rust - 如何强制 `async_std`任务在rust中的单个线程上运行？

Question

我是使用rust 的新手，并且首次尝试在 rust 中进行异步。在此示例中，我使用async-std运行cpu密集的递归斐波那契数列操作，并且无论我运行一个任务还是两个任务，这大约花费了相同的时间(〜7秒)。

异步运行时似乎正在多个线程上执行任务。

有什么方法可以强制异步任务在同一线程上运行？

use std::time;
use async_std::task;

fn main() {

    let start = time::Instant::now();

    let s1 = task::spawn(fib_async(46));
    let s2 = task::spawn(fib_async(46));

    task::block_on(s1);
    task::block_on(s2);
    let elap = time::Instant::now().duration_since(start);

    println!("time took for tasks to finish: {} seconds", elap.as_secs());

}


async fn fib_async(n: i32) -> i32 {
    fib(n)
}

fn fib(n: i32) -> i32 {
    if n == 0 || n == 1 {
        1
    } else {
        fib(n-1) + fib(n-2)
    }
}

Answer 1

根据 async_std::task 的文档，您不能。

This module is similar to std::thread, except it uses asynchronous tasks in place of threads.

因此，它旨在启动新线程，就像使用 std::thread::spawn启动新线程一样。

如果要顺序处理数据，可以实现一个工作程序生产者模式:

use async_std::task;
use std::time;

#[async_std::main]
async fn main() {
    let (tx, rx) = flume::unbounded();

    let s = task::spawn(async move {
        for n in rx.into_iter() {
            println!("Calc fib of {}", n);
            let fib = fib_async(n).await;
            println!("{}", fib);
        }
    });

    let start = time::Instant::now();


    tx.send(40).unwrap();
    tx.send(40).unwrap();

    drop(tx);
    s.await;

    let elap = time::Instant::now().duration_since(start);
    println!("time took for tasks to finish: {} seconds", elap.as_secs());
}

async fn fib_async(n: i32) -> i32 {
    fib(n)
}

fn fib(n: i32) -> i32 {
    if n == 0 || n == 1 {
        1
    } else {
        fib(n - 1) + fib(n - 2)
    }
}

具有 cargo 依赖关系:

[dependencies]
async-std = { version = "1.5", features = ["attributes"] }
flume = "0.7"