未找到实现`Future`的类型的名为`poll`的方法 [英] No method named `poll` found for a type that implements `Future`
问题描述
我正在尝试创建一个允许某人调用.shutdown()的结构，该结构将解决将来的问题(否则该问题将待处理).它只能被调用一次.在Future特征的实现中，尽管
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 once. In the implementation of the Future trait, I receive an error that poll is not defined, despite it being present in the documentation (under impl Future).
尽管我将std::future::Future用作impl，但是我尝试添加use futures::prelude::*，这会将预览特征纳入范围. RLS和rustc都告诉我，导入尚未使用，所以这不是问题.
Though I am using std::future::Future as the impl, I tried adding use futures::prelude::*, which would bring the preview trait into scope. Both RLS and rustc inform me that the import is unused, so that's not the issue.
请注意，我不是使用简单的布尔标志，因为我希望它可以从任何线程中调用，这是与此处无关的实现细节.
Note that I am not using a simple boolean flag, as I intend for this to be able to be callable from any thread — that's an implementation detail that is irrelevant here.
use futures::channel::oneshot; // futures-preview@0.3.0-alpha.17
use std::{
future::Future,
pin::Pin,
task::{Context, Poll},
};
pub struct ShutdownHandle {
sender: oneshot::Sender<()>,
receiver: oneshot::Receiver<()>,
}
impl ShutdownHandle {
pub fn new() -> Self {
let (sender, receiver) = oneshot::channel();
Self { sender, receiver }
}
pub fn shutdown(self) -> Result<(), ()> {
self.sender.send(())
}
}
impl Future for ShutdownHandle {
type Output = ();
fn poll(self: Pin<&mut Self>, cx: &mut Context) -> Poll<Self::Output> {
self.receiver.poll(&mut cx).map(|_| ())
}
}
fn main() {
let runner = ShutdownHandle::new();
assert!(runner.shutdown().is_ok());
}
我收到以下错误:
error[E0599]: no method named `poll` found for type `futures_channel::oneshot::Receiver<()>` in the current scope
--> src/main.rs:28:23
|
28 | self.receiver.poll(&mut cx).map(|_| ())
| ^^^^
我想念什么?当然，有某种方式可以通过"轮询.我每天晚上使用(2019-07-18).
What am I missing? Surely there's some way to "pass through" the polling. I am using nightly (2019-07-18).
推荐答案
是的，Receiver没有实现Future；只有 Pin<&mut Receiver> 可以.您需要投影从类型到字段的固定.
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.
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(|_| ())
}
}
您必须阅读 pin模块以便全面了解在此处使用unsafe的要求.
You must read the pin module to thoroughly understand the requirements to use unsafe here.
我喜欢使用帮助程序库(例如 pin_project )来处理更复杂的投影类型:
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(|_| ())
}
}
当基础类型实现Unpin
时
ÖmerErden指出，期货预览板条箱提供了 Unpin 并使用它创建一个全新的Pin.
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.
由于oneshot::Receiver确实实现了Unpin，因此可以在此处使用它:
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(|_| ())
}
}
另请参见
- 什么时候可以安全地将会员价值移出固定的未来?
- 为什么对于明显已实现的类型未实现特征?
- 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?
See also
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