Unable to infer the type of a ~fn inside an Option

[MicahChalmer]

There seems to be no way to use a ~fn inside an Option without rewriting the type every time you use it. This line:

let _:Option<~fn()> = Some(|| { });

gives this error message:

error: mismatched types: expected `std::option::Option<~fn:Send()>` but found `std::option::Option<&fn<no-bounds>()>` (expected ~ closure, found & closure)

The problem is that the function inside the Some is interpreted to be a &fn rather than a ~fn, and then the compiler complains about a type mismatch. The only way to make it work is this to make it Some::<~fn()>(|| { }) instead. But when the function isn't just ~fn() and has parameters and a return type, that becomes extremely annoying.

[MicahChalmer]

This does not appear to be an issue anymore, ever since the change to proc instead of fn. The syntax is different than a &fn, so the compiler can no longer get confused. The equivalent code after the change: let _:Option<proc()> = Some(proc() {}) compiles just fine. Closing this.