I have a trait with specific bounds, for which I want to provide a default implementation of bar(), as shown in the minimal example below. However, I’m getting an unexpected error E0308. It seems that Rust either forgets that the primitive type i32 implements Mul<i32> and ignores the explicit type of the local variable a, or ignores the Output = i32 part of trait Foo: Sized + Add<i32, Output = i32>...

Rust playground

use std::ops::{Add, Mul};

trait Foo: Sized + Add<i32, Output = i32>
where
    i32: Mul<Self>,
{
    fn bar(self) -> i32 {
        let a: i32 = self + 0;

        a * 1
    }
}

The error I’m getting:

error[E0308]: mismatched types
  --> src/lib.rs:10:13
   |
3  | trait Foo: Sized + Add<i32, Output = i32>
   | ----------------------------------------- expected this type parameter
...
10 |         a * 1
   |             ^ expected type parameter `Self`, found integer
   |
   = note: expected type parameter `Self`
                        found type `{integer}`

error[E0308]: mismatched types
  --> src/lib.rs:10:9
   |
7  |     fn bar(self) -> i32 {
   |                     --- expected `i32` because of return type
...
10 |         a * 1
   |         ^^^^^ expected `i32`, found associated type
   |
   = note:         expected type `i32`
           found associated type `<i32 as Mul<Self>>::Output`
   = help: consider constraining the associated type `<i32 as Mul<Self>>::Output` to `i32`
   = note: for more information, visit https://doc.rust-lang.org/book/ch19-03-advanced-traits.html

For more information about this error, try `rustc --explain E0308`.

Adding Mul<i32> to the trait bounds solves the issue, but I don’t understand why since i32 implements it anyway.

use std::ops::{Add, Mul};

trait Foo: Sized + Add<i32, Output = i32>
where
    i32: Mul<Self> + Mul<i32>,
{
    fn bar(self) -> i32 {
        let a: i32 = self + 0;

        a * 1
    }
}