我有一个 gRPC 服务器方法，我想模拟一些工作，为此我有以下代码：
async fn write(&self, request: Request<WriteRequest>)
                 -> Result<Response<WriteResponse>, Status> {
    let mut rng = rand::thread_rng();
    let random = rng.gen_range(1000..5000);
    // Sleep for the random duration
    sleep(Duration::from_millis(random)).await;
    return Ok(Response::new(WriteResponse { ... }));
}

未能安全地在线程之间发送 future 由 async 块创建的 future 不是 `Send`

<details>
<summary>英文:</summary>
I have a gRPC server method for which I want to simulate some work for this I have this
    async fn write(&amp;self, request: Request&lt;WriteRequest&gt;)
                         -&gt; Result&lt;Response&lt;WriteResponse&gt;, Status&gt; {
        let mut rng = rand::thread_rng();
        let random = rng.gen_range(1000..5000);
        // Sleep for the random duration
        sleep(Duration::from_millis(random)).await;
        return Ok(Response::new(WriteResponse { ... }));
    }
but the code is not happy about the random value, I get the error

I know I can add + Send in the function but in this case, I cannot modify the signature of the method because it is an interface generated by the gRPC compiler
When I used hardcoded value within the Duration in works.
How can I call the random number generation? or how can I sleep for a random value using sleep? is there another way?
</details>
# 答案1
**得分**: 4
这是因为 `rng` 超过了一个等待点而发生的。您可以通过将其封装在一个块中来解决这个问题，这样可以确保它不会存储在异步函数的状态机中。
```rs
let random_secs = {
    let mut rng = rand::thread_rng();
    rng.gen_range(1000..5000)
};

use rand::{Rng, SeedableRng, rngs::StdRng};
    
let mut rng = {
    let rng = rand::thread_rng();
    StdRng::from_rng(rng).unwrap()
};
    
loop {
    let millis = rng.gen_range(1..10);
    tokio::time::sleep(std::time::Duration::from_millis(millis)).await;
    println!("{millis}");
}