是否有可能 Future::and_then 有条件地 return 不同的期货？

Question

在我的代码的这个简化版本中，我想有时执行标记的行，有时不执行，可能会返回一个错误：

extern crate futures; // 0.1.26
extern crate hyper; // 0.12.25

use hyper::rt::{Future, Stream};
use std::str::FromStr;

struct MyStream {}

impl Stream for MyStream {
    type Item = hyper::Uri;
    type Error = ();

    fn poll(&mut self) -> Result<futures::Async<Option<Self::Item>>, Self::Error> {
        Ok(futures::Async::Ready(Some(
            hyper::Uri::from_str("http://www.google.com/").unwrap(),
        )))
    }
}

fn main() {
    let client = hyper::Client::new();
    let futs = MyStream {}
        .map(move |uri| {
            client
                .get(uri)
                .and_then(|res| {
                    res.into_body().concat2() // <----------------
                })
                .map(|body| {
                    println!("len is {}.", body.len());
                })
                .map_err(|e| {
                    println!("Error: {:?}", e);
                })
        })
        .buffer_unordered(2)
        .for_each(|_| Ok(()));

    hyper::rt::run(futs);
}

我想我可以用这样的东西替换这行：

let do_i_want_to_get_the_full_page = true;
if do_i_want_to_get_the_full_page {
    res.into_body().concat2().map_err(|_| ())
} else {
    futures::future::err(())
}

因为期货的Error部分是相同的，所以Item部分可以推断出来。但是，它不起作用。我该怎么做？

这是我得到的错误：

error[E0308]: if and else have incompatible types
  --> src/main.rs:31:25
   |
28 | /                     if do_i_want_to_get_the_full_page {
29 | |                         res.into_body().concat2().map_err(|_| ())
   | |                         ----------------------------------------- expected because of this
30 | |                     } else {
31 | |                         futures::future::err(())
   | |                         ^^^^^^^^^^^^^^^^^^^^^^^^ expected struct `futures::MapErr`, found struct `futures::FutureResult`
32 | |                     }
   | |_____________________- if and else have incompatible types
   |
   = note: expected type `futures::MapErr<futures::stream::Concat2<hyper::Body>, [closure@src/main.rs:29:59: 29:65]>`
              found type `futures::FutureResult<_, ()>`

Answer 1

问题是 map_err returns a struct MapErr, while err returns a struct FutureResult.

解决您的问题的一种方法是像这样统一它们：

let f = if do_i_want_to_get_the_full_page {
    futures::future::ok(())
} else {
    futures::future::err(())
};
f.and_then (|_| { res.into_body().concat2().map_err(|_| ()) })

另一个解决方案是将您的 return 值装箱：

if do_i_want_to_get_the_full_page {
    Box::<Future<_, _>>::new (res.into_body().concat2().map_err(|_| ()))
} else {
    Box::<Future<_, _>>::new (futures::future::err(()))
}

第三个解决方案是总是return一个MapErr:

if do_i_want_to_get_the_full_page {
    res.into_body().concat2().map_err(|_| ())
} else {
    futures::future::err(()).map_err(|_| ())
}

但是，您会遇到一个问题，因为 client.get(…).and_then(…) 的错误类型必须实现 From<hyper::Error>:

error[E0271]: type mismatch resolving `<futures::AndThen<futures::FutureResult<(), ()>, futures::MapErr<futures::stream::Concat2<hyper::Body>, [closure@src/main.rs:34:70: 34:76]>, [closure@src/main.rs:34:32: 34:77 res:_]> as futures::IntoFuture>::Error == hyper::Error`
  --> src/main.rs:28:18
   |
28 |                 .and_then(|res| {
   |                  ^^^^^^^^ expected (), found struct `hyper::Error`
   |
   = note: expected type `()`
              found type `hyper::Error`

如果你不关心这个错误，你可以在 and_then:

之前把它映射掉

client
    .get(uri)
    .map_err (|_| ())
    .and_then(|res| {
        let f = if do_i_want_to_get_the_full_page {
            futures::future::ok(())
        } else {
            futures::future::err(())
        };
        f.and_then(|_| res.into_body().concat2().map_err(|_| ()))
    }).map(|body| {
        println!("len is {}.", body.len());
    }).map_err(|e| {
        println!("Error: {:?}", e);
    })

playground

或者在对 futures::future::err.

的调用中使用实现 From<hyper::Error> 的类型

是否有可能 Future::and_then 有条件地 return 不同的期货？

Is it possible to have Future::and_then conditionally return different futures?

future

rust

hyper

rust-tokio