asynchronous - 如何接受异步函数作为参数？

Question

我想复制将闭包/函数作为参数的行为和人体工程学，很像 map做:iterator.map(|x| ...) .

我注意到一些库代码允许传入异步功能，但此方法不允许我传入参数:

pub fn spawn<F, T>(future: F) -> JoinHandle<T>
where
    F: Future<Output = T> + Send + 'static,
    T: Send + 'static,

spawn(async { foo().await });

我希望执行以下操作之一:

iterator.map(async |x| {...});

async fn a(x: _) {}
iterator.map(a)

Answer 1

async函数被有效地脱糖为返回 impl Future .一旦你知道了这一点，那就是结合现有的 Rust 技术来接受一个函数/闭包，从而产生一个具有两种泛型类型的函数:

use std::future::Future;

async fn example<F, Fut>(f: F)
where
    F: FnOnce(i32, i32) -> Fut,
    Fut: Future<Output = bool>,
{
    f(1, 2).await;
}

这也可以写成

use std::future::Future;

async fn example<Fut>(f: impl FnOnce(i32, i32) -> Fut)
where
    Fut: Future<Output = bool>,
{
    f(1, 2).await;
}

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