Understanding Rust Error E0792

Explanation of Rust Error E0792

Rust error E0792 occurs when a type alias with trait implementation is used incorrectly in a non-generic function. To fix this error, the function must be made generic, ensuring that the hidden type is always assigned.

Problematic Code Example

Here's an example of code that would produce the E0792 error:

#![feature(type_alias_impl_trait)]

type Foo<T> = impl Default;

fn foo() -> Foo<u32> {
    5u32
}

fn main() {
    let x = Foo::<&'static mut String>::default();
}

In this example, the function foo() returns a Foo<u32> type, which is not generic. This conflicts with the definition of Foo, which requires a generic type.

Solution

To resolve this issue, we need to make the foo() function generic, as shown in the following revised example:

#![feature(type_alias_impl_trait)]

type Foo<T> = impl Default;

fn foo<U>() -> Foo<U> {
    5u32
}

fn main() {
}

In this revision, we changed the foo() function to accept a generic type U as an argument. Now, the foo() function can return a Foo instance with the correct generic type.

Advanced Usage

You can further refine your code by linking the generic parameter to the hidden type using an additional trait constraint. Here's an example:

#![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() {
}

In this example, we added the Debug trait constraint to T, allowing us to use the Debug trait on U. Consequently, the foo() function now returns a new Vec<U> instance that implements the Debug trait.