采用静音的异步递归函数

Question

您如何创建一个sync递归函数，以占用静音？ Rust声称该代码在等待点上含有互惠码。但是，该值在.await之前已删除。

#[async_recursion]
async fn f(mutex: &Arc<Mutex<u128>>) {
    let mut unwrapped = mutex.lock().unwrap();
    *unwrapped += 1;
    let value = *unwrapped;
    drop(unwrapped);
    tokio::time::sleep(tokio::time::Duration::from_millis(1000)).await;
    if value < 100 {
        f(mutex);
    }
}

错误

future cannot be sent between threads safely
within `impl futures::Future<Output = ()>`, the trait `std::marker::Send` is not implemented for `std::sync::MutexGuard<'_, u128>`
required for the cast to the object type `dyn futures::Future<Output = ()> + std::marker::Send`rustc
lib.rs(251, 65): future is not `Send` as this value is used across an await

Answer 1

在这种情况下，您可以对代码进行重组以使其未包装无法在上使用：

let value = {
    let mut unwrapped = mutex.lock().unwrap();
    *unwrapped += 1;
    *unwrapped
};
tokio::time::sleep(tokio::time::Duration::from_millis(1000)).await;
if value < 100 {
    f(mutex);
}

如果您无法执行此操作，那么您' d需要做到这一点，这样您就不会返回实现发送的未来。 async_recursion docs指定选项您可以传递到宏以禁用send限制它添加：

#[async_recursion(?Send)]
async fn f(mutex: &Arc<Mutex<u128>>) {
    ...

​