使用tokio::spawn和Box<dyn Error>时，如何修复跨线程发送不安全错误？

问题代码 #

#[tokio::main]
async fn main() {
    tokio::spawn(async {
        foo().await;
    });
}

async fn foo() {
    let f1 = bar();
    let f2 = bar();

    tokio::join!(f1, f2);
}

async fn bar() -> Result<(), Box<dyn std::error::Error>> {
    println!("Hello world");
    Ok(())
}

编译错误信息 #

error[E0277]: `(dyn std::error::Error + 'static)` cannot be sent between threads safely
   --> src/main.rs:5:18
    |
5   |       tokio::spawn(async {
    |  _____------------_^
    | |     |
    | |     required by a bound introduced by this call
6   | |         foo().await;
7   | |     });
    | |_____^ `(dyn std::error::Error + 'static)` cannot be sent between threads safely
    |
    = help: the trait `Send` is not implemented for `(dyn std::error::Error + 'static)`
    = note: required for `Unique<(dyn std::error::Error + 'static)>` to implement `Send`
    = note: required because it appears within the type `Box<dyn Error>`
    = note: required because it appears within the type `Result<(), Box<dyn Error>>`
    = note: required because it appears within the type `MaybeDone<impl Future<Output = Result<(), Box<dyn Error>>>>`
    = note: required because it appears within the type `(MaybeDone<impl Future<Output = Result<(), Box<dyn Error>>>>, MaybeDone<impl Future<Output = Result<(), Box<dyn Error>>>>)`
    = note: required because it captures the following types: `ResumeTy`, `impl Future<Output = Result<(), Box<dyn Error>>>`, `(MaybeDone<impl Future<Output = Result<(), Box<dyn Error>>>>, MaybeDone<impl Future<Output = Result<(), Box<dyn Error>>>>)`, `&mut (MaybeDone<impl Future<Output = Result<(), Box<dyn Error>>>>, MaybeDone<impl Future<Output = Result<(), Box<dyn Error>>>>)`, `u32`, `[closure@join.rs:95:17]`, `PollFn<[closure@join.rs:95:17]>`, `()`
note: required because it's used within this `async fn` body

错误原因 #

Tokio的tokio::spawn函数要求传入的异步任务（future）必须实现Send trait——因为Tokio的多线程运行时会在不同线程间调度任务，只有实现Send的类型才能安全跨线程传递。

你的代码里，bar返回Result<(), Box<dyn std::error::Error>>，但std::error::Error本身没有强制要求实现Send，所以Box<dyn std::error::Error>默认不保证是Send的。当foo用tokio::join!组合两个bar的future时，生成的future会包含这个非Send的错误类型，导致整个被spawn包裹的异步块不满足约束，最终触发编译错误。

移除bar的返回类型后，返回值变成()（天然实现Send），整个future满足约束，所以编译通过。

解决方法 #

修改bar的返回类型，给dyn Error加上Send约束，确保错误类型可以安全跨线程传递：

async fn bar() -> Result<(), Box<dyn std::error::Error + Send>> {
    println!("Hello world");
    Ok(())
}

如果场景需要，还可以进一步加上Sync约束：Box<dyn std::error::Error + Send + Sync>。

内容的提问来源于stack exchange，提问作者guenhter