借用的值在异步函数中的寿命不够长

Question

我是 Rust 的新手，正在尝试使一些代码与运行并行的一堆任务异步。这是一个简化的例子：

use futures::future::join_all;

#[tokio::main]
async fn main() {
    let mut list = Vec::new();
    for i in 1..10 {
        let my_str = format!("Value is: {:?}", &i);
        let future = do_something(&my_str);
        list.push(future);
    }
    join_all(list).await;
}

async fn do_something(value: &str)
{
    println!("value is: {:?}", value);
}

这在 do_something(&my_str) 调用中失败，并显示“借用的值寿命不够长”。我可以通过更改 do_something 以接受字符串而不是 &str 来编译代码。但是，当 &str 可以工作时要求 String 似乎有点奇怪。这里有更好的模式吗？谢谢！

Answer 1

“但是，当 &str 可以工作时要求字符串似乎有点奇怪。” 但是 &str 不能在这里工作，因为它只借用my_str，它在未来完成之前被销毁：

for i in 1..10 {
    // Create a new `String` and store it in `my_str`
    let my_str = format!("Value is: {:?}", &i);
    // Create a future that borrows `my_str`. Note that the future is not
    // yet started
    let future = do_something(&my_str);
    // Store the future in `list`
    list.push(future);
    // Destroy `my_str` since it goes out of scope and wasn't moved.
}
// Run the futures from `list` until they complete. At this point each
// future will try to access the string that they have borrowed, but those
// strings have already been freed!
join_all(list).await;

相反，您的 do_something 应该取得字符串的所有权并负责释放它：

use futures::future::join_all;

#[tokio::main]
async fn main() {
    let mut list = Vec::new();
    for i in 1..10 {
        // Create a new `String` and store it in `my_str`
        let my_str = format!("Value is: {:?}", &i);
        // Create a future and _move_ `my_str` into it.
        let future = do_something(my_str);
        // Store the future in `list`
        list.push(future);
        // `my_str` is not destroyed since it was moved into the future.
    }
    join_all(list).await;
}

async fn do_something(value: String)
{
    println!("value is: {:?}", value);
    // Destroy `value` since it goes out of scope and wasn't moved.
}

借用的值在异步函数中的寿命不够长

Borrowed value does not live long enough with async funtion

rust

rust-tokio