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我想知道为什么将代码从Future直接使用更改tokio::spawn为使用async move块可以编译代码。

直接使用：

struct ClientMsg {
    ...
    resp: oneshot::Sender<Bytes>,
}

async fn client_thread(
    mut rx: Receiver<ClientMsg>,
    client: Client,
) -> Result<(), Box<dyn Error>> {
    while let Some(msg) = rx.recv().await {
        ...
        let response = client.get(url).send().await?.bytes().await?;
        msg.resp.send(response).unwrap();
    }
    Ok(())
}

#[tokio::main]
async fn main() -> Result<(), Box<dyn Error>> {
    ...
    let (tx, rx) = mpsc::channel(5);
    tokio::spawn(client_thread(rx, client)); // <--- Difference is here

    Ok(())
}

异步块：

struct ClientMsg {
    ...
    resp: oneshot::Sender<Bytes>,
}

async fn client_thread(
    mut rx: Receiver<ClientMsg>,
    client: Client,
) -> Result<(), Box<dyn Error>> {
    while let Some(msg) = rx.recv().await {
        ...
        let response = client.get(url).send().await?.bytes().await?;
        msg.resp.send(response).unwrap();
    }
    Ok(())
}

#[tokio::main]
async fn main() -> Result<(), Box<dyn Error>> {
    ...
    let (tx, rx) = mpsc::channel(5);
    tokio::spawn(async move { client_thread(rx, client) }); // <-- Difference is here

    Ok(())
}

请注意以下事项use：

use bytes::Bytes;
use reqwest::Client;

use tokio::sync::{
    mpsc::{self, Receiver},
    oneshot,
};
use url::Url;

直接使用代码失败并显示：

error[E0277]: `(dyn StdError + 'static)` cannot be sent between threads safely
   --> src/main.rs:44:5
    |
44  |     tokio::spawn(client_thread(rx, client, base_url));
    |     ^^^^^^^^^^^^ `(dyn StdError + 'static)` cannot be sent between threads safely
    | 
   ::: /home/jeanluc/.cargo/registry/src/github.com-1ecc6299db9ec823/tokio-1.2.0/src/task/spawn.rs:130:20
    |
130 |         T::Output: Send + 'static,
    |                    ---- required by this bound in `tokio::spawn`
    |
    = help: the trait `Send` is not implemented for `(dyn StdError + 'static)`
    = note: required because of the requirements on the impl of `Send` for `Unique<(dyn StdError + 'static)>`
    = note: required because it appears within the type `Box<(dyn StdError + 'static)>`
    = note: required because it appears within the type `std::result::Result<(), Box<(dyn StdError + 'static)>>`

的返回类型与函数client_thread完全相同main，但是，在 Tokio 上运行没有任何问题。此外，来自implements的错误类型reqwestSend。

4

1 回答 1

1

像你所做的那样将函数调用包装在一个async {}块中是行不通的。运行时，它调用创建未来的函数，然后返回它。它永远不会被轮询，所以它永远不会取得进展。直接使用它或与它一起使用.await确保它被轮询。

use futures::executor::block_on; // 0.3.12

async fn f(s: &str) {
    println!("ran {}", s);
}

fn main() {
    block_on(f("first"));
    block_on(async { f("second") }); // this doesn't print
    block_on(async { f("third").await });
}

ran first
ran third

由于未来没有持久化，它最终不会影响async {}块的特征，因此可以是Send. 你会再次遇到同样的问题.await。

在您的情况下，您需要做的就是确保未来实现Send，以便可以使用tokio::spawn. 解决方法是规定Error返回的特征实现Send：

async fn client_thread(
    mut rx: Receiver<ClientMsg>,
    client: Client,
) -> Result<(), Box<dyn Error + Send>> {
    // ...                    ^^^^^^
}
于 2021-02-20T21:07:17.920 回答