`Foo::<>` does not use type parameter defaults

[bluss]

This is about the stable Rust feature that types can have type parameter defaults.

Example code below. I expected that if Test::<T>::method specifies one type parameter, and leaves the second to the defaulted, that Test::<>::method should leave both type parameters to be defaulted. Instead it is equivalent to Test::method (both type parameters inferred).

#[derive(Default)]
pub struct Test<X = String, Y = String> {
 x: X,
 y: Y,
}

fn main() {
 // both explicit
 let s = Test::<f32, f32>::default();

 // use default for second
 let t = Test::<i32>::default();

 // use default for both? 
 // error[E0282]: unable to infer enough type information about `_`
 let u = Test::<>::default();

 // This uses the default for both.
 let u2 = <Test>::default();
}

playground link

[petrochenkov]

This works as expected right now, but the rules are... counterintuitive.
Basically <Test>, Test (aka Test::<>) and Test::<A, B, C> use different rules - 1) defaults are used, nothing is inferred, 2) everything is inferred, defaults are not used 3) defaults are used, nothing is inferred.
I hope this can be fixed somehow by using inference and falling back to defaults, but any inaccurate step from the current rules is a breaking change unfortunately.
@eddyb should know more details.

[bluss]

@petrochenkov Ok, I'm reminded that ::<> is valid on all types, I had forgot about that. Maybe this issue is a bit silly! It's still surprising that one can't reach the all defaults case using the regular path syntax.

[eddyb]

::<> is a noop, not sure if it should be or not. It is quite silly either way 😞.

[steveklabnik]

Closing in favor of rust-lang/reference#24