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type X = ::Y
use ::Y as X
To reproduce, try to compile this:
mod foo { pub enum Bar { A } } // use foo::Bar as Baz; // this works type Baz = foo::Bar; // this doesn't fn main() { let _ = Baz::A; }
This seems... counterintuitive.
The text was updated successfully, but these errors were encountered:
This is a duplicate of #26264
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To reproduce, try to compile this:
This seems... counterintuitive.
The text was updated successfully, but these errors were encountered: