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|
The compiler could not infer a type and asked for a type annotation.
Erroneous code example:
```compile_fail,E0282
let x = Vec::new();
```
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.
In the example above, type `Vec` has a type parameter `T`. When calling
`Vec::new`, barring any other later usage of the variable `x` that allows the
compiler to infer what type `T` is, the compiler needs to be told what it is.
The type can be specified on the variable:
```
let x: Vec<i32> = Vec::new();
```
The type can also be specified in the path of the expression:
```
let x = Vec::<i32>::new();
```
In cases with more complex types, 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:
```compile_fail,E0282
struct Foo<T> {
num: T,
}
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.
|
