什么是C ++ 11的方式来触发异步任务并忘记它？ [英] What's the C++ 11 way to fire off an asynchronous task and forget about it?

问题描述

我需要这样的东西：

void launch_task()
{
    std::thread([](){ run_async_task(); });
}

除了线程的析构函数会终止我的任务。我不需要任何控制任务，也不需要一个返回值。它只需要运行它的过程，然后线程应该终止，并且C ++线程对象应该被丢弃。我需要什么C ++ 11设施？

Except thread's destructor will terminate my task. I don't need any control over the task, don't need a return value either. It just has to run its course and then the thread should terminate and C++ thread object should be disposed of. What C++ 11 facility do I need?

我看过 std :: async 找到我的case的用法的例子。它似乎是一个非常复杂的系统，我需要以某种方式存储和操纵 std :: future 或它会变成同步的（如果我的理解是正确的;我没有找到关于 std :: async 的好清楚的文章。

I've looked at std::async, but couldn't find an example of usage for my case. It seems to be a pretty complicated system, and I'd need to somehow store and manipulate std::future or it'd become synchronous (if my understanding is correct; I didn't find a good clear article on std::async).

推荐答案

只要在创建后立即分离即可。

Just detach it immediately after creation.

std::thread([](){ run_async_task(); }).detach();

一旦分离，线程将不再是可连接的，所以〜thread ）将没有效果。
此回答讨论此行为的更多详细信息。

Once detached, the thread will no longer be joinable, so ~thread() will have no effect. This answer discusses more details of this behavior.

正如WB所提到的下面， std :: async 将无法工作，原因如下：引用。

As mentioned by W.B. below, std::async will not work for the following reason, pulled from this reference.

如果从std :: async获取的std :: future有临时对象
生存期（不移动或绑定到变量），
std :: future的析构函数将在完整表达式的结尾处阻塞，直到
异步操作完成

If the std::future obtained from std::async has temporary object lifetime (not moved or bound to a variable), the destructor of the std::future will block at the end of the full expression until the asynchronous operation completes

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