为了 API 方便，我如何包装动态类型的流？

Question

我希望为任何 returns 特定类型的流实现包装器结构，以减少乱丢我的应用程序的动态关键字。我遇到过 BoxStream，但不知道如何在 Stream::poll_next 中使用它。这是我目前所拥有的：

use std::pin::Pin;
use std::task::{Context, Poll};
use futures::prelude::stream::BoxStream;
use futures::Stream;

pub struct Row;

pub struct RowCollection<'a> {
    stream: BoxStream<'a, Row>,
}

impl RowCollection<'_> {
    pub fn new<'a>(stream: BoxStream<Row>) -> RowCollection {
        RowCollection { stream }
    }
}

impl Stream for RowCollection<'_> {
    type Item = Row;
    fn poll_next(self: Pin<&mut Self>, cx: &mut Context<'_>) -> Poll<Option<Self::Item>> {
        // I have no idea what to put here, but it needs to get the information from self.stream and return appropriate value
    }
}

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期货=“0.3”

Answer 1

由于 Box implements Unpin，BoxStream 实施 Unpin，因此 RowCollection。

因此，您可以利用 Pin::get_mut which will give you a &mut RowCollection. From that, you can get a &mut BoxStream. You can re-pin that via Pin::new and then call poll_next on it. This is called pin-projection。

impl Stream for RowCollection<'_> {
    type Item = Row;
    
    fn poll_next(self: Pin<&mut Self>, cx: &mut Context<'_>) -> Poll<Option<Self::Item>> {
        Pin::new(&mut self.get_mut().stream).poll_next(cx)
    }
}

另请参阅：

为了 API 方便，我如何包装动态类型的流？

How can I wrap a dynamically typed stream for API convenience?

rust

rust-futures

依赖关系：