Error message for T <binop> U worsens if T has exactly one implementation of the relevant trait

[SNCPlay42]

This code

struct S;

fn main() {
    let _ = S == ();
}

Emits the error message

error[E0369]: binary operation `==` cannot be applied to type `S`
 --> src/main.rs:4:15
  |
4 |     let _ = S == ();
  |             - ^^ -- ()
  |             |
  |             S
  |
  = note: an implementation of `std::cmp::PartialEq` might be missing for `S`

However, if the type S has any implementation of PartialEq<T> for a single type T that is not (), the error message becomes less helpful:

#[derive(PartialEq)]
struct S;

fn main() {
    let _ = S == ();
}

error[E0308]: mismatched types
 --> src/main.rs:5:18
  |
5 |     let _ = S == ();
  |                  ^^ expected struct `S`, found `()`

If there are two or more impls of PartialEq, the error message becomes helpful again:

#[derive(PartialEq)]
struct S;

impl PartialEq<u8> for S {
    fn eq(&self, _: &u8) -> bool {
        unimplemented!()
    }
}

fn main() {
    let _ = S == ();
}

error[E0277]: can't compare `S` with `()`
  --> src/main.rs:11:15
   |
11 |     let _ = S == ();
   |               ^^ no implementation for `S == ()`
   |
   = help: the trait `std::cmp::PartialEq<()>` is not implemented for `S`

This also applies to other binary operation traits like PartialOrd and Add.

[SNCPlay42]

This is related to #40660 (which concerns &T == T), and is probably the root cause (I think? &T actually has two PartialEq implementations - for &T and for &mut T - but maybe something's going on with coercions), but the problem is more general than described in that issue.

[SNCPlay42]

So what's going on here is that when type-checking a binop, the compiler:

1. type-checks the LHS, determining its type
2. creates a type inference variable T1 for the type of the RHS
3. looks for an impl of Eq<T1> (or Add/PartialOrd/etc.) on the LHS, recording what this allows it to infer about T1
4. type-checks the RHS and demands that it be coercable to T1

So how exactly this fails depends on how many impls of Eq there are:

• If there are no impls of Eq for the LHS, step 3 fails
• If there is exactly one impl of Eq for the LHS, step 3 succeeds and the compiler can now infer exactly what T1 must be (e.g. S in the above case). Step 4 then fails when the RHS does not have the expected type. The compiler only complains that RHS != S without giving its reasoning for why it expected S.
• If there are more than one impl of Eq, step 3 succeeds but the compiler can't determine exactly what T1 is, only that it must satisfy LHS: Eq<T1>. Step 4 then tries to coerce the RHS to T1 and gives the error that LHS: Eq<RHS> doesn't hold.

Ideally one code path would be responsible for the diagnostics in all three cases.

[nagisa]

See the issue above for a reproduction of this which arguably is much worse. I've also been able to reproduce this without binops too.