asynchronous - Rust 预期类型找到的结构

Question

我有以下代码:

use actix_service::Service;
use actix_web::{web, App, HttpServer, Responder};

use actix_router::{Path, Url};
use actix_web::dev::{ServiceRequest, ServiceResponse};
use actix_web::error::ResponseError;
use actix_web::{http, http::StatusCode, Error, HttpRequest, HttpResponse};

async fn greet(req: HttpRequest) -> impl Responder {
    let name = req.match_info().get("name").unwrap_or("World");
    format!("Hello {}!", &name)
}

#[actix_rt::main]
async fn main() -> std::io::Result<()> {
    HttpServer::new(|| {
        let app = App::new()
            .wrap_fn(|req, srv| {
                let passed: bool;

                // change to false to simulate a failed check
                let check = true;

                if *&req.path().contains("/item/") {
                    passed = check;
                } else {
                    passed = true;
                }

                let fresh_result = match passed {
                    true => {
                        let fut = srv.call(req);
                        Box::pin(async {
                            let result = fut.await?;
                            Ok(result)
                        })
                    }
                    false => Box::pin(async {
                        let result = req.into_response(
                            HttpResponse::Found()
                                .header(http::header::LOCATION, "/login")
                                .finish()
                                .into_body(),
                        );
                        Ok(result)
                    }),
                };
                async {
                    let last_outcome = fresh_result.await?;
                    Ok(last_outcome)
                }
            })
            .route("/", web::get().to(greet));
        return app;
    })
    .bind("127.0.0.1:8000")?
    .run()
    .await
}

但是，我收到以下错误:

110 |                         let fresh_result = match passed {
    |    ________________________________________-
111 |   |                         true => {
112 |   |                             let fut = srv.call(req);
113 |   |                             Box::pin(
    |  _|_____________________________-
114 | | |                                 async {
115 | | |                             let result = fut.await?;
116 | | |                             Ok(result)
117 | | |                             }
118 | | |                             )
    | |_|_____________________________- this is found to be of type `std::pin::Pin<std::boxed::Box<impl core::future::future::Future>>`
...     |
121 | / |                             Box::pin(
122 | | |                             async {
123 | | |                             let result = req.into_response(
124 | | |                                 HttpResponse::Found()
...   | |
129 | | |                             }
130 | | |                             )
    | |_|_____________________________^ expected generator, found a different generator
131 |   |                         }
132 |   |                     };
    |   |_____________________- `match` arms have incompatible types
    | 
   ::: /Users/maxwellflitton/.rustup/toolchains/stable-x86_64-apple-darwin/lib/rustlib/src/rust/src/libcore/future/mod.rs:55:43
    |
55  |     pub const fn from_generator<T>(gen: T) -> impl Future<Output = T::Return>
    |                                               ------------------------------- the found opaque type
    |
    = note: expected type `std::pin::Pin<std::boxed::Box<impl core::future::future::Future>>` (generator)
             found struct `std::pin::Pin<std::boxed::Box<impl core::future::future::Future>>` (generator)

我完全坚持这一点。我不知道如何确保它是一种类型。如果没有 match 语句并且只有一个异步 block ，那么一切运行良好。这适用于 Actix-web 服务器中的中间件。如果凭据不起作用，我正在尝试重定向查看器。

Answer 1

您使用 Box::pin创造盒装 future ，
但是 Rust 仍然认为你想要 impl Future<...>版本(意味着 future 被装箱到堆中，但不使用动态调度)。
这就是错误中的类型为 Pin<Box<impl Future<...>>> 的原因。 ，并且因为任意两个 impl Future<...> s 是不同的类型，你得到 expected/found错误。
你需要告诉 Rust 你对动态调度感兴趣(因为你有两个不同的 future 可以存储在 Box 中，而哪个真正存储在那里只能在运行时知道)，例如通过显式指定 return像这样输入:

use std::pin::Pin;
use std::future::Future;

let fresh_result: Pin<Box<dyn Future<Output=_>>> = match passed {
    // ...
}

现在你会得到一个关于 cannot infer type for type parameter E 的新错误。这可以通过指定 Output 来解决 future 也一样，
所以新行应该是:

let fresh_result: Pin<Box<dyn Future<Output=Result<ServiceResponse, Error>>>> = match passed {
    // ...
}

这将成功编译!