如何从 actix 处理程序调用固定的未来？

Question

我有一个异步特质方法，它返回一个std Future：

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

< code>ResultType是trait的关联类型，它是< code>Sync Send。

请注意，此类型不是Unpin类型。

我想从actix处理程序调用它，然后对结果做些什么。

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);
    }
}

这会失败，因为into_actor拥有未来，这是Pin不允许的。清除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<_>`

我该怎么做？

Answer 1

实际的问题不是未来被钉住了，而是它实现了错误的未来特性。

关于固定的本节解释了轮询是为Pin实现的

因此，into_actor签名自我：impl Future -

调用＜code＞。compat（）在调用之前，先查看未来。into_actor解决了这个问题。

有关转换期货的更多详细信息，请参阅此帖子。