where clause hides f64: Div<f64> impl

[m-ou-se]

This code:

use std::ops::Div;

pub fn f<T>(_: T) -> f64
where
    f64: Div<T, Output = f64>,
{
    1.0 / 2.0
}

Results in this error:

error[E0308]: mismatched types
 --> src/lib.rs:7:11
  |
3 | pub fn f<T>(_: T) -> f64
  |          - this type parameter
...
7 |     1.0 / 2.0
  |           ^^^ expected type parameter `T`, found floating-point number
  |
  = note: expected type parameter `T`
                       found type `{float}`

But I expected it to compile fine, since f64: Div<f64, Output = f64> also exists unconditionally. It seems that the bound on f64 hides the existing f64: Div<f64>.

This seems to happen on all versions of Rust since 1.0.

[m-ou-se]

Some more data points:

• Div::div(1f64, 2f64) inside that function breaks with the same error.

• Div::<f64>::div(1.0, 2.0) works.

[fbstj]

playground link

• it also happens similarly with std::ops::Add
• it also happens if you constrain the RHS to 2.0f64 (found type 64)
• it happens to every {float} / ? operation in the function body, not just the return value

[cuviper]

I think this is part of what #41756 is talking about.

[cuviper]

Chasing that conversation leads to a more direct dupe, #24066.