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A type alias impl trait can only have its hidden type assigned
when used fully generically (and within their defining scope).
This means
```compile_fail,E0792
#![feature(type_alias_impl_trait)]
type Foo<T> = impl std::fmt::Debug;
fn foo() -> Foo<u32> {
5u32
}
```
is not accepted. If it were accepted, one could create unsound situations like
```compile_fail,E0792
#![feature(type_alias_impl_trait)]
type Foo<T> = impl Default;
fn foo() -> Foo<u32> {
5u32
}
fn main() {
let x = Foo::<&'static mut String>::default();
}
```
Instead you need to make the function generic:
```
#![feature(type_alias_impl_trait)]
type Foo<T> = impl std::fmt::Debug;
fn foo<U>() -> Foo<U> {
5u32
}
```
This means that no matter the generic parameter to `foo`,
the hidden type will always be `u32`.
If you want to link the generic parameter to the hidden type,
you can do that, too:
```
#![feature(type_alias_impl_trait)]
use std::fmt::Debug;
type Foo<T: Debug> = impl Debug;
fn foo<U: Debug>() -> Foo<U> {
Vec::<U>::new()
}
```
|
