# Error code E0792

A type alias impl trait can only have its hidden type assigned when used fully generically (and within their defining scope). This means

    #![allow(unused)]
    #![feature(type_alias_impl_trait)]

    fn main() {
    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

    #![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
    }

    fn main() {}

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()
    }

    fn main() {}