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我有一个异步特征方法，它返回一个标准未来：

Pin<Box<dyn Future<Output = Result<Vec<ResultType>, Box<(dyn Error + 'static)>>> + Send>>

ResultType是 trait 的关联类型Sync + Send。

请注意，此类型不是 Unpin。

我想从一个 actix 处理程序中调用它，然后对结果做一些事情。

例如：

impl StreamHandler<ws::Message, ws::ProtocolError> for MyActor {
    fn handle(&mut self, msg: ws::Message) {
        let fut = get_future();
        let actor_fut = fut
            .into_actor(&self)
            .map(|r, _actor, ctx| {
                ctx.text(r.map(|| ...))
            });

        ctx.spawn(actor_fut);
    }
}

这失败了，因为into_actor取得了未来的所有权，这是不允许的Pin。清理后的错误消息如下所示：

error[E0599]: no method named `into_actor` found for type `Pin<Box<dyn Future<Output = Result<Vec<ResultType>, Box<dyn Error>>> + Send>>` in the current scope
   --> src/app_socket.rs:194:26
    |
194 |                         .into_actor(&self)
    |                          ^^^^^^^^^^ method not found in `Pin<Box<dyn Future<Output = Result<Vec<ResultType>, Box<dyn Error>>> + Send>>`
    |
    = note: the method `into_actor` exists but the following trait bounds were not satisfied:
            `&dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send : WrapFuture<_>`
            `&dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send : WrapStream<_>`
            `&mut dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send : WrapFuture<_>`
            `&mut dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send : WrapStream<_>`
            `&mut Pin<std::boxed::Box<dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send>> : WrapFuture<_>`
            `&mut Pin<std::boxed::Box<dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send>> : WrapStream<_>`
            `&Pin<std::boxed::Box<dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send>> : WrapFuture<_>`
            `&Pin<std::boxed::Box<dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send>> : WrapStream<_>`
            `dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send : WrapFuture<_>`
            `dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send : WrapStream<_>`
            `Pin<std::boxed::Box<dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send>> : WrapFuture<_>`
            `Pin<std::boxed::Box<dyn Future<Output = Result<std::vec::Vec<ResultType>, std::boxed::Box<dyn std::error::Error>>> + Send>> : WrapStream<_>`

我怎样才能做到这一点？

4

1 回答 1

1

实际的问题不是未来被固定，而是它实现了错误的未来特征。

Pinning 的这一部分解释了poll为Pin<ref T: Future>.

因此，into_actor签名self: impl Future -> WrapFuture<_>很好。问题是该方法在预期的情况下async返回了未来的实现。std::future::Futureinto_actorfutures01::future::Future

在调用.compat()之前调用未来.into_actor可以解决问题。

有关转换期货的更多详细信息，请参阅这篇文章。

于 2019-10-13T23:54:55.440 回答