rustc macro: unexpected compilation error when `ty` designator is used within where clause

[0xBACD]

This code doesn't seem to compile with the following error [playground]:

macro_rules! foo {
    ($t:ty) => {
        fn bar<T>(_: T) where T: $t {}
    }
}

foo!(std::fmt::Debug);

error: expected one of `!`, `(`, `,`, `?`, `for`, `{`, `~`, lifetime, or path, found `std::fmt::Debug`
 --> src/main.rs:3:34
  |
3 |         fn bar<T>(_: T) where T: $t {}
  |                                  ^^ expected one of 9 possible tokens
...
7 | foo!(std::fmt::Debug);
  | --------------------- in this macro invocation
  |
  = note: this error originates in the macro `foo` (in Nightly builds, run with -Z macro-backtrace for more info)

There should be no error since ty designator stands for type, std::fmt::Debug is definitely a type and where clause accepts trait types.
This problem persists on any channel I've tested it - stable, beta and nightly.

[compiler-errors]

In this position, std::fmt::Debug is not meant to be a type. It's meant to be a trait, which is parsed differently than a type.

Have you tried using the path macro fragment?

[0xBACD]

Yeah, it seems I just misread the reference. Actually it says that ty only matches to trait objects and impl trait so compiler behaviour here seems legit. Thank you, @compiler-errors!

[petrochenkov]

FWIW, this will work once #96724 and subsequent PRs land (maybe not very soon).
std::fmt::Debug will be treated as just an "untyped" token stream then.