Rust error E0792: expected generic type parameter, found `u32`

The compiler team's explanation

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

#![feature(type_alias_impl_trait)]

type Foo<T> = impl std::fmt::Debug;

#[define_opaque(Foo)]
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;

#[define_opaque(Foo)]
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;

#[define_opaque(Foo)]
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;

#[define_opaque(Foo)]
fn foo<U: Debug>() -> Foo<U> {
    Vec::<U>::new()
}

fn main() {}

From rustc --explain E0792 (1.99.0), part of the Rust project, MIT or Apache-2.0. Official page.