使用await时异步阻塞吗

本文介绍了使用await时异步阻塞吗的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

在以下代码中，运行两个异步:

In the following code two asyncs are run:

import kotlinx.coroutines.* import kotlin.system.* fun main() = runBlocking<Unit> { val time = measureTimeMillis { val one = async { doSomethingUsefulOne() } val two = async { doSomethingUsefulTwo() } two.await() one.await() println("Finished") } } suspend fun doSomethingUsefulOne(): Int { delay(3000L) println("first") return 13 } suspend fun doSomethingUsefulTwo(): Int { delay(1000L) println("second") return 29 }

为什么不首先打印完成"?文档说了关于await:

Why is "Finished" not printed first? The docs says this about await:

等待该值完成而不会阻塞线程

Awaits for completion of this value without blocking a thread

推荐答案

通过稍微修改代码，人们将获得一些见解.让我们将经过的时间和当前线程添加到打印语句中，以查看发生了什么:

By slightly modifying the code, one will gain some insights. Let's add the elapsed time and the current thread to the print statements to see, what's going on:

在科特林游乐场执行

import kotlinx.coroutines.* import kotlin.system.* private val started = System.currentTimeMillis() private fun debug(s: String) { val elapsed = System.currentTimeMillis() - started println("[$elapsed] $s -- ${Thread.currentThread()}") } fun main() = runBlocking<Unit> { val time = measureTimeMillis { val one = async { doSomethingUsefulOne() } val two = async { doSomethingUsefulTwo() } debug("awaiting") two.await() one.await() debug("Finished") } debug("time = $time") } suspend fun doSomethingUsefulOne(): Int { delay(3000L) debug("first") return 13 } suspend fun doSomethingUsefulTwo(): Int { delay(1000L) debug("second") return 29 }

与-Dkotlinx.coroutines.debug一起运行时，应该会看到类似于

When run with -Dkotlinx.coroutines.debug one should see something similar to

[46] awaiting -- Thread[main @coroutine#1,5,main] [1056] second -- Thread[main @coroutine#3,5,main] [3054] first -- Thread[main @coroutine#2,5,main] [3054] Finished -- Thread[main @coroutine#1,5,main] [3054] time = 3013 -- Thread[main @coroutine#1,5,main]

只有一个线程.

如果await阻止了线程，则程序可能不会终止.

If await had blocked the thread, the program could not have terminated.