std :: async无法异步工作 [英] std::async doesn't work asynchronously

问题描述

我有以下非常简单的代码：

I have the following very simple code:

void TestSleep()
{
    std::cout << "TestSleep " << std::endl;
    sleep(10);
    std::cout << "TestSleep Ok" << std::endl;

}

void TestAsync()
{
    std::cout << "TestAsync" << std::endl;
    std::async(std::launch::async, TestSleep);
    std::cout << "TestAsync ok!!!" << std::endl;

}

int main()
{
    TestAsync();
    return 0;
}

因为我使用 std :: launch :: async 我希望 TestSleep（）将异步运行，并且我将获得以下输出：

Since I use std::launch::async I expect that TestSleep() will be run asynchronously and I will have the following output:

TestAsync
TestAsync ok!!!
TestSleep 
TestSleep Ok

但实际上我有同步运行的输出：

But really I have the output for synchronous run:

TestAsync
TestSleep 
TestSleep Ok
TestAsync ok!!!

您能否解释为什么以及如何制作 TestSleep 真的是异步调用。

Could you explain why and how to make TestSleep call really asynchronously.

推荐答案

来自此 std :: async 参考注释部分

From this std::async reference notes section

如果从 std :: async 获得的 std :: future 从引用或绑定到引用， std :: future 的析构函数将在完整表达式的末尾阻塞，直到异步操作完成，从本质上使代码...同步

If the std::future obtained from std::async is not moved from or bound to a reference, the destructor of the std::future will block at the end of the full expression until the asynchronous operation completes, essentially making code ... synchronous

这就是这里发生的情况。由于您不存储 std :: async 返回的未来，因此它将在表达式末尾（即 std： ：async 调用），它将阻塞直到线程结束。

This is what happens here. Since you don't store the future that std::async returns, it will be destructed at the end of the expression (which is the std::async call) and that will block until the thread finishes.

如果您这样做，例如

auto f = std::async(...);

然后销毁 f TestAsync 的行将被阻止，并且文本 TestAsync ok !!! 应该在 TestSleep OK 。

then the destruction of f at the end of TestAsync will block, and the text "TestAsync ok!!!" should be printed before "TestSleep Ok".

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