No method named `poll` found for a type that implements `Future`

Question

I am attempting to create a struct that will allow someone to call .shutdown(), which will resolve a future (that is otherwise pending). It can only be called o

Answer 1

It's true, Receiver does not implement Future; only Pin<&mut Receiver> does. You need to project the pinning from your type to the field.

When the underlying type may not implement Unpin

impl Future for ShutdownHandle {
    type Output = ();

    fn poll(self: Pin<&mut Self>, cx: &mut Context) -> Poll<Self::Output> {
        // I copied this code from Stack Overflow without reading the text that
        // told me how to verify that this code uses `unsafe` correctly.
        unsafe { self.map_unchecked_mut(|s| &mut s.receiver) }.poll(cx).map(|_| ())
    }
}

You must read the pin module to thoroughly understand the requirements to use unsafe here.

A cleaner solution

I like to use a helper library, such as pin_project, to handle more complicated types of projection:

#[unsafe_project(Unpin)]
pub struct ShutdownHandle {
    #[pin]
    sender: oneshot::Sender<()>,
    #[pin]
    receiver: oneshot::Receiver<()>,
}

impl Future for ShutdownHandle {
    type Output = ();

    fn poll(self: Pin<&mut Self>, cx: &mut Context) -> Poll<Self::Output> {
        let this = self.project();
        this.receiver.poll(cx).map(|_| ())
    }
}

When the underlying type implements Unpin

Ömer Erden points out that the futures-preview crate provides FutureExt::poll_unpin. This method takes a mutable reference to a type that implements Unpin and creates a brand new Pin with it.

Since oneshot::Receiver does implement Unpin, this can be used here:

impl Future for ShutdownHandle {
    type Output = ();

    fn poll(mut self: Pin<&mut Self>, cx: &mut Context) -> Poll<Self::Output> {
        self.receiver.poll_unpin(cx).map(|_| ())
    }
}

See also

• When is it safe to move a member value out of a pinned future?
• Why is a trait not implemented for a type that clearly has it implemented?