为异步函数定义Handler trait时出现Rust编译错误

你的代码中，为异步函数类型实现Handler trait时，缺少对返回future的Send约束，导致编译报错。

原代码及错误 #

#[async_trait]
trait Handler: Send + Sync + 'static {
    async fn handle(&self, req: Request<Body>) -> Result<Response<Body>, GenericError>;
}

#[async_trait]
impl<F: Send + Sync + 'static, Fut> Handler for F
where 
    F: Fn(Request<Body>) -> Fut,
    Fut: Future<Output = Result<Response<Body>, GenericError>>,
{
    async fn handle(&self, req: Request<Body>) -> Result<Response<Body>, GenericError> {
        self(req).await
    }
}

编译错误：

future cannot be sent between threads safely
required for the cast to the object type `dyn futures::Future<Output = Result<Response<Body>, Box<(dyn std::error::Error + Sync + std::marker::Send + 'static)>>> + std::marker::Send`rustc
router.rs(21, 9): future is not `Send` as it awaits another future which is not `Send`
router.rs(18, 63): consider further restricting this bound: ` + std::marker::Send`

解决方案 #

在Fut的约束中添加Send trait限定，修改后的代码如下：

#[async_trait]
trait Handler: Send + Sync + 'static {
    async fn handle(&self, req: Request<Body>) -> Result<Response<Body>, GenericError>;
}

#[async_trait]
impl<F: Send + Sync + 'static, Fut> Handler for F
where 
    F: Fn(Request<Body>) -> Fut,
    Fut: Future<Output = Result<Response<Body>, GenericError>> + Send,
{
    async fn handle(&self, req: Request<Body>) -> Result<Response<Body>, GenericError> {
        self(req).await
    }
}

原因说明 #

Handler trait本身要求实现者满足Send + Sync，意味着它可能在多线程环境中被调用和传递。async_trait宏会将异步方法的返回future包装成dyn Future + Send类型，因此必须保证F返回的Fut也实现了Send，才能满足跨线程安全传递的要求。错误提示已经明确建议添加+ std::marker::Send约束，按照提示补充即可解决问题。

内容的提问来源于stack exchange，提问作者Scalalang2