Foo { type Output = u8; fn index<'a>(&'a self, idx: u8) -> &'a u8 { &self.a } } const a: Foo = Foo { a: 0u8 }; const b: u8 = a[0]; // Index trait is defined by the user, bad!
Only operators on builtin types are allowed.
Example:
const a: &'static [i32] = &[1, 2, 3]; const b: i32 = a[0]; // Good!
Static and const variables can refer to other const variables. But a const
variable cannot refer to a static variable. For example, Y cannot refer to X
here:
static X: i32 = 42; const Y: i32 = X;
To fix this, the value can be extracted as a const and then used:
const A: i32 = 42; static X: i32 = A; const Y: i32 = A;
Constants can only be initialized by a constant value or, in a future version of Rust, a call to a const function. This error indicates the use of a path (like a::b, or x) denoting something other than one of these allowed items. Example:
const FOO: i32 = { let x = 0; x }; // 'x' isn't a constant nor a function!
To avoid it, you have to replace the non-constant value:
const FOO: i32 = { const X : i32 = 0; X }; // or even: const FOO: i32 = { 0 }; // but brackets are useless here
The only functions that can be called in static or constant expressions are
const functions. Rust currently does not support more general compile-time
function execution.
See RFC 911 for more details on the design of const fns.
Blocks in constants may only contain items (such as constant, function definition, etc...) and a tail expression. Example:
const FOO: i32 = { let x = 0; x }; // 'x' isn't an item!
To avoid it, you have to replace the non-item object:
const FOO: i32 = { const X : i32 = 0; X };
References in statics and constants may only refer to immutable values. Example:
static X: i32 = 1; const C: i32 = 2; // these three are not allowed: const CR: &'static mut i32 = &mut C; static STATIC_REF: &'static mut i32 = &mut X; static CONST_REF: &'static mut i32 = &mut C;
Statics are shared everywhere, and if they refer to mutable data one might violate memory safety since holding multiple mutable references to shared data is not allowed.
If you really want global mutable state, try using static mut or a global
UnsafeCell.
The value of static and const variables must be known at compile time. You can't cast a pointer as an integer because we can't know what value the address will take.
However, pointers to other constants' addresses are allowed in constants, example:
const X: u32 = 50; const Y: *const u32 = &X;
Therefore, casting one of these non-constant pointers to an integer results in a non-constant integer which lead to this error. Example:
const X: u32 = 1; const Y: usize = &X as *const u32 as usize; println!("{}", Y);
A function call isn't allowed in the const's initialization expression because the expression's value must be known at compile-time. Example of erroneous code:
enum Test { V1 } impl Test { fn test(&self) -> i32 { 12 } } fn main() { const FOO: Test = Test::V1; const A: i32 = FOO.test(); // You can't call Test::func() here ! }
Remember: you can't use a function call inside a const's initialization expression! However, you can totally use it anywhere else:
fn main() { const FOO: Test = Test::V1; FOO.func(); // here is good let x = FOO.func(); // or even here! }
This error indicates that an attempt was made to divide by zero (or take the remainder of a zero divisor) in a static or constant expression.
Constant functions are not allowed to mutate anything. Thus, binding to an argument with a mutable pattern is not allowed. For example,
const fn foo(mut x: u8) { // do stuff }
is bad because the function body may not mutate x.
Remove any mutable bindings from the argument list to fix this error. In case
you need to mutate the argument, try lazily initializing a global variable
instead of using a const fn, or refactoring the code to a functional style to
avoid mutation if possible.
A pattern used to match against an enum variant must provide a sub-pattern for each field of the enum variant. This error indicates that a pattern attempted to extract an incorrect number of fields from a variant.
enum Fruit { Apple(String, String) Pear(u32) }
Here the Apple variant has two fields, and should be matched against like so:
// Correct. match x { Apple(a, b) => ... }
Matching with the wrong number of fields has no sensible interpretation:
// Incorrect. match x { Apple(a) => ..., Apple(a, b, c) => ... }
Check how many fields the enum was declared with and ensure that your pattern uses the same number.
This error indicates that a pattern attempted to extract the fields of an enum variant with no fields. Here's a tiny example of this error:
// This enum has two variants. enum Number { // This variant has no fields. Zero, // This variant has one field. One(u32) } // Assuming x is a Number we can pattern match on its contents. match x { Zero(inside) => ..., One(inside) => ... }
The pattern match Zero(inside) is incorrect because the Zero variant
contains no fields, yet the inside name attempts to bind the first field of
the enum.
Each field of a struct can only be bound once in a pattern. Each occurrence of a field name binds the value of that field, so to fix this error you will have to remove or alter the duplicate uses of the field name. Perhaps you misspelt another field name?
This error indicates that a struct pattern attempted to extract a non-existant
field from a struct. Struct fields are identified by the name used before the
colon : so struct patterns should resemble the declaration of the struct type
being matched.
// Correct matching. struct Thing { x: u32, y: u32 } let thing = Thing { x: 1, y: 2 }; match thing { Thing { x: xfield, y: yfield } => ... }
If you are using shorthand field patterns but want to refer to the struct field by a different name, you should rename it explicitly.
// Change this: match thing { Thing { x, z } => ... } // To this: match thing { Thing { x, y: z } => ... }
This error indicates that a pattern for a struct fails to specify a sub-pattern
for every one of the struct's fields. Ensure that each field from the struct's
definition is mentioned in the pattern, or use .. to ignore unwanted fields.
For example:
struct Dog { name: String, age: u32 } let d = Dog { name: "Rusty".to_string(), age: 8 }; // This is incorrect. match d { Dog { age: x } => ... } // This is correct (explicit). match d { Dog { name: n, age: x } => ... } // This is also correct (ignore unused fields). match d { Dog { age: x, .. } => ... }
In a match expression, only numbers and characters can be matched against a range. This is because the compiler checks that the range is non-empty at compile-time, and is unable to evaluate arbitrary comparison functions. If you want to capture values of an orderable type between two end-points, you can use a guard.
// The ordering relation for strings can't be evaluated at compile time, // so this doesn't work: match string { "hello" ... "world" => ... _ => ... } // This is a more general version, using a guard: match string { s if s >= "hello" && s <= "world" => ... _ => ... }
When matching against a range, the compiler verifies that the range is non-empty. Range patterns include both end-points, so this is equivalent to requiring the start of the range to be less than or equal to the end of the range.
For example:
match 5u32 { // This range is ok, albeit pointless. 1 ... 1 => ... // This range is empty, and the compiler can tell. 1000 ... 5 => ... }
This error indicates that a pointer to a trait type cannot be implicitly dereferenced by a pattern. Every trait defines a type, but because the size of trait implementors isn't fixed, this type has no compile-time size. Therefore, all accesses to trait types must be through pointers. If you encounter this error you should try to avoid dereferencing the pointer.
let trait_obj: &SomeTrait = ...; // This tries to implicitly dereference to create an unsized local variable. let &invalid = trait_obj; // You can call methods without binding to the value being pointed at. trait_obj.method_one(); trait_obj.method_two();
You can read more about trait objects in the Trait Object section of the Reference:
The compiler doesn't know what method to call because more than one method has the same prototype. Example:
struct Test; trait Trait1 { fn foo(); } trait Trait2 { fn foo(); } impl Trait1 for Test { fn foo() {} } impl Trait2 for Test { fn foo() {} } fn main() { Test::foo() // error, which foo() to call? }
To avoid this error, you have to keep only one of them and remove the others. So let's take our example and fix it:
struct Test; trait Trait1 { fn foo(); } impl Trait1 for Test { fn foo() {} } fn main() { Test::foo() // and now that's good! }
However, a better solution would be using fully explicit naming of type and trait:
struct Test; trait Trait1 { fn foo(); } trait Trait2 { fn foo(); } impl Trait1 for Test { fn foo() {} } impl Trait2 for Test { fn foo() {} } fn main() { <Test as Trait1>::foo() }
You tried to give a type parameter where it wasn't needed. Bad example:
struct Test; impl Test { fn method(&self) {} } fn main() { let x = Test; x.method::<i32>(); // Error: Test::method doesn't need type parameter! }
To fix this error, just remove the type parameter:
struct Test; impl Test { fn method(&self) {} } fn main() { let x = Test; x.method(); // OK, we're good! }
This error occurrs when you pass too many or not enough type parameters to a method. Example:
struct Test; impl Test { fn method<T>(&self, v: &[T]) -> usize { v.len() } } fn main() { let x = Test; let v = &[0i32]; x.method::<i32, i32>(v); // error: only one type parameter is expected! }
To fix it, just specify a correct number of type parameters:
struct Test; impl Test { fn method<T>(&self, v: &[T]) -> usize { v.len() } } fn main() { let x = Test; let v = &[0i32]; x.method::<i32>(v); // OK, we're good! }
Please note on the last example that we could have called method like this:
x.method(v);
Trait objects like Box<Trait> can only be constructed when certain
requirements are satisfied by the trait in question.
Trait objects are a form of dynamic dispatch and use a dynamically sized type
for the inner type. So, for a given trait Trait, when Trait is treated as a
type, as in Box<Trait>, the inner type is 'unsized'. In such cases the boxed
pointer is a 'fat pointer' that contains an extra pointer to a table of methods
(among other things) for dynamic dispatch. This design mandates some
restrictions on the types of traits that are allowed to be used in trait
objects, which are collectively termed as 'object safety' rules.
Attempting to create a trait object for a non object-safe trait will trigger this error.
There are various rules:
Self: SizedWhen Trait is treated as a type, the type does not implement the special
Sized trait, because the type does not have a known size at compile time and
can only be accessed behind a pointer. Thus, if we have a trait like the
following:
trait Foo where Self: Sized { }
we cannot create an object of type Box<Foo> or &Foo since in this case
Self would not be Sized.
Generally, Self : Sized is used to indicate that the trait should not be used
as a trait object. If the trait comes from your own crate, consider removing
this restriction.
Self type in its arguments or return typeThis happens when a trait has a method like the following:
trait Trait { fn foo(&self) -> Self; } impl Trait for String { fn foo(&self) -> Self { "hi".to_owned() } } impl Trait for u8 { fn foo(&self) -> Self { 1 } }
(Note that &self and &mut self are okay, it's additional Self types which
cause this problem)
In such a case, the compiler cannot predict the return type of foo() in a
situation like the following:
fn call_foo(x: Box<Trait>) { let y = x.foo(); // What type is y? // ... }
If only some methods aren't object-safe, you can add a where Self: Sized bound
on them to mark them as explicitly unavailable to trait objects. The
functionality will still be available to all other implementers, including
Box<Trait> which is itself sized (assuming you impl Trait for Box<Trait>).
trait Trait { fn foo(&self) -> Self where Self: Sized; // more functions }
Now, foo() can no longer be called on a trait object, but you will now be
allowed to make a trait object, and that will be able to call any object-safe
methods". With such a bound, one can still call foo() on types implementing
that trait that aren't behind trait objects.
As mentioned before, trait objects contain pointers to method tables. So, if we have:
trait Trait { fn foo(&self); } impl Trait for String { fn foo(&self) { // implementation 1 } } impl Trait for u8 { fn foo(&self) { // implementation 2 } } // ...
At compile time each implementation of Trait will produce a table containing
the various methods (and other items) related to the implementation.
This works fine, but when the method gains generic parameters, we can have a problem.
Usually, generic parameters get monomorphized. For example, if I have
fn foo<T>(x: T) { // ... }
the machine code for foo::<u8>(), foo::<bool>(), foo::<String>(), or any
other type substitution is different. Hence the compiler generates the
implementation on-demand. If you call foo() with a bool parameter, the
compiler will only generate code for foo::<bool>(). When we have additional
type parameters, the number of monomorphized implementations the compiler
generates does not grow drastically, since the compiler will only generate an
implementation if the function is called with unparametrized substitutions
(i.e., substitutions where none of the substituted types are themselves
parametrized).
However, with trait objects we have to make a table containing every object that implements the trait. Now, if it has type parameters, we need to add implementations for every type that implements the trait, and there could theoretically be an infinite number of types.
For example, with:
trait Trait { fn foo<T>(&self, on: T); // more methods } impl Trait for String { fn foo<T>(&self, on: T) { // implementation 1 } } impl Trait for u8 { fn foo<T>(&self, on: T) { // implementation 2 } } // 8 more implementations
Now, if we have the following code:
fn call_foo(thing: Box<Trait>) { thing.foo(true); // this could be any one of the 8 types above thing.foo(1); thing.foo("hello"); }
we don't just need to create a table of all implementations of all methods of
Trait, we need to create such a table, for each different type fed to
foo(). In this case this turns out to be (10 types implementing Trait)*(3
types being fed to foo()) = 30 implementations!
With real world traits these numbers can grow drastically.
To fix this, it is suggested to use a where Self: Sized bound similar to the
fix for the sub-error above if you do not intend to call the method with type
parameters:
trait Trait { fn foo<T>(&self, on: T) where Self: Sized; // more methods }
If this is not an option, consider replacing the type parameter with another
trait object (e.g. if T: OtherTrait, use on: Box<OtherTrait>). If the number
of types you intend to feed to this method is limited, consider manually listing
out the methods of different types.
Methods that do not take a self parameter can't be called since there won't be
a way to get a pointer to the method table for them
trait Foo { fn foo() -> u8; }
This could be called as <Foo as Foo>::foo(), which would not be able to pick
an implementation.
Adding a Self: Sized bound to these methods will generally make this compile.
trait Foo { fn foo() -> u8 where Self: Sized; }
Self as a type parameter in the supertrait listingThis is similar to the second sub-error, but subtler. It happens in situations like the following:
trait Super<A> {} trait Trait: Super<Self> { } struct Foo; impl Super<Foo> for Foo{} impl Trait for Foo {}
Here, the supertrait might have methods as follows:
trait Super<A> { fn get_a(&self) -> A; // note that this is object safe! }
If the trait Foo was deriving from something like Super<String> or
Super<T> (where Foo itself is Foo<T>), this is okay, because given a type
get_a() will definitely return an object of that type.
However, if it derives from Super<Self>, even though Super is object safe,
the method get_a() would return an object of unknown type when called on the
function. Self type parameters let us make object safe traits no longer safe,
so they are forbidden when specifying supertraits.
There's no easy fix for this, generally code will need to be refactored so that
you no longer need to derive from Super<Self>.
It is not allowed to manually call destructors in Rust. It is also not
necessary to do this since drop is called automatically whenever a value goes
out of scope.
Here's an example of this error:
struct Foo { x: i32, } impl Drop for Foo { fn drop(&mut self) { println!("kaboom"); } } fn main() { let mut x = Foo { x: -7 }; x.drop(); // error: explicit use of destructor method }
You can't use type parameters on foreign items. Example of erroneous code:
extern { fn some_func<T>(x: T); }
To fix this, replace the type parameter with the specializations that you need:
extern { fn some_func_i32(x: i32); } extern { fn some_func_i64(x: i64); }
Rust only supports variadic parameters for interoperability with C code in its FFI. As such, variadic parameters can only be used with functions which are using the C ABI. Examples of erroneous code:
extern "rust-call" { fn foo(x: u8, ...); } // or fn foo(x: u8, ...) {}
To fix such code, put them in an extern "C" block:
extern "C" fn foo (x: u8, ...); // or: extern "C" { fn foo (x: u8, ...); }
When trying to make some type implement a trait Foo, you must, at minimum,
provide implementations for all of Foo's required methods (meaning the
methods that do not have default implementations), as well as any required
trait items like associated types or constants.
This error indicates that an attempted implementation of a trait method has the wrong number of type parameters.
For example, the trait below has a method foo with a type parameter T,
but the implementation of foo for the type Bar is missing this parameter:
trait Foo { fn foo<T: Default>(x: T) -> Self; } struct Bar; // error: method `foo` has 0 type parameters but its trait declaration has 1 // type parameter impl Foo for Bar { fn foo(x: bool) -> Self { Bar } }
This error indicates that an attempted implementation of a trait method has the wrong number of function parameters.
For example, the trait below has a method foo with two function parameters
(&self and u8), but the implementation of foo for the type Bar omits
the u8 parameter:
trait Foo { fn foo(&self, x: u8) -> bool; } struct Bar; // error: method `foo` has 1 parameter but the declaration in trait `Foo::foo` // has 2 impl Foo for Bar { fn foo(&self) -> bool { true } }
The parameters of any trait method must match between a trait implementation and the trait definition.
Here are a couple examples of this error:
trait Foo { fn foo(x: u16); fn bar(&self); } struct Bar; impl Foo for Bar { // error, expected u16, found i16 fn foo(x: i16) { } // error, values differ in mutability fn bar(&mut self) { } }
It is not allowed to cast to a bool. If you are trying to cast a numeric type to a bool, you can compare it with zero instead:
let x = 5; // Ok let x_is_nonzero = x != 0; // Not allowed, won't compile let x_is_nonzero = x as bool;
During a method call, a value is automatically dereferenced as many times as
needed to make the value's type match the method's receiver. The catch is that
the compiler will only attempt to dereference a number of times up to the
recursion limit (which can be set via the recursion_limit attribute).
For a somewhat artificial example:
#![recursion_limit="2"] struct Foo; impl Foo { fn foo(&self) {} } fn main() { let foo = Foo; let ref_foo = &&Foo; // error, reached the recursion limit while auto-dereferencing &&Foo ref_foo.foo(); }
One fix may be to increase the recursion limit. Note that it is possible to create an infinite recursion of dereferencing, in which case the only fix is to somehow break the recursion.
When invoking closures or other implementations of the function traits Fn,
FnMut or FnOnce using call notation, the number of parameters passed to the
function must match its definition.
An example using a closure:
let f = |x| x * 3; let a = f(); // invalid, too few parameters let b = f(4); // this works! let c = f(2, 3); // invalid, too many parameters
A generic function must be treated similarly:
fn foo<F: Fn()>(f: F) { f(); // this is valid, but f(3) would not work }
The built-in function traits are generic over a tuple of the function arguments.
If one uses angle-bracket notation (Fn<(T,), Output=U>) instead of parentheses
(Fn(T) -> U) to denote the function trait, the type parameter should be a
tuple. Otherwise function call notation cannot be used and the trait will not be
implemented by closures.
The most likely source of this error is using angle-bracket notation without wrapping the function argument type into a tuple, for example:
fn foo<F: Fn<i32>>(f: F) -> F::Output { f(3) }
It can be fixed by adjusting the trait bound like this:
fn foo<F: Fn<(i32,)>>(f: F) -> F::Output { f(3) }
Note that (T,) always denotes the type of a 1-tuple containing an element of
type T. The comma is necessary for syntactic disambiguation.
External C functions are allowed to be variadic. However, a variadic function
takes a minimum number of arguments. For example, consider C's variadic printf
function:
extern crate libc; use libc::{ c_char, c_int }; extern "C" { fn printf(_: *const c_char, ...) -> c_int; }
Using this declaration, it must be called with at least one argument, so
simply calling printf() is invalid. But the following uses are allowed:
unsafe { use std::ffi::CString; printf(CString::new("test\n").unwrap().as_ptr()); printf(CString::new("number = %d\n").unwrap().as_ptr(), 3); printf(CString::new("%d, %d\n").unwrap().as_ptr(), 10, 5); }
The number of arguments passed to a function must match the number of arguments specified in the function signature.
For example, a function like
fn f(a: u16, b: &str) {}
must always be called with exactly two arguments, e.g. f(2, "test").
Note, that Rust does not have a notion of optional function arguments or variadic functions (except for its C-FFI).
This error indicates that during an attempt to build a struct or struct-like enum variant, one of the fields was specified more than once. Each field should be specified exactly one time.
This error indicates that during an attempt to build a struct or struct-like enum variant, one of the fields was not provided. Each field should be specified exactly once.
Box placement expressions (like C++'s "placement new") do not yet support any
place expression except the exchange heap (i.e. std::boxed::HEAP).
Furthermore, the syntax is changing to use in instead of box. See RFC 470
and RFC 809 for more details.
The left-hand side of a compound assignment expression must be an lvalue expression. An lvalue expression represents a memory location and includes item paths (ie, namespaced variables), dereferences, indexing expressions, and field references.
Let's start with some bad examples:
use std::collections::LinkedList; // Bad: assignment to non-lvalue expression LinkedList::new() += 1; // ... fn some_func(i: &mut i32) { i += 12; // Error : '+=' operation cannot be applied on a reference ! }
And now some good examples:
let mut i : i32 = 0; i += 12; // Good ! // ... fn some_func(i: &mut i32) { *i += 12; // Good ! }
No description.
The compiler found a function whose body contains a return; statement but
whose return type is not (). An example of this is:
// error fn foo() -> u8 { return; }
Since return; is just like return ();, there is a mismatch between the
function's return type and the value being returned.
The left-hand side of an assignment operator must be an lvalue expression. An lvalue expression represents a memory location and can be a variable (with optional namespacing), a dereference, an indexing expression or a field reference.
More details can be found here: https://doc.rust-lang.org/reference.html#lvalues,-rvalues-and-temporaries
Now, we can go further. Here are some bad examples:
struct SomeStruct { x: i32, y: i32 } const SOME_CONST : i32 = 12; fn some_other_func() {} fn some_function() { SOME_CONST = 14; // error : a constant value cannot be changed! 1 = 3; // error : 1 isn't a valid lvalue! some_other_func() = 4; // error : we can't assign value to a function! SomeStruct.x = 12; // error : SomeStruct a structure name but it is used // like a variable! }
And now let's give good examples:
struct SomeStruct { x: i32, y: i32 } let mut s = SomeStruct {x: 0, y: 0}; s.x = 3; // that's good ! // ... fn some_func(x: &mut i32) { *x = 12; // that's good ! }
You tried to use a structure initialization with a non-structure type. Example of erroneous code:
enum Foo { FirstValue }; let u = Foo::FirstValue { value: 0i32 }; // error: Foo::FirstValue // isn't a structure! // or even simpler, if the structure wasn't defined at all: let u = RandomName { random_field: 0i32 }; // error: RandomName // isn't a structure!
To fix this, please check: * Did you spell it right? * Did you accidentaly used an enum as a struct? * Did you accidentaly make an enum when you intended to use a struct?
Here is the previous code with all missing information:
struct Inner { value: i32 } enum Foo { FirstValue(Inner) } fn main() { let u = Foo::FirstValue(Inner { value: 0i32 }); let t = Inner { value: 0i32 }; }
When defining a recursive struct or enum, any use of the type being defined
from inside the definition must occur behind a pointer (like Box or &).
This is because structs and enums must have a well-defined size, and without
the pointer the size of the type would need to be unbounded.
Consider the following erroneous definition of a type for a list of bytes:
// error, invalid recursive struct type struct ListNode { head: u8, tail: Option<ListNode>, }
This type cannot have a well-defined size, because it needs to be arbitrarily
large (since we would be able to nest ListNodes to any depth). Specifically,
size of `ListNode` = 1 byte for `head`
+ 1 byte for the discriminant of the `Option`
+ size of `ListNode`
One way to fix this is by wrapping ListNode in a Box, like so:
struct ListNode { head: u8, tail: Option<Box<ListNode>>, }
This works because Box is a pointer, so its size is well-known.
You cannot define a struct (or enum) Foo that requires an instance of Foo
in order to make a new Foo value. This is because there would be no way a
first instance of Foo could be made to initialize another instance!
Here's an example of a struct that has this problem:
struct Foo { x: Box<Foo> } // error
One fix is to use Option, like so:
struct Foo { x: Option<Box<Foo>> }
Now it's possible to create at least one instance of Foo: Foo { x: None }.
When using the #[simd] attribute on a tuple struct, the components of the
tuple struct must all be of a concrete, nongeneric type so the compiler can
reason about how to use SIMD with them. This error will occur if the types
are generic.
#[simd] struct Bad<T>(T, T, T); // This will cause an error #[simd] struct Good(u32, u32, u32); // This will not
The #[simd] attribute can only be applied to non empty tuple structs, because
it doesn't make sense to try to use SIMD operations when there are no values to
operate on.
#[simd] struct Bad; // This will cause an error #[simd] struct Good(u32); // This will not
When using the #[simd] attribute to automatically use SIMD operations in tuple
struct, the types in the struct must all be of the same type, or the compiler
will trigger this error.
#[simd] struct Bad(u16, u32, u32); // This will cause an error #[simd] struct Good(u32, u32, u32); // This will not
When using the #[simd] attribute on a tuple struct, the elements in the tuple
must be machine types so SIMD operations can be applied to them.
#[simd] struct Bad(String); // This will cause an error #[simd] struct Good(u32, u32, u32); // This will not
Enum variants which contain no data can be given a custom integer representation. This error indicates that the value provided is not an integer literal and is therefore invalid.
For example, in the following code,
enum Foo { Q = "32" }
we try to set the representation to a string.
There's no general fix for this; if you can work with an integer then just set it to one:
enum Foo { Q = 32 }
however if you actually wanted a mapping between variants and non-integer objects, it may be preferable to use a method with a match instead:
enum Foo { Q } impl Foo { fn get_str(&self) -> &'static str { match *self { Foo::Q => "32", } } }
This error indicates that the compiler was unable to sensibly evaluate an integer expression provided as an enum discriminant. Attempting to divide by 0 or causing integer overflow are two ways to induce this error. For example:
enum Enum { X = (1 << 500), Y = (1 / 0) }
Ensure that the expressions given can be evaluated as the desired integer type. See the FFI section of the Reference for more information about using a custom integer type:
Enum discriminants are used to differentiate enum variants stored in memory. This error indicates that the same value was used for two or more variants, making them impossible to tell apart.
// Good. enum Enum { P, X = 3, Y = 5 } // Bad. enum Enum { P = 3, X = 3, Y = 5 }
Note that variants without a manually specified discriminant are numbered from top to bottom starting from 0, so clashes can occur with seemingly unrelated variants.
enum Bad { X, Y = 0 }
Here X will have already been assigned the discriminant 0 by the time Y is
encountered, so a conflict occurs.
The default type for enum discriminants is isize, but it can be adjusted by
adding the repr attribute to the enum declaration. This error indicates that
an integer literal given as a discriminant is not a member of the discriminant
type. For example:
#[repr(u8)] enum Thing { A = 1024, B = 5 }
Here, 1024 lies outside the valid range for u8, so the discriminant for A is
invalid. You may want to change representation types to fix this, or else change
invalid discriminant values so that they fit within the existing type.
Note also that without a representation manually defined, the compiler will optimize by using the smallest integer type possible.
At present, it's not possible to define a custom representation for an enum with
a single variant. As a workaround you can add a Dummy variant.
It is impossible to define an integer type to be used to represent zero-variant enum values because there are no zero-variant enum values. There is no way to construct an instance of the following type using only safe code:
enum Empty {}
No description.
No description.
Too many type parameters were supplied for a function. For example:
fn foo<T>() {} fn main() { foo::<f64, bool>(); // error, expected 1 parameter, found 2 parameters }
The number of supplied parameters much exactly match the number of defined type parameters.
You gave too many lifetime parameters. Erroneous code example:
fn f() {} fn main() { f::<'static>() // error: too many lifetime parameters provided }
Please check you give the right number of lifetime parameters. Example:
fn f() {} fn main() { f() // ok! }
It's also important to note that the Rust compiler can generally determine the lifetime by itself. Example:
struct Foo { value: String } impl Foo { // it can be written like this fn get_value<'a>(&'a self) -> &'a str { &self.value } // but the compiler works fine with this too: fn without_lifetime(&self) -> &str { &self.value } } fn main() { let f = Foo { value: "hello".to_owned() }; println!("{}", f.get_value()); println!("{}", f.without_lifetime()); }
Not enough type parameters were supplied for a function. For example:
fn foo<T, U>() {} fn main() { foo::<f64>(); // error, expected 2 parameters, found 1 parameter }
Note that if a function takes multiple type parameters but you want the compiler to infer some of them, you can use type placeholders:
fn foo<T, U>(x: T) {} fn main() { let x: bool = true; foo::<f64>(x); // error, expected 2 parameters, found 1 parameter foo::<_, f64>(x); // same as `foo::<bool, f64>(x)` }
No description.
You gave an unnecessary type parameter in a type alias. Erroneous code example:
type Foo<T> = u32; // error: type parameter `T` is unused // or: type Foo<A,B> = Box<A>; // error: type parameter `B` is unused
Please check you didn't write too many type parameters. Example:
type Foo = u32; // ok! type Foo<A> = Box<A>; // ok!
You tried to declare an undefined atomic operation function. Erroneous code example:
#![feature(intrinsics)] extern "rust-intrinsic" { fn atomic_foo(); // error: unrecognized atomic operation // function }
Please check you didn't make a mistake in the function's name. All intrinsic functions are defined in librustc_trans/trans/intrinsic.rs and in libcore/intrinsics.rs in the Rust source code. Example:
#![feature(intrinsics)] extern "rust-intrinsic" { fn atomic_fence(); // ok! }
You declared an unknown intrinsic function. Erroneous code example:
#![feature(intrinsics)] extern "rust-intrinsic" { fn foo(); // error: unrecognized intrinsic function: `foo` } fn main() { unsafe { foo(); } }
Please check you didn't make a mistake in the function's name. All intrinsic functions are defined in librustc_trans/trans/intrinsic.rs and in libcore/intrinsics.rs in the Rust source code. Example:
#![feature(intrinsics)] extern "rust-intrinsic" { fn atomic_fence(); // ok! } fn main() { unsafe { atomic_fence(); } }
You gave an invalid number of type parameters to an intrinsic function. Erroneous code example:
#![feature(intrinsics)] extern "rust-intrinsic" { fn size_of<T, U>() -> usize; // error: intrinsic has wrong number // of type parameters }
Please check that you provided the right number of lifetime parameters and verify with the function declaration in the Rust source code. Example:
#![feature(intrinsics)] extern "rust-intrinsic" { fn size_of<T>() -> usize; // ok! }
You hit this error because the compiler the compiler lacks information to determine a type for this expression. Erroneous code example:
fn main() { let x = |_| {}; // error: cannot determine a type for this expression }
You have two possibilities to solve this situation: * Give an explicit definition of the expression * Infer the expression
Examples:
fn main() { let x = |_ : u32| {}; // ok! // or: let x = |_| {}; x(0u32); }
You hit this error because the compiler lacks information to determine a type for this variable. Erroneous code example:
fn demo(devil: fn () -> !) { let x: &_ = devil(); // error: cannot determine a type for this local variable } fn oh_no() -> ! { panic!("the devil is in the details") } fn main() { demo(oh_no); }
To solve this situation, constrain the type of the variable. Examples:
fn some_func(x: &u32) { // some code } fn demo(devil: fn () -> !) { let x: &u32 = devil(); // Here we defined the type at the variable creation let x: &_ = devil(); some_func(x); // Here, the type is determined by the function argument type } fn oh_no() -> ! { panic!("the devil is in the details") } fn main() { demo(oh_no); }
No description.
No description.
This error indicates that a lifetime is missing from a type. If it is an error inside a function signature, the problem may be with failing to adhere to the lifetime elision rules (see below).
Here are some simple examples of where you'll run into this error:
struct Foo { x: &bool } // error struct Foo<'a> { x: &'a bool } // correct enum Bar { A(u8), B(&bool), } // error enum Bar<'a> { A(u8), B(&'a bool), } // correct type MyStr = &str; // error type MyStr<'a> = &'a str; //correct
Lifetime elision is a special, limited kind of inference for lifetimes in function signatures which allows you to leave out lifetimes in certain cases. For more background on lifetime elision see the book.
The lifetime elision rules require that any function signature with an elided output lifetime must either have
&self or &mut self receiverIn the first case, the output lifetime is inferred to be the same as the unique
input lifetime. In the second case, the lifetime is instead inferred to be the
same as the lifetime on &self or &mut self.
Here are some examples of elision errors:
// error, no input lifetimes fn foo() -> &str { ... } // error, `x` and `y` have distinct lifetimes inferred fn bar(x: &str, y: &str) -> &str { ... } // error, `y`'s lifetime is inferred to be distinct from `x`'s fn baz<'a>(x: &'a str, y: &str) -> &str { ... }
This error means that an incorrect number of lifetime parameters were provided for a type (like a struct or enum) or trait.
Some basic examples include:
struct Foo<'a>(&'a str); enum Bar { A, B, C } struct Baz<'a> { foo: Foo, // error: expected 1, found 0 bar: Bar<'a>, // error: expected 0, found 1 }
Here's an example that is currently an error, but may work in a future version of Rust:
struct Foo<'a>(&'a str); trait Quux { } impl Quux for Foo { } // error: expected 1, found 0
Lifetime elision in implementation headers was part of the lifetime elision RFC. It is, however, currently unimplemented.
You tried to give a type parameter to a type which doesn't need it. Erroneous code example:
type X = u32<i32>; // error: type parameters are not allowed on this type
Please check that you used the correct type and recheck its definition. Perhaps it doesn't need the type parameter.
Example:
type X = u32; // this compiles
Note that type parameters for enum-variant constructors go after the variant,
not after the enum (Option::None::
You tried to give a lifetime parameter to a type which doesn't need it. Erroneous code example:
type X = u32<'static>; // error: lifetime parameters are not allowed on // this type
Please check that the correct type was used and recheck its definition; perhaps it doesn't need the lifetime parameter. Example:
type X = u32; // ok!
You can only define an inherent implementation for a type in the same crate
where the type was defined. For example, an impl block as below is not allowed
since Vec is defined in the standard library:
impl Vec<u8> { ... } // error
To fix this problem, you can do either of these things:
Note that using the type keyword does not work here because type only
introduces a type alias:
type Bytes = Vec<u8>; impl Bytes { ... } // error, same as above
This error indicates a violation of one of Rust's orphan rules for trait implementations. The rule prohibits any implementation of a foreign trait (a trait defined in another crate) where
Here's one example of this error:
impl Drop for u32 {}
To avoid this kind of error, ensure that at least one local type is referenced
by the impl:
pub struct Foo; // you define your type in your crate impl Drop for Foo { // and you can implement the trait on it! // code of trait implementation here } impl From<Foo> for i32 { // or you use a type from your crate as // a type parameter fn from(i: Foo) -> i32 { 0 } }
Alternatively, define a trait locally and implement that instead:
trait Bar { fn get(&self) -> usize; } impl Bar for u32 { fn get(&self) -> usize { 0 } }
For information on the design of the orphan rules, see RFC 1023.
No description.
There are conflicting trait implementations for the same type. Example of erroneous code:
trait MyTrait { fn get(&self) -> usize; } impl<T> MyTrait for T { fn get(&self) -> usize { 0 } } struct Foo { value: usize } impl MyTrait for Foo { // error: conflicting implementations for trait // `MyTrait` fn get(&self) -> usize { self.value } }
When looking for the implementation for the trait, the compiler finds
both the impl<T> MyTrait for T where T is all types and the impl MyTrait for Foo. Since a trait cannot be implemented multiple times,
this is an error. So, when you write:
impl<T> MyTrait for T { fn get(&self) -> usize { 0 } }
This makes the trait implemented on all types in the scope. So if you try to implement it on another one after that, the implementations will conflict. Example:
trait MyTrait { fn get(&self) -> usize; } impl<T> MyTrait for T { fn get(&self) -> usize { 0 } } struct Foo; fn main() { let f = Foo; f.get(); // the trait is implemented so we can use it }
An attempt was made to implement Drop on a trait, which is not allowed: only structs and enums can implement Drop. An example causing this error:
trait MyTrait {} impl Drop for MyTrait { fn drop(&mut self) {} }
A workaround for this problem is to wrap the trait up in a struct, and implement Drop on that. An example is shown below:
trait MyTrait {} struct MyWrapper<T: MyTrait> { foo: T } impl <T: MyTrait> Drop for MyWrapper<T> { fn drop(&mut self) {} }
Alternatively, wrapping trait objects requires something like the following:
trait MyTrait {} //or Box<MyTrait>, if you wanted an owned trait object struct MyWrapper<'a> { foo: &'a MyTrait } impl <'a> Drop for MyWrapper<'a> { fn drop(&mut self) {} }
In order to be consistent with Rust's lack of global type inference, type placeholders are disallowed by design in item signatures.
Examples of this error include:
fn foo() -> _ { 5 } // error, explicitly write out the return type instead static BAR: _ = "test"; // error, explicitly write out the type instead
No description.
No description.
You declared two fields of a struct with the same name. Erroneous code example:
struct Foo { field1: i32, field1: i32 // error: field is already declared }
Please verify that the field names have been correctly spelled. Example:
struct Foo { field1: i32, field2: i32 // ok! }
No description.
Type parameter defaults can only use parameters that occur before them. Erroneous code example:
pub struct Foo<T=U, U=()> { field1: T, filed2: U, } // error: type parameters with a default cannot use forward declared // identifiers
Since type parameters are evaluated in-order, you may be able to fix this issue by doing:
pub struct Foo<U=(), T=U> { field1: T, filed2: U, }
Please also verify that this wasn't because of a name-clash and rename the type parameter if so.
No description.
You declared a pattern as an argument in a foreign function declaration. Erroneous code example:
extern { fn foo((a, b): (u32, u32)); // error: patterns aren't allowed in foreign // function declarations }
Please replace the pattern argument with a regular one. Example:
struct SomeStruct { a: u32, b: u32, } extern { fn foo(s: SomeStruct); // ok! } // or extern { fn foo(a: (u32, u32)); // ok! }
It is not possible to define main with type parameters, or even with function
parameters. When main is present, it must take no arguments and return ().
It is not possible to declare type parameters on a function that has the start
attribute. Such a function must have the following type signature:
fn(isize, *const *const u8) -> isize
Using unsafe functionality, such as dereferencing raw pointers and calling
functions via FFI or marked as unsafe, is potentially dangerous and disallowed
by safety checks. These safety checks can be relaxed for a section of the code
by wrapping the unsafe instructions with an unsafe block. For instance:
unsafe fn f() { return; } fn main() { unsafe { f(); } }
A binary can only have one entry point, and by default that entry point is the
function main(). If there are multiple such functions, please rename one.
This error indicates that the compiler found multiple functions with the
#[main] attribute. This is an error because there must be a unique entry
point into a Rust program.
This error indicates that the compiler found multiple functions with the
#[start] attribute. This is an error because there must be a unique entry
point into a Rust program.
There are various restrictions on transmuting between types in Rust; for example types being transmuted must have the same size. To apply all these restrictions, the compiler must know the exact types that may be transmuted. When type parameters are involved, this cannot always be done.
So, for example, the following is not allowed:
struct Foo<T>(Vec<T>) fn foo<T>(x: Vec<T>) { // we are transmuting between Vec<T> and Foo<T> here let y: Foo<T> = unsafe { transmute(x) }; // do something with y }
In this specific case there's a good chance that the transmute is harmless (but
this is not guaranteed by Rust). However, when alignment and enum optimizations
come into the picture, it's quite likely that the sizes may or may not match
with different type parameter substitutions. It's not possible to check this for
all possible types, so transmute() simply only accepts types without any
unsubstituted type parameters.
If you need this, there's a good chance you're doing something wrong. Keep in mind that Rust doesn't guarantee much about the layout of different structs (even two structs with identical declarations may have different layouts). If there is a solution that avoids the transmute entirely, try it instead.
If it's possible, hand-monomorphize the code by writing the function for each possible type substitution. It's possible to use traits to do this cleanly, for example:
trait MyTransmutableType { fn transmute(Vec<Self>) -> Foo<Self> } impl MyTransmutableType for u8 { fn transmute(x: Foo<u8>) -> Vec<u8> { transmute(x) } } impl MyTransmutableType for String { fn transmute(x: Foo<String>) -> Vec<String> { transmute(x) } } // ... more impls for the types you intend to transmute fn foo<T: MyTransmutableType>(x: Vec<T>) { let y: Foo<T> = <T as MyTransmutableType>::transmute(x); // do something with y }
Each impl will be checked for a size match in the transmute as usual, and since there are no unbound type parameters involved, this should compile unless there is a size mismatch in one of the impls.
It is also possible to manually transmute:
let result: SomeType = mem::uninitialized(); unsafe { copy_nonoverlapping(&v, &result) }; result // `v` transmuted to type `SomeType`
No description.
Lang items are already implemented in the standard library. Unless you are writing a free-standing application (e.g. a kernel), you do not need to provide them yourself.
You can build a free-standing crate by adding #![no_std] to the crate
attributes:
#![feature(no_std)] #![no_std]
No description.
Imports (use statements) are not allowed after non-item statements, such as
variable declarations and expression statements.
Here is an example that demonstrates the error:
fn f() { // Variable declaration before import let x = 0; use std::io::Read; ... }
The solution is to declare the imports at the top of the block, function, or file.
Here is the previous example again, with the correct order:
fn f() { use std::io::Read; let x = 0; ... }
See the Declaration Statements section of the reference for more information about what constitutes an Item declaration and what does not:
No description.
const and static mean different things. A const is a compile-time
constant, an alias for a literal value. This property means you can match it
directly within a pattern.
The static keyword, on the other hand, guarantees a fixed location in memory.
This does not always mean that the value is constant. For example, a global
mutex can be declared static as well.
If you want to match against a static, consider using a guard instead:
static FORTY_TWO: i32 = 42; match Some(42) { Some(x) if x == FORTY_TWO => ... ... }
You tried to use a trait as a struct constructor. Erroneous code example:
trait TraitNotAStruct {} TraitNotAStruct{ value: 0 }; // error: use of trait `TraitNotAStruct` as a // struct constructor
Please verify you used the correct type name or please implement the trait on a struct and use this struct constructor. Example:
trait TraitNotAStruct {} struct Foo { value: i32 } Foo{ value: 0 }; // ok!
In Rust, you can only move a value when its size is known at compile time.
To work around this restriction, consider "hiding" the value behind a reference:
either &x or &mut x. Since a reference has a fixed size, this lets you move
it around as usual.
An if-let pattern attempts to match the pattern, and enters the body if the
match was successful. If the match is irrefutable (when it cannot fail to
match), use a regular let-binding instead. For instance:
struct Irrefutable(i32); let irr = Irrefutable(0); // This fails to compile because the match is irrefutable. if let Irrefutable(x) = irr { // This body will always be executed. foo(x); } // Try this instead: let Irrefutable(x) = irr; foo(x);
No description.
No description.
A while-let pattern attempts to match the pattern, and enters the body if the
match was successful. If the match is irrefutable (when it cannot fail to
match), use a regular let-binding inside a loop instead. For instance:
struct Irrefutable(i32); let irr = Irrefutable(0); // This fails to compile because the match is irrefutable. while let Irrefutable(x) = irr { ... } // Try this instead: loop { let Irrefutable(x) = irr; ... }
This error means that the compiler found a return expression in a function
marked as diverging. A function diverges if it has ! in the place of the
return type in its signature. For example:
fn foo() -> ! { return; } // error
For a function that diverges, every control path in the function must never
return, for example with a loop that never breaks or a call to another
diverging function (such as panic!()).
No description.
No description.
Enum variants are qualified by default. For example, given this type:
enum Method { GET, POST }
you would match it using:
match m { Method::GET => ... Method::POST => ... }
If you don't qualify the names, the code will bind new variables named "GET" and
"POST" instead. This behavior is likely not what you want, so rustc warns when
that happens.
Qualified names are good practice, and most code works well with them. But if you prefer them unqualified, you can import the variants into scope:
use Method::*; enum Method { GET, POST }
If you want others to be able to import variants from your module directly, use
pub use:
pub use Method::*; enum Method { GET, POST }
This error means that an attempt was made to specify the type of a variable with a combination of a concrete type and a trait. Consider the following example:
fn foo(bar: i32+std::fmt::Display) {}
The code is trying to specify that we want to receive a signed 32-bit integer
which also implements Display. This doesn't make sense: when we pass i32, a
concrete type, it implicitly includes all of the traits that it implements.
This includes Display, Debug, Clone, and a host of others.
If i32 implements the trait we desire, there's no need to specify the trait
separately. If it does not, then we need to impl the trait for i32 before
passing it into foo. Either way, a fixed definition for foo will look like
the following:
fn foo(bar: i32) {}
To learn more about traits, take a look at the Book:
No description.
No description.
In types, the + type operator has low precedence, so it is often necessary
to use parentheses.
For example:
trait Foo {} struct Bar<'a> { w: &'a Foo + Copy, // error, use &'a (Foo + Copy) x: &'a Foo + 'a, // error, use &'a (Foo + 'a) y: &'a mut Foo + 'a, // error, use &'a mut (Foo + 'a) z: fn() -> Foo + 'a, // error, use fn() -> (Foo + 'a) }
More details can be found in RFC 438.
No description.
No description.
Explicitly implementing both Drop and Copy for a type is currently disallowed. This feature can make some sense in theory, but the current implementation is incorrect and can lead to memory unsafety (see issue #20126), so it has been disabled for now.
An associated function for a trait was defined to be static, but an
implementation of the trait declared the same function to be a method (i.e. to
take a self parameter).
Here's an example of this error:
trait Foo { fn foo(); } struct Bar; impl Foo for Bar { // error, method `foo` has a `&self` declaration in the impl, but not in // the trait fn foo(&self) {} }
An associated function for a trait was defined to be a method (i.e. to take a
self parameter), but an implementation of the trait declared the same function
to be static.
Here's an example of this error:
trait Foo { fn foo(&self); } struct Bar; impl Foo for Bar { // error, method `foo` has a `&self` declaration in the trait, but not in // the impl fn foo() {} }
No description.
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Trait objects need to have all associated types specified. Erroneous code example:
trait Trait { type Bar; } type Foo = Trait; // error: the value of the associated type `Bar` (from // the trait `Trait`) must be specified
Please verify you specified all associated types of the trait and that you used the right trait. Example:
trait Trait { type Bar; } type Foo = Trait<Bar=i32>; // ok!
Negative impls are only allowed for traits with default impls. For more information see the opt-in builtin traits RFC.
where clauses must use generic type parameters: it does not make sense to use
them otherwise. An example causing this error:
trait Foo { fn bar(&self); } #[derive(Copy,Clone)] struct Wrapper<T> { Wrapped: T } impl Foo for Wrapper<u32> where Wrapper<u32>: Clone { fn bar(&self) { } }
This use of a where clause is strange - a more common usage would look
something like the following:
impl <T> Foo for Wrapper<T> where Wrapper<T>: Clone { fn bar(&self) { } }
Here, we're saying that the implementation exists on Wrapper only when the
wrapped type T implements Clone. The where clause is important because
some types will not implement Clone, and thus will not get this method.
In our erroneous example, however, we're referencing a single concrete type.
Since we know for certain that Wrapperwhere clause.
A type parameter was declared which shadows an existing one. An example of this error:
trait Foo<T> { fn do_something(&self) -> T; fn do_something_else<T: Clone>(&self, bar: T); }
In this example, the trait Foo and the trait method do_something_else both
define a type parameter T. This is not allowed: if the method wishes to
define a type parameter, it must use a different name for it.
Your method's lifetime parameters do not match the trait declaration. Erroneous code example:
trait Trait { fn bar<'a,'b:'a>(x: &'a str, y: &'b str); } struct Foo; impl Trait for Foo { fn bar<'a,'b>(x: &'a str, y: &'b str) { // error: lifetime parameters or bounds on method `bar` // do not match the trait declaration } }
The lifetime constraint 'b for bar() implementation does not match the
trait declaration. Ensure lifetime declarations match exactly in both trait
declaration and implementation. Example:
trait Trait { fn t<'a,'b:'a>(x: &'a str, y: &'b str); } struct Foo; impl Trait for Foo { fn t<'a,'b:'a>(x: &'a str, y: &'b str) { // ok! } }
No description.
Inherent implementations (one that do not implement a trait but provide
methods associated with a type) are always safe because they are not
implementing an unsafe trait. Removing the unsafe keyword from the inherent
implementation will resolve this error.
struct Foo; // this will cause this error unsafe impl Foo { } // converting it to this will fix it impl Foo { }
A negative implementation is one that excludes a type from implementing a particular trait. Not being able to use a trait is always a safe operation, so negative implementations are always safe and never need to be marked as unsafe.
struct Foo; // unsafe is unnecessary unsafe impl !Clone for Foo { } // this will compile impl !Clone for Foo { }
Safe traits should not have unsafe implementations, therefore marking an implementation for a safe trait unsafe will cause a compiler error. Removing the unsafe marker on the trait noted in the error will resolve this problem.
struct Foo; trait Bar { } // this won't compile because Bar is safe unsafe impl Bar for Foo { } // this will compile impl Bar for Foo { }
Unsafe traits must have unsafe implementations. This error occurs when an implementation for an unsafe trait isn't marked as unsafe. This may be resolved by marking the unsafe implementation as unsafe.
struct Foo; unsafe trait Bar { } // this won't compile because Bar is unsafe and impl isn't unsafe impl Bar for Foo { } // this will compile unsafe impl Bar for Foo { }
It is an error to define two associated items (like methods, associated types, associated functions, etc.) with the same identifier.
For example:
struct Foo(u8); impl Foo { fn bar(&self) -> bool { self.0 > 5 } fn bar() {} // error: duplicate associated function } trait Baz { type Quux; fn baz(&self) -> bool; } impl Baz for Foo { type Quux = u32; fn baz(&self) -> bool { true } // error: duplicate method fn baz(&self) -> bool { self.0 > 5 } // error: duplicate associated type type Quux = u32; }
Inherent associated types were part of RFC 195 but are not yet implemented. See the tracking issue for the status of this implementation.
No description.
An attempt to implement the Copy trait for a struct failed because one of the
fields does not implement Copy. To fix this, you must implement Copy for the
mentioned field. Note that this may not be possible, as in the example of
struct Foo { foo : Vec<u32>, } impl Copy for Foo { }
This fails because Vec<T> does not implement Copy for any T.
Here's another example that will fail:
#[derive(Copy)] struct Foo<'a> { ty: &'a mut bool, }
This fails because &mut T is not Copy, even when T is Copy (this
differs from the behavior for &T, which is always Copy).
An attempt to implement the Copy trait for an enum failed because one of the
variants does not implement Copy. To fix this, you must implement Copy for
the mentioned variant. Note that this may not be possible, as in the example of
enum Foo { Bar(Vec<u32>), Baz, } impl Copy for Foo { }
This fails because Vec<T> does not implement Copy for any T.
Here's another example that will fail:
#[derive(Copy)] enum Foo<'a> { Bar(&'a mut bool), Baz }
This fails because &mut T is not Copy, even when T is Copy (this
differs from the behavior for &T, which is always Copy).
You can only implement Copy for a struct or enum. Both of the following
examples will fail, because neither i32 (primitive type) nor &'static Bar
(reference to Bar) is a struct or enum:
type Foo = i32; impl Copy for Foo { } // error #[derive(Copy, Clone)] struct Bar; impl Copy for &'static Bar { } // error
You declared an unused type parameter when implementing a trait on an object. Erroneous code example:
trait MyTrait { fn get(&self) -> usize; } struct Foo; impl<T> MyTrait for Foo { fn get(&self) -> usize { 0 } }
Please check your object definition and remove unused type parameter(s). Example:
trait MyTrait { fn get(&self) -> usize; } struct Foo; impl MyTrait for Foo { fn get(&self) -> usize { 0 } }
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This error indicates a violation of one of Rust's orphan rules for trait implementations. The rule concerns the use of type parameters in an implementation of a foreign trait (a trait defined in another crate), and states that type parameters must be "covered" by a local type. To understand what this means, it is perhaps easiest to consider a few examples.
If ForeignTrait is a trait defined in some external crate foo, then the
following trait impl is an error:
extern crate foo; use foo::ForeignTrait; impl<T> ForeignTrait for T { ... } // error
To work around this, it can be covered with a local type, MyType:
struct MyType<T>(T); impl<T> ForeignTrait for MyType<T> { ... } // Ok
For another example of an error, suppose there's another trait defined in foo
named ForeignTrait2 that takes two type parameters. Then this impl results
in the same rule violation:
struct MyType2; impl<T> ForeignTrait2<T, MyType<T>> for MyType2 { ... } // error
The reason for this is that there are two appearances of type parameter T in
the impl header, both as parameters for ForeignTrait2. The first appearance
is uncovered, and so runs afoul of the orphan rule.
Consider one more example:
impl<T> ForeignTrait2<MyType<T>, T> for MyType2 { ... } // Ok
This only differs from the previous impl in that the parameters T and
MyType<T> for ForeignTrait2 have been swapped. This example does not
violate the orphan rule; it is permitted.
To see why that last example was allowed, you need to understand the general
rule. Unfortunately this rule is a bit tricky to state. Consider an impl:
impl<P1, ..., Pm> ForeignTrait<T1, ..., Tn> for T0 { ... }
where P1, ..., Pm are the type parameters of the impl and T0, ..., Tn
are types. One of the types T0, ..., Tn must be a local type (this is another
orphan rule, see the explanation for E0117). Let i be the smallest integer
such that Ti is a local type. Then no type parameter can appear in any of the
Tj for j < i.
For information on the design of the orphan rules, see RFC 1023.
You used an intrinsic function which doesn't correspond to its definition. Erroneous code example:
#![feature(intrinsics)] extern "rust-intrinsic" { fn size_of<T>(); // error: intrinsic has wrong type }
Please check the function definition. Example:
#![feature(intrinsics)] extern "rust-intrinsic" { fn size_of<T>() -> usize; }
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You used an associated type which isn't defined in the trait. Erroneous code example:
trait Trait { type Bar; } type Foo = Trait<F=i32>; // error: associated type `F` not found for // `Trait`
Please verify you used the right trait or you didn't misspell the associated type name. Example:
trait Trait { type Bar; } type Foo = Trait<Bar=i32>; // ok!
An attempt was made to retrieve an associated type, but the type was ambiguous. For example:
trait T1 {} trait T2 {} trait Foo { type A: T1; } trait Bar : Foo { type A: T2; fn do_something() { let _: Self::A; } }
In this example, Foo defines an associated type A. Bar inherits that type
from Foo, and defines another associated type of the same name. As a result,
when we attempt to use Self::A, it's ambiguous whether we mean the A defined
by Foo or the one defined by Bar.
There are two options to work around this issue. The first is simply to rename one of the types. Alternatively, one can specify the intended type using the following syntax:
fn do_something() { let _: <Self as Bar>::A; }
An attempt was made to retrieve an associated type, but the type was ambiguous. For example:
trait MyTrait {type X; } fn main() { let foo: MyTrait::X; }
The problem here is that we're attempting to take the type of X from MyTrait. Unfortunately, the type of X is not defined, because it's only made concrete in implementations of the trait. A working version of this code might look like:
trait MyTrait {type X; } struct MyStruct; impl MyTrait for MyStruct { type X = u32; } fn main() { let foo: <MyStruct as MyTrait>::X; }
This syntax specifies that we want the X type from MyTrait, as made concrete in
MyStruct. The reason that we cannot simply use MyStruct::X is that MyStruct
might implement two different traits with identically-named associated types.
This syntax allows disambiguation between the two.
No description.
You attempted to use multiple types as bounds for a closure or trait object. Rust does not currently support this. A simple example that causes this error:
fn main() { let _: Box<std::io::Read+std::io::Write>; }
Builtin traits are an exception to this rule: it's possible to have bounds of one non-builtin type, plus any number of builtin types. For example, the following compiles correctly:
fn main() { let _: Box<std::io::Read+Copy+Sync>; }
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The attribute must have a value. Erroneous code example:
#[rustc_on_unimplemented] // error: this attribute must have a value trait Bar {}
Please supply the missing value of the attribute. Example:
#[rustc_on_unimplemented = "foo"] // ok! trait Bar {}
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This error indicates that not enough type parameters were found in a type or trait.
For example, the Foo struct below is defined to be generic in T, but the
type parameter is missing in the definition of Bar:
struct Foo<T> { x: T } struct Bar { x: Foo }
This error indicates that too many type parameters were found in a type or trait.
For example, the Foo struct below has no type parameters, but is supplied
with two in the definition of Bar:
struct Foo { x: bool } struct Bar<S, T> { x: Foo<S, T> }
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This error indicates an attempt to use a value where a type is expected. For example:
enum Foo { Bar(u32) } fn do_something(x: Foo::Bar) { }
In this example, we're attempting to take a type of Foo::Bar in the
do_something function. This is not legal: Foo::Bar is a value of type Foo,
not a distinct static type. Likewise, it's not legal to attempt to
impl Foo::Bar: instead, you must impl Foo and then pattern match to specify
behaviour for specific enum variants.
This error indicates a constant expression for the array length was found, but it was not an integer (signed or unsigned) expression.
Some examples of code that produces this error are:
const A: [u32; "hello"] = []; // error const B: [u32; true] = []; // error const C: [u32; 0.0] = []; // error
There was an error while evaluating the expression for the length of a fixed- size array type.
Some examples of this error are:
// divide by zero in the length expression const A: [u32; 1/0] = []; // Rust currently will not evaluate the function `foo` at compile time fn foo() -> usize { 12 } const B: [u32; foo()] = []; // it is an error to try to add `u8` and `f64` use std::{f64, u8}; const C: [u32; u8::MAX + f64::EPSILON] = [];
Two items of the same name cannot be imported without rebinding one of the items under a new local name.
An example of this error:
use foo::baz; use bar::*; // error, do `use foo::baz as quux` instead on the previous line fn main() {} mod foo { pub struct baz; } mod bar { pub mod baz {} }
Two items of the same name cannot be imported without rebinding one of the items under a new local name.
An example of this error:
use foo::baz; use bar::baz; // error, do `use bar::baz as quux` instead fn main() {} mod foo { pub struct baz; } mod bar { pub mod baz {} }
Attempt was made to import an unimportable value. This can happen when trying to import a method from a trait. An example of this error:
mod foo { pub trait MyTrait { fn do_something(); } } use foo::MyTrait::do_something;
It's invalid to directly import methods belonging to a trait or concrete type.
No description.
You can't import a value whose name is the same as another value defined in the module.
An example of this error:
use bar::foo; // error, do `use bar::foo as baz` instead fn foo() {} mod bar { pub fn foo() {} } fn main() {}
You can't import a type or module when the name of the item being imported is the same as another type or submodule defined in the module.
An example of this error:
use foo::Bar; // error type Bar = u32; mod foo { pub mod Bar { } } fn main() {}
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The name chosen for an external crate conflicts with another external crate that has been imported into the current module.
Wrong example:
extern crate a; extern crate crate_a as a;
The solution is to choose a different name that doesn't conflict with any external crate imported into the current module.
Correct example:
extern crate a; extern crate crate_a as other_name;
The name for an item declaration conflicts with an external crate's name.
For instance,
extern crate abc; struct abc;
There are two possible solutions:
Solution #1: Rename the item.
extern crate abc; struct xyz;
Solution #2: Import the crate with a different name.
extern crate abc as xyz; struct abc;
See the Declaration Statements section of the reference for more information about what constitutes an Item declaration and what does not:
When using a lifetime like 'a in a type, it must be declared before being
used.
These two examples illustrate the problem:
// error, use of undeclared lifetime name `'a` fn foo(x: &'a str) { } struct Foo { // error, use of undeclared lifetime name `'a` x: &'a str, }
These can be fixed by declaring lifetime parameters:
fn foo<'a>(x: &'a str) { } struct Foo<'a> { x: &'a str, }
Declaring certain lifetime names in parameters is disallowed. For example,
because the 'static lifetime is a special built-in lifetime name denoting
the lifetime of the entire program, this is an error:
// error, invalid lifetime parameter name `'static` fn foo<'static>(x: &'static str) { }
A lifetime name cannot be declared more than once in the same scope. For example:
// error, lifetime name `'a` declared twice in the same scope fn foo<'a, 'b, 'a>(x: &'a str, y: &'b str) { }
No description.
This error indicates that a static or constant references itself. All statics and constants need to resolve to a value in an acyclic manner.
For example, neither of the following can be sensibly compiled:
const X: u32 = X;
const X: u32 = Y; const Y: u32 = X;
This error indicates the use of a loop keyword (break or continue) inside a
closure but outside of any loop. Erroneous code example:
let w = || { break; }; // error: `break` inside of a closure
break and continue keywords can be used as normal inside closures as long as
they are also contained within a loop. To halt the execution of a closure you
should instead use a return statement. Example:
let w = || { for _ in 0..10 { break; } }; w();
This error indicates the use of a loop keyword (break or continue) outside
of a loop. Without a loop to break out of or continue in, no sensible action can
be taken. Erroneous code example:
fn some_func() { break; // error: `break` outside of loop }
Please verify that you are using break and continue only in loops. Example:
fn some_func() { for _ in 0..10 { break; // ok! } }
Functions must eventually return a value of their return type. For example, in the following function
fn foo(x: u8) -> u8 { if x > 0 { x // alternatively, `return x` } // nothing here }
if the condition is true, the value x is returned, but if the condition is
false, control exits the if block and reaches a place where nothing is being
returned. All possible control paths must eventually return a u8, which is not
happening here.
An easy fix for this in a complicated function is to specify a default return value, if possible:
fn foo(x: u8) -> u8 { if x > 0 { x // alternatively, `return x` } // lots of other if branches 0 // return 0 if all else fails }
It is advisable to find out what the unhandled cases are and check for them, returning an appropriate value or panicking if necessary.
Rust lets you define functions which are known to never return, i.e. are
'diverging', by marking its return type as !.
For example, the following functions never return:
fn foo() -> ! { loop {} } fn bar() -> ! { foo() // foo() is diverging, so this will diverge too } fn baz() -> ! { panic!(); // this macro internally expands to a call to a diverging function }
Such functions can be used in a place where a value is expected without returning a value of that type, for instance:
let y = match x { 1 => 1, 2 => 4, _ => foo() // diverging function called here }; println!("{}", y)
If the third arm of the match block is reached, since foo() doesn't ever
return control to the match block, it is fine to use it in a place where an
integer was expected. The match block will never finish executing, and any
point where y (like the print statement) is needed will not be reached.
However, if we had a diverging function that actually does finish execution
fn foo() -> { loop {break;} }
then we would have an unknown value for y in the following code:
let y = match x { 1 => 1, 2 => 4, _ => foo() }; println!("{}", y);
In the previous example, the print statement was never reached when the wildcard
match arm was hit, so we were okay with foo() not returning an integer that we
could set to y. But in this example, foo() actually does return control, so
the print statement will be executed with an uninitialized value.
Obviously we cannot have functions which are allowed to be used in such
positions and yet can return control. So, if you are defining a function that
returns !, make sure that there is no way for it to actually finish executing.
This is because of a type mismatch between the associated type of some
trait (e.g. T::Bar, where T implements trait Quux { type Bar; })
and another type U that is required to be equal to T::Bar, but is not.
Examples follow.
Here is a basic example:
trait Trait { type AssociatedType; } fn foo<T>(t: T) where T: Trait<AssociatedType=u32> { println!("in foo"); } impl Trait for i8 { type AssociatedType = &'static str; } foo(3_i8);
Here is that same example again, with some explanatory comments:
trait Trait { type AssociatedType; } fn foo<T>(t: T) where T: Trait<AssociatedType=u32> { // ~~~~~~~~ ~~~~~~~~~~~~~~~~~~ // | | // This says `foo` can | // only be used with | // some type that | // implements `Trait`. | // | // This says not only must // `T` be an impl of `Trait` // but also that the impl // must assign the type `u32` // to the associated type. println!("in foo"); } impl Trait for i8 { type AssociatedType = &'static str; } ~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ // | | // `i8` does have | // implementation | // of `Trait`... | // ... but it is an implementation // that assigns `&'static str` to // the associated type. foo(3_i8); // Here, we invoke `foo` with an `i8`, which does not satisfy // the constraint `<i8 as Trait>::AssociatedType=u32`, and // therefore the type-checker complains with this error code.
Here is a more subtle instance of the same problem, that can arise with for-loops in Rust:
let vs: Vec<i32> = vec![1, 2, 3, 4]; for v in &vs { match v { 1 => {} _ => {} } }
The above fails because of an analogous type mismatch, though may be harder to see. Again, here are some explanatory comments for the same example:
{
let vs = vec![1, 2, 3, 4];
// `for`-loops use a protocol based on the `Iterator`
// trait. Each item yielded in a `for` loop has the
// type `Iterator::Item` -- that is,I `Item` is the
// associated type of the concrete iterator impl.
for v in &vs {
// ~ ~~~
// | |
// | We borrow `vs`, iterating over a sequence of
// | *references* of type `&Elem` (where `Elem` is
// | vector's element type). Thus, the associated
// | type `Item` must be a reference `&`-type ...
// |
// ... and `v` has the type `Iterator::Item`, as dictated by
// the `for`-loop protocol ...
match v {
1 => {}
// ~
// |
// ... but *here*, `v` is forced to have some integral type;
// only types like `u8`,`i8`,`u16`,`i16`, et cetera can
// match the pattern `1` ...
_ => {}
}
// ... therefore, the compiler complains, because it sees
// an attempt to solve the equations
// `some integral-type` = type-of-`v`
// = `Iterator::Item`
// = `&Elem` (i.e. `some reference type`)
//
// which cannot possibly all be true.
}
}
To avoid those issues, you have to make the types match correctly. So we can fix the previous examples like this:
// Basic Example: trait Trait { type AssociatedType; } fn foo<T>(t: T) where T: Trait<AssociatedType = &'static str> { println!("in foo"); } impl Trait for i8 { type AssociatedType = &'static str; } foo(3_i8); // For-Loop Example: let vs = vec![1, 2, 3, 4]; for v in &vs { match v { &1 => {} _ => {} } }
The #[rustc_on_unimplemented] attribute lets you specify a custom error
message for when a particular trait isn't implemented on a type placed in a
position that needs that trait. For example, when the following code is
compiled:
fn foo<T: Index<u8>>(x: T){} #[rustc_on_unimplemented = "the type `{Self}` cannot be indexed by `{Idx}`"] trait Index<Idx> { ... } foo(true); // `bool` does not implement `Index<u8>`
there will be an error about bool not implementing Index<u8>, followed by a
note saying "the type bool cannot be indexed by u8".
As you can see, you can specify type parameters in curly braces for substitution
with the actual types (using the regular format string syntax) in a given
situation. Furthermore, {Self} will substitute to the type (in this case,
bool) that we tried to use.
This error appears when the curly braces contain an identifier which doesn't
match with any of the type parameters or the string Self. This might happen if
you misspelled a type parameter, or if you intended to use literal curly braces.
If it is the latter, escape the curly braces with a second curly brace of the
same type; e.g. a literal { is {{
The #[rustc_on_unimplemented] attribute lets you specify a custom error
message for when a particular trait isn't implemented on a type placed in a
position that needs that trait. For example, when the following code is
compiled:
fn foo<T: Index<u8>>(x: T){} #[rustc_on_unimplemented = "the type `{Self}` cannot be indexed by `{Idx}`"] trait Index<Idx> { ... } foo(true); // `bool` does not implement `Index<u8>`
there will be an error about bool not implementing Index<u8>, followed by a
note saying "the type bool cannot be indexed by u8".
As you can see, you can specify type parameters in curly braces for substitution
with the actual types (using the regular format string syntax) in a given
situation. Furthermore, {Self} will substitute to the type (in this case,
bool) that we tried to use.
This error appears when the curly braces do not contain an identifier. Please
add one of the same name as a type parameter. If you intended to use literal
braces, use {{ and }} to escape them.
The #[rustc_on_unimplemented] attribute lets you specify a custom error
message for when a particular trait isn't implemented on a type placed in a
position that needs that trait. For example, when the following code is
compiled:
fn foo<T: Index<u8>>(x: T){} #[rustc_on_unimplemented = "the type `{Self}` cannot be indexed by `{Idx}`"] trait Index<Idx> { ... } foo(true); // `bool` does not implement `Index<u8>`
there will be an error about bool not implementing Index<u8>, followed by a
note saying "the type bool cannot be indexed by u8".
For this to work, some note must be specified. An empty attribute will not do anything, please remove the attribute or add some helpful note for users of the trait.
This error occurs when there was a recursive trait requirement that overflowed before it could be evaluated. Often this means that there is unbounded recursion in resolving some type bounds.
For example, in the following code
trait Foo {} struct Bar<T>(T); impl<T> Foo for T where Bar<T>: Foo {}
to determine if a T is Foo, we need to check if Bar<T> is Foo. However,
to do this check, we need to determine that Bar<Bar<T>> is Foo. To determine
this, we check if Bar<Bar<Bar<T>>> is Foo, and so on. This is clearly a
recursive requirement that can't be resolved directly.
Consider changing your trait bounds so that they're less self-referential.
This error occurs when a bound in an implementation of a trait does not match the bounds specified in the original trait. For example:
trait Foo { fn foo<T>(x: T); } impl Foo for bool { fn foo<T>(x: T) where T: Copy {} }
Here, all types implementing Foo must have a method foo<T>(x: T) which can
take any type T. However, in the impl for bool, we have added an extra
bound that T is Copy, which isn't compatible with the original trait.
Consider removing the bound from the method or adding the bound to the original method definition in the trait.
You tried to use a type which doesn't implement some trait in a place which expected that trait. Erroneous code example:
// here we declare the Foo trait with a bar method trait Foo { fn bar(&self); } // we now declare a function which takes an object implementing the Foo trait fn some_func<T: Foo>(foo: T) { foo.bar(); } fn main() { // we now call the method with the i32 type, which doesn't implement // the Foo trait some_func(5i32); // error: the trait `Foo` is not implemented for the // type `i32` }
In order to fix this error, verify that the type you're using does implement the trait. Example:
trait Foo { fn bar(&self); } fn some_func<T: Foo>(foo: T) { foo.bar(); // we can now use this method since i32 implements the // Foo trait } // we implement the trait on the i32 type impl Foo for i32 { fn bar(&self) {} } fn main() { some_func(5i32); // ok! }
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This error indicates that type inference did not result in one unique possible type, and extra information is required. In most cases this can be provided by adding a type annotation. Sometimes you need to specify a generic type parameter manually.
A common example is the collect method on Iterator. It has a generic type
parameter with a FromIterator bound, which for a char iterator is
implemented by Vec and String among others. Consider the following snippet
that reverses the characters of a string:
let x = "hello".chars().rev().collect();
In this case, the compiler cannot infer what the type of x should be:
Vec<char> and String are both suitable candidates. To specify which type to
use, you can use a type annotation on x:
let x: Vec<char> = "hello".chars().rev().collect();
It is not necessary to annotate the full type. Once the ambiguity is resolved, the compiler can infer the rest:
let x: Vec<_> = "hello".chars().rev().collect();
Another way to provide the compiler with enough information, is to specify the generic type parameter:
let x = "hello".chars().rev().collect::<Vec<char>>();
Again, you need not specify the full type if the compiler can infer it:
let x = "hello".chars().rev().collect::<Vec<_>>();
Apart from a method or function with a generic type parameter, this error can occur when a type parameter of a struct or trait cannot be inferred. In that case it is not always possible to use a type annotation, because all candidates have the same return type. For instance:
struct Foo<T> { // Some fields omitted. } impl<T> Foo<T> { fn bar() -> i32 { 0 } fn baz() { let number = Foo::bar(); } }
This will fail because the compiler does not know which instance of Foo to
call bar on. Change Foo::bar() to Foo::<T>::bar() to resolve the error.
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This error indicates that the given recursion limit could not be parsed. Ensure that the value provided is a positive integer between quotes, like so:
#![recursion_limit="1000"]
Patterns used to bind names must be irrefutable. That is, they must guarantee
that a name will be extracted in all cases. Instead of pattern matching the
loop variable, consider using a match or if let inside the loop body. For
instance:
// This fails because `None` is not covered. for Some(x) in xs { ... } // Match inside the loop instead: for item in xs { match item { Some(x) => ... None => ... } } // Or use `if let`: for item in xs { if let Some(x) = item { ... } }
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Mutable borrows are not allowed in pattern guards, because matching cannot have side effects. Side effects could alter the matched object or the environment on which the match depends in such a way, that the match would not be exhaustive. For instance, the following would not match any arm if mutable borrows were allowed:
match Some(()) { None => { }, option if option.take().is_none() => { /* impossible, option is `Some` */ }, Some(_) => { } // When the previous match failed, the option became `None`. }
Assignments are not allowed in pattern guards, because matching cannot have side effects. Side effects could alter the matched object or the environment on which the match depends in such a way, that the match would not be exhaustive. For instance, the following would not match any arm if assignments were allowed:
match Some(()) { None => { }, option if { option = None; false } { }, Some(_) => { } // When the previous match failed, the option became `None`. }
In certain cases it is possible for sub-bindings to violate memory safety. Updates to the borrow checker in a future version of Rust may remove this restriction, but for now patterns must be rewritten without sub-bindings.
// Before. match Some("hi".to_string()) { ref op_string_ref @ Some(ref s) => ... None => ... } // After. match Some("hi".to_string()) { Some(ref s) => { let op_string_ref = &Some(s); ... } None => ... }
The op_string_ref binding has type &Option<&String> in both cases.
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In an array literal [x; N], N is the number of elements in the array. This
number cannot be negative.
The length of an array is part of its type. For this reason, this length must be a compile-time constant.
This error occurs when the compiler was unable to infer the concrete type of a variable. It can occur for several cases, the most common of which is a mismatch in the expected type that the compiler inferred for a variable's initializing expression, and the actual type explicitly assigned to the variable.
For example:
let x: i32 = "I am not a number!"; // ~~~ ~~~~~~~~~~~~~~~~~~~~ // | | // | initializing expression; // | compiler infers type `&str` // | // type `i32` assigned to variable `x`
Types in type definitions have lifetimes associated with them that represent how long the data stored within them is guaranteed to be live. This lifetime must be as long as the data needs to be alive, and missing the constraint that denotes this will cause this error.
// This won't compile because T is not constrained, meaning the data // stored in it is not guaranteed to last as long as the reference struct Foo<'a, T> { foo: &'a T } // This will compile, because it has the constraint on the type parameter struct Foo<'a, T: 'a> { foo: &'a T }
Types in type definitions have lifetimes associated with them that represent how long the data stored within them is guaranteed to be live. This lifetime must be as long as the data needs to be alive, and missing the constraint that denotes this will cause this error.
// This won't compile because T is not constrained to the static lifetime // the reference needs struct Foo<T> { foo: &'static T } // This will compile, because it has the constraint on the type parameter struct Foo<T: 'static> { foo: &'static T }
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User-defined types or type parameters cannot shadow the primitive types. This error indicates you tried to define a type, struct or enum with the same name as an existing primitive type.
See the Types section of the reference for more information about the primitive types:
Default impls for a trait must be located in the same crate where the trait was defined. For more information see the opt-in builtin traits RFC.
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The Sized trait is a special trait built-in to the compiler for types with a
constant size known at compile-time. This trait is automatically implemented
for types as needed by the compiler, and it is currently disallowed to
explicitly implement it for a type.
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The types of any associated constants in a trait implementation must match the types in the trait definition. This error indicates that there was a mismatch.
Here's an example of this error:
trait Foo { const BAR: bool; } struct Bar; impl Foo for Bar { const BAR: u32 = 5; // error, expected bool, found u32 }
You cannot use associated items other than constant items as patterns. This includes method items. Example of erroneous code:
enum B {} impl B { fn bb() -> i32 { 0 } } fn main() { match 0 { B::bb => {} // error: associated items in match patterns must // be constants } }
Please check that you're not using a method as a pattern. Example:
enum B { ba, bb } fn main() { match B::ba { B::bb => {} // ok! _ => {} } }
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Private items cannot be publicly re-exported. This error indicates that
you attempted to pub use a type or value that was not itself public.
Here is an example that demonstrates the error:
mod foo { const X: u32 = 1; } pub use foo::X;
The solution to this problem is to ensure that the items that you are
re-exporting are themselves marked with pub:
mod foo { pub const X: u32 = 1; } pub use foo::X;
See the 'Use Declarations' section of the reference for more information on this topic:
Private modules cannot be publicly re-exported. This error indicates
that you attempted to pub use a module that was not itself public.
Here is an example that demonstrates the error:
mod foo { pub const X: u32 = 1; } pub use foo as foo2;
The solution to this problem is to ensure that the module that you are
re-exporting is itself marked with pub:
pub mod foo { pub const X: u32 = 1; } pub use foo as foo2;
See the 'Use Declarations' section of the reference for more information on this topic:
An attempt was made to implement Drop on a concrete specialization of a
generic type. An example is shown below:
struct Foo<T> { t: T } impl Drop for Foo<u32> { fn drop(&mut self) {} }
This code is not legal: it is not possible to specialize Drop to a subset of
implementations of a generic type. One workaround for this is to wrap the
generic type, as shown below:
struct Foo<T> { t: T } struct Bar { t: Foo<u32> } impl Drop for Bar { fn drop(&mut self) {} }
An attempt was made to implement Drop on a specialization of a generic type.
An example is shown below:
trait Foo{} struct MyStruct<T> { t: T } impl<T: Foo> Drop for MyStruct<T> { fn drop(&mut self) {} }
This code is not legal: it is not possible to specialize Drop to a subset of
implementations of a generic type. In order for this code to work, MyStruct
must also require that T implements Foo. Alternatively, another option is
to wrap the generic type in another that specializes appropriately:
trait Foo{} struct MyStruct<T> { t: T } struct MyStructWrapper<T: Foo> { t: MyStruct<T> } impl <T: Foo> Drop for MyStructWrapper<T> { fn drop(&mut self) {} }
This error indicates that a binary assignment operator like += or ^= was
applied to the wrong types. For example:
let mut x: u16 = 5; x ^= true; // error, `^=` cannot be applied to types `u16` and `bool` x += (); // error, `+=` cannot be applied to types `u16` and `()`
Another problem you might be facing is this: suppose you've overloaded the +
operator for some type Foo by implementing the std::ops::Add trait for
Foo, but you find that using += does not work, as in this example:
use std::ops::Add; struct Foo(u32); impl Add for Foo { type Output = Foo; fn add(self, rhs: Foo) -> Foo { Foo(self.0 + rhs.0) } } fn main() { let mut x: Foo = Foo(5); x += Foo(7); // error, `+= cannot be applied to types `Foo` and `Foo` }
This is because the binary assignment operators currently do not work off of traits, so it is not possible to overload them. See RFC 953 for a proposal to change this.
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When Trait2 is a subtrait of Trait1 (for example, when Trait2 has a
definition like trait Trait2: Trait1 { ... }), it is not allowed to implement
Trait1 for Trait2. This is because Trait2 already implements Trait1 by
definition, so it is not useful to do this.
Example:
trait Foo { fn foo(&self) { } } trait Bar: Foo { } trait Baz: Bar { } impl Bar for Baz { } // error, `Baz` implements `Bar` by definition impl Foo for Baz { } // error, `Baz` implements `Bar` which implements `Foo` impl Baz for Baz { } // error, `Baz` (trivially) implements `Baz` impl Baz for Bar { } // Note: This is OK
Trying to implement a trait for a trait object (as in impl Trait1 for Trait2 { ... }) does not work if the trait is not object-safe. Please see the
RFC 255 for more details on object safety rules.
This error occurs when an attempt is made to use data captured by a closure, when that data may no longer exist. It's most commonly seen when attempting to return a closure:
fn foo() -> Box<Fn(u32) -> u32> { let x = 0u32; Box::new(|y| x + y) }
Notice that x is stack-allocated by foo(). By default, Rust captures
closed-over data by reference. This means that once foo() returns, x no
longer exists. An attempt to access x within the closure would thus be unsafe.
Another situation where this might be encountered is when spawning threads:
fn foo() { let x = 0u32; let y = 1u32; let thr = std::thread::spawn(|| { x + y }); }
Since our new thread runs in parallel, the stack frame containing x and y
may well have disappeared by the time we try to use them. Even if we call
thr.join() within foo (which blocks until thr has completed, ensuring the
stack frame won't disappear), we will not succeed: the compiler cannot prove
that this behaviour is safe, and so won't let us do it.
The solution to this problem is usually to switch to using a move closure.
This approach moves (or copies, where possible) data into the closure, rather
than taking references to it. For example:
fn foo() -> Box<Fn(u32) -> u32> { let x = 0u32; Box::new(move |y| x + y) }
Now that the closure has its own copy of the data, there's no need to worry about safety.
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Method calls that aren't calls to inherent const methods are disallowed
in statics, constants, and constant functions.
For example:
const BAZ: i32 = Foo(25).bar(); // error, `bar` isn't `const` struct Foo(i32); impl Foo { const fn foo(&self) -> i32 { self.bar() // error, `bar` isn't `const` } fn bar(&self) -> i32 { self.0 } }
For more information about const fn's, see RFC 911.
Default impls are only allowed for traits with no methods or associated items. For more information see the opt-in builtin traits RFC.
It is not allowed to use or capture an uninitialized variable. For example:
fn main() { let x: i32; let y = x; // error, use of possibly uninitialized variable
To fix this, ensure that any declared variables are initialized before being used.
This error occurs when an attempt is made to use a variable after its contents have been moved elsewhere. For example:
struct MyStruct { s: u32 } fn main() { let mut x = MyStruct{ s: 5u32 }; let y = x; x.s = 6; println!("{}", x.s); }
Since MyStruct is a type that is not marked Copy, the data gets moved out
of x when we set y. This is fundamental to Rust's ownership system: outside
of workarounds like Rc, a value cannot be owned by more than one variable.
If we own the type, the easiest way to address this problem is to implement
Copy and Clone on it, as shown below. This allows y to copy the
information in x, while leaving the original version owned by x. Subsequent
changes to x will not be reflected when accessing y.
#[derive(Copy, Clone)] struct MyStruct { s: u32 } fn main() { let mut x = MyStruct{ s: 5u32 }; let y = x; x.s = 6; println!("{}", x.s); }
Alternatively, if we don't control the struct's definition, or mutable shared
ownership is truly required, we can use Rc and RefCell:
use std::cell::RefCell; use std::rc::Rc; struct MyStruct { s: u32 } fn main() { let mut x = Rc::new(RefCell::new(MyStruct{ s: 5u32 })); let y = x.clone(); x.borrow_mut().s = 6; println!("{}", x.borrow.s); }
With this approach, x and y share ownership of the data via the Rc (reference
count type). RefCell essentially performs runtime borrow checking: ensuring
that at most one writer or multiple readers can access the data at any one time.
If you wish to learn more about ownership in Rust, start with the chapter in the Book:
This error occurs when an attempt is made to partially reinitialize a structure that is currently uninitialized.
For example, this can happen when a drop has taken place:
let mut x = Foo { a: 1 }; drop(x); // `x` is now uninitialized x.a = 2; // error, partial reinitialization of uninitialized structure `t`
This error can be fixed by fully reinitializing the structure in question:
let mut x = Foo { a: 1 }; drop(x); x = Foo { a: 2 };
This error occurs when an attempt is made to reassign an immutable variable. For example:
fn main(){ let x = 3; x = 5; // error, reassignment of immutable variable }
By default, variables in Rust are immutable. To fix this error, add the keyword
mut after the keyword let when declaring the variable. For example:
fn main(){ let mut x = 3; x = 5; }
No description.
This error occurs when an attempt is made to mutate the target of a mutable reference stored inside an immutable container.
For example, this can happen when storing a &mut inside an immutable Box:
let mut x: i64 = 1; let y: Box<_> = Box::new(&mut x); **y = 2; // error, cannot assign to data in an immutable container
This error can be fixed by making the container mutable:
let mut x: i64 = 1; let mut y: Box<_> = Box::new(&mut x); **y = 2;
It can also be fixed by using a type with interior mutability, such as Cell or
RefCell:
let x: i64 = 1; let y: Box<Cell<_>> = Box::new(Cell::new(x)); y.set(2);
This error occurs when an attempt is made to mutate or mutably reference data that a closure has captured immutably. Examples of this error are shown below:
// Accepts a function or a closure that captures its environment immutably. // Closures passed to foo will not be able to mutate their closed-over state. fn foo<F: Fn()>(f: F) { } // Attempts to mutate closed-over data. Error message reads: // `cannot assign to data in a captured outer variable...` fn mutable() { let mut x = 0u32; foo(|| x = 2); } // Attempts to take a mutable reference to closed-over data. Error message // reads: `cannot borrow data mutably in a captured outer variable...` fn mut_addr() { let mut x = 0u32; foo(|| { let y = &mut x; }); }
The problem here is that foo is defined as accepting a parameter of type Fn.
Closures passed into foo will thus be inferred to be of type Fn, meaning that
they capture their context immutably.
If the definition of foo is under your control, the simplest solution is to
capture the data mutably. This can be done by defining foo to take FnMut
rather than Fn:
fn foo<F: FnMut()>(f: F) { }
Alternatively, we can consider using the Cell and RefCell types to achieve
interior mutability through a shared reference. Our example's mutable function
could be redefined as below:
use std::cell::Cell; fn mutable() { let x = Cell::new(0u32); foo(|| x.set(2)); }
You can read more about cell types in the API documentation:
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This error indicates that some types or traits depend on each other and therefore cannot be constructed.
The following example contains a circular dependency between two traits:
trait FirstTrait : SecondTrait { } trait SecondTrait : FirstTrait { }
This error indicates that a type or lifetime parameter has been declared but not actually used. Here is an example that demonstrates the error:
enum Foo<T> { Bar }
If the type parameter was included by mistake, this error can be fixed by simply removing the type parameter, as shown below:
enum Foo { Bar }
Alternatively, if the type parameter was intentionally inserted, it must be used. A simple fix is shown below:
enum Foo<T> { Bar(T) }
This error may also commonly be found when working with unsafe code. For example, when using raw pointers one may wish to specify the lifetime for which the pointed-at data is valid. An initial attempt (below) causes this error:
struct Foo<'a, T> { x: *const T }
We want to express the constraint that Foo should not outlive 'a, because
the data pointed to by T is only valid for that lifetime. The problem is
that there are no actual uses of 'a. It's possible to work around this
by adding a PhantomData type to the struct, using it to tell the compiler
to act as if the struct contained a borrowed reference &'a T:
use std::marker::PhantomData; struct Foo<'a, T: 'a> { x: *const T, phantom: PhantomData<&'a T> }
PhantomData can also be used to express information about unused type parameters. You can read more about it in the API documentation:
https://doc.rust-lang.org/std/marker/struct.PhantomData.html
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From RFC 246:
It is invalid for a static to reference another static by value. It is required that all references be borrowed.
The value assigned to a constant expression must be known at compile time, which is not the case when comparing raw pointers. Erroneous code example:
static foo: i32 = 42; static bar: i32 = 43; static baz: bool = { (&foo as *const i32) == (&bar as *const i32) }; // error: raw pointers cannot be compared in statics!
Please check that the result of the comparison can be determined at compile time or isn't assigned to a constant expression. Example:
static foo: i32 = 42; static bar: i32 = 43; let baz: bool = { (&foo as *const i32) == (&bar as *const i32) }; // baz isn't a constant expression so it's ok
The value assigned to a constant expression must be known at compile time, which is not the case when dereferencing raw pointers. Erroneous code example:
const foo: i32 = 42; const baz: *const i32 = (&foo as *const i32); const deref: i32 = *baz; // error: raw pointers cannot be dereferenced in constants
To fix this error, please do not assign this value to a constant expression. Example:
const foo: i32 = 42; const baz: *const i32 = (&foo as *const i32); unsafe { let deref: i32 = *baz; } // baz isn't a constant expression so it's ok
You'll also note that this assignment must be done in an unsafe block!
It is not allowed for a mutable static to allocate or have destructors. For example:
// error: mutable statics are not allowed to have boxes static mut FOO: Option<Box<usize>> = None; // error: mutable statics are not allowed to have destructors static mut BAR: Option<Vec<i32>> = None;
In Rust 1.3, the default object lifetime bounds are expected to change, as described in RFC #1156 1. You are getting a warning because the compiler thinks it is possible that this change will cause a compilation error in your code. It is possible, though unlikely, that this is a false alarm.
The heart of the change is that where &'a Box<SomeTrait> used to
default to &'a Box<SomeTrait+'a>, it now defaults to &'a Box<SomeTrait+'static> (here, SomeTrait is the name of some trait
type). Note that the only types which are affected are references to
boxes, like &Box<SomeTrait> or &[Box<SomeTrait>]. More common
types like &SomeTrait or Box<SomeTrait> are unaffected.
To silence this warning, edit your code to use an explicit bound.
Most of the time, this means that you will want to change the
signature of a function that you are calling. For example, if
the error is reported on a call like foo(x), and foo is
defined as follows:
fn foo(arg: &Box<SomeTrait>) { ... }
you might change it to:
fn foo<'a>(arg: &Box<SomeTrait+'a>) { ... }
This explicitly states that you expect the trait object SomeTrait to
contain references (with a maximum lifetime of 'a).
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Some type parameters have the same name. Example of erroneous code:
fn foo<T, T>(s: T, u: T) {} // error: the name `T` is already used for a type // parameter in this type parameter list
Please verify that none of the type parameterss are misspelled, and rename any clashing parameters. Example:
fn foo<T, Y>(s: T, u: Y) {} // ok!
You tried to implement something which was not a trait on an object. Example of erroneous code:
struct Foo; struct Bar; impl Foo for Bar {} // error: `Foo` is not a trait
Please verify that you didn't misspell the trait's name or otherwise use the wrong identifier. Example:
trait Foo { // some functions } struct Bar; impl Foo for Bar { // ok! // functions implementation }
An unknown trait was implemented. Example of erroneous code:
struct Foo; impl SomeTrait for Foo {} // error: use of undeclared trait name `SomeTrait`
Please verify that the name of the trait wasn't misspelled and ensure that it was imported. Example:
// solution 1: use some_file::SomeTrait; // solution 2: trait SomeTrait { // some functions } struct Foo; impl SomeTrait for Foo { // ok! // implements functions }
No description.
A definition of a method not in the implemented trait was given in a trait implementation. Example of erroneous code:
trait Foo { fn a(); } struct Bar; impl Foo for Bar { fn a() {} fn b() {} // error: method `b` is not a member of trait `Foo` }
Please verify you didn't misspell the method name and you used the correct trait. First example:
trait Foo { fn a(); fn b(); } struct Bar; impl Foo for Bar { fn a() {} fn b() {} // ok! }
Second example:
trait Foo { fn a(); } struct Bar; impl Foo for Bar { fn a() {} } impl Bar { fn b() {} }
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An undeclared type name was used. Example of erroneous codes:
impl Something {} // error: use of undeclared type name `Something` // or: trait Foo { fn bar(N); // error: use of undeclared type name `N` } // or: fn foo(x: T) {} // error: use of undeclared type name `T`
To fix this error, please verify you didn't misspell the type name, you did declare it or imported it into the scope. Examples:
struct Something; impl Something {} // ok! // or: trait Foo { type N; fn bar(Self::N); // ok! } //or: fn foo<T>(x: T) {} // ok!
A declaration shadows an enum variant or unit-like struct in scope. Example of erroneous code:
struct Foo; let Foo = 12i32; // error: declaration of `Foo` shadows an enum variant or // unit-like struct in scope
To fix this error, rename the variable such that it doesn't shadow any enum variable or structure in scope. Example:
struct Foo; let foo = 12i32; // ok!
Or:
struct FooStruct; let Foo = 12i32; // ok!
The goal here is to avoid a conflict of names.
No description.
More than one function parameter have the same name. Example of erroneous code:
fn foo(f: i32, f: i32) {} // error: identifier `f` is bound more than // once in this parameter list
Please verify you didn't misspell parameters' name. Example:
fn foo(f: i32, g: i32) {} // ok!
An identifier is bound more than once in a pattern. Example of erroneous code:
match (1, 2) { (x, x) => {} // error: identifier `x` is bound more than once in the // same pattern }
Please verify you didn't misspell identifiers' name. Example:
match (1, 2) { (x, y) => {} // ok! }
Or maybe did you mean to unify? Consider using a guard:
match (A, B, C) { (x, x2, see) if x == x2 => { /* A and B are equal, do one thing */ } (y, z, see) => { /* A and B unequal; do another thing */ } }
A static variable was referenced in a pattern. Example of erroneous code:
static FOO : i32 = 0; match 0 { FOO => {} // error: static variables cannot be referenced in a // pattern, use a `const` instead _ => {} }
The compiler needs to know the value of the pattern at compile time; compile-time patterns can defined via const or enum items. Please verify that the identifier is spelled correctly, and if so, use a const instead of static to define it. Example:
const FOO : i32 = 0; match 0 { FOO => {} // ok! _ => {} }
No description.
An unknown enum variant, struct or const was used. Example of erroneous code:
match 0 { Something::Foo => {} // error: unresolved enum variant, struct // or const `Foo` }
Please verify you didn't misspell it and the enum variant, struct or const has been declared and imported into scope. Example:
enum Something { Foo, NotFoo, } match Something::NotFoo { Something::Foo => {} // ok! _ => {} }
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A struct variant name was used like a function name. Example of
erroneous code:
struct Foo { a: bool}; let f = Foo(); // error: `Foo` is a struct variant name, but this expression uses // it like a function name
Please verify you didn't misspell the name of what you actually wanted to use here. Example:
fn Foo() -> u32 { 0 } let f = Foo(); // ok!
The self keyword was used in a static method. Example of erroneous code:
struct Foo; impl Foo { fn bar(self) {} fn foo() { self.bar(); // error: `self` is not available in a static method. } }
Please check if the method's argument list should have contained self,
&self, or &mut self (in case you didn't want to create a static
method), and add it if so. Example:
struct Foo; impl Foo { fn bar(self) {} fn foo(self) { self.bar(); // ok! } }
An unresolved name was used. Example of erroneous codes:
something_that_doesnt_exist::foo; // error: unresolved name `something_that_doesnt_exist::foo` // or: trait Foo { fn bar() { Self; // error: unresolved name `Self` } } // or: let x = unknown_variable; // error: unresolved name `unknown_variable`
Please verify that the name wasn't misspelled and ensure that the identifier being referred to is valid for the given situation. Example:
enum something_that_does_exist { foo } // or: mod something_that_does_exist { pub static foo : i32 = 0i32; } something_that_does_exist::foo; // ok! // or: let unknown_variable = 12u32; let x = unknown_variable; // ok!
An undeclared label was used. Example of erroneous code:
loop { break 'a; // error: use of undeclared label `'a` }
Please verify you spelt or declare the label correctly. Example:
'a: loop { break 'a; // ok! }
No description.
A type or module has been defined more than once. Example of erroneous code:
struct Bar; struct Bar; // error: duplicate definition of value `Bar`
Please verify you didn't misspell the type/module's name or remove/rename the duplicated one. Example:
struct Bar; struct Bar2; // ok!
No description.
The self import appears more than once in the list. Erroneous code example:
use something::{self, self}; // error: `self` import can only appear once in // the list
Please verify you didn't misspell the import name or remove the duplicated
self import. Example:
use something::self; // ok!
self import was made. Erroneous code example:
use {self}; // error: `self` import can only appear in an import list with a // non-empty prefix
You cannot import the current module into itself, please remove this import or verify you didn't misspell it.
An import was unresolved. Erroneous code example:
use something::Foo; // error: unresolved import `something::Foo`.
Please verify you didn't misspell the import name or the import does exist in the module from where you tried to import it. Example:
use something::Foo; // ok! mod something { pub struct Foo; }
Invalid import. Example of erroneous code:
use something_which_doesnt_exist; // error: unresolved import `something_which_doesnt_exist`
Please verify you didn't misspell the import's name.
No description.
A non-constant value was used to initialise a constant. Example of erroneous code:
let foo = 42u32; const FOO : u32 = foo; // error: attempt to use a non-constant value in a // constant
To fix this error, please replace the value with a constant. Example:
const FOO : u32 = 42u32; // ok! // or: const OTHER_FOO : u32 = 42u32; const FOO : u32 = OTHER_FOO; // ok!
No description.
Trait implementations can only implement associated types that are members of the trait in question. This error indicates that you attempted to implement an associated type whose name does not match the name of any associated type in the trait.
Here is an example that demonstrates the error:
trait Foo {} impl Foo for i32 { type Bar = bool; }
The solution to this problem is to remove the extraneous associated type:
trait Foo {} impl Foo for i32 {}
Trait implementations can only implement associated constants that are members of the trait in question. This error indicates that you attempted to implement an associated constant whose name does not match the name of any associated constant in the trait.
Here is an example that demonstrates the error:
#![feature(associated_consts)] trait Foo {} impl Foo for i32 { const BAR: bool = true; }
The solution to this problem is to remove the extraneous associated constant:
trait Foo {} impl Foo for i32 {}