Lifetime issue when `#[pin_project]` is used about pin-project HOT 6 CLOSED

And here is the result of reducing that expanded code:

trait Bar<X> {
    type Y;
}
struct Example<A>(A);
impl<X, T> Bar<X> for Example<T> {
    type Y = Option<T>;
}
struct Foo<A, B> {
    x: <Example<A> as Bar<B>>::Y,
}
struct __FooProjection<'pin, A, B> {
    x: &'pin mut <Example<A> as Bar<B>>::Y,
}

from pin-project.

It looks like the compiler is giving the correct error here. From the infer_outlives rfc:

In turn, if this associated type were used in a struct, where-clauses would be required. As we'll see in the reference-level explanation, this is a consequence of the fact that we do inference without regard for associated type normalization, but it makes for a relatively simple rule -- explicit where clauses are needed in the preseence of impls like the one above:

struct Foo<'a, T>
  where T: 'a // still required, not inferred from `field`
{
  field: <Vec<T> as MakeRef<'a>>::Type
}    

I think the solution is for pin-project to always generate T: 'pin for each type parameter T.

from pin-project.

@Aaron1011 it might be safer to add this bound instead:
where Foo<A, B>: 'pin

This also seems to fix the problem.

trait Bar<X> {
    type Y;
}
struct Example<A>(A);
impl<X, T> Bar<X> for Example<T> {
    type Y = Option<T>;
}
struct Foo<A, B> {
    x: <Example<A> as Bar<B>>::Y,
}
struct __FooProjection<'pin, A, B> where Foo<A, B>: 'pin {
    x: &'pin mut <Example<A> as Bar<B>>::Y,
}

Just in case there are parameters which don't need to have that constraint, we can leave that to the compiler to figure out.

from pin-project.

@Diggsey: That's a good idea - I'll do that instead.

from pin-project.

Published 0.4.9 which fixes this issue.

from pin-project.

Nice! 👍

from pin-project.