Non-exported type in exported type signature does not error

[jethrogb]

mod m1 {
    struct A;
    pub fn x() -> A { A }
}

fn main() {
    let x=m1::x();
    println!("{:?}",&x as *const _ as *const u8);
}

This generates an error:

<anon>:3:16: 3:17 error: private type in exported type signature [E0446]
<anon>:3    pub fn x() -> A { A }
                          ^

Yet this compiles fine:

mod m1 {
    mod m2 { pub struct A; }
    use self::m2::A as A;
    pub fn x() -> A { A }
}

fn main() {
    let x=m1::x();
    println!("{:?}",&x as *const _ as *const u8);
}

However, one is unable to specify the return type of m1::x() outside of m1.

[jethrogb]

Related/potential dupe of #28450, #28325, #18082

[jethrogb]

There was some discussion on IRC that since struct A is marked pub, it might be exported, which might make this not-a-bug. I think it should be possible to definitively prove whether the type is exported in an accessible way. In particular, one might require for exported functions that when traversing up the path, at some point there must be a valid exported path for the function and all types in the function type signature.

[jethrogb]

Actually that can be circumvented by later re-exporting a module containing exporting a function but not a module exporting the appropriate type.

[petrochenkov]

According to RFC 136 a type is exported automatically if it's mentioned in any kind of public interface, with one minor sanity check - it should be marked as pub. It's not yet implemented though.
So, yeah, unnameable types is a thing. (But not necessarily a good one)

[DemiMarie]

The D language has Voldemort types, which are types that cannot be named. In that language they are widely used. However, I don't know how useful they are in Rust, which uses generics instead of templates.