如何仅使用C ++ 11永久睡眠 [英] How to sleep forever only using C++11

问题描述

是的，我可以在Windows中使用Sleep（）或在POSIX中使用pause（），然后继续。
但是如何仅使用C ++ 11进行睡眠？我认为有一种方法，可以使用std :: this_thread加入调用线程，但与pthread函数不同，std :: this_thread没有join（）方法。
更不用说我们不能使用C ++ 11处理信号，而且我知道如何像下面这样永远迭代睡眠：

Yes, I could use Sleep() in Windows or pause() in POSIX, and goes on. But how do I sleep only using C++11? I thought there was a way, which is joining calling thread using std::this_thread but std::this_thread has no join() method unlike pthread functions. Not to mention that we can't handle signals with C++11 and I know how to iterate sleep forever like below:

while(true)
   std::this_thread::sleep_for(std::chrono::seconds(1));

但是，正如您所看到的，它一点也不优雅。这段代码仍然消耗CPU时间。调度程序必须照顾这个过程。我也可以使用条件变量或promise，但是再一次它会占用一些内存或无法在某些操作系统上运行（它将抛出异常以避免死锁）。

However, as you can see, It's not elegant at all. This code still consumes CPU time. The scheduler has to care for this process. I could also use conditional variable or promise, but then again it takes up a bit of memory or wouldn't work on certain OS(it would throw an exception to avoid deadlock).

也许这等效于Windows的Sleep（INFINITE）：

Maybe this could be the equivalent of Sleep(INFINITE) of Windows:

while(true)
   std::this_thread::sleep_for(std::chrono::hours::max());

但是很多人说这不切实际。

But many say it's not practical.

有人能想到出色的方法吗？

Could anyone think of brilliant way?

推荐答案

我可以想到两种方法。

一个，@ Guiroux提到的是 sleep_until 不可到达的时间：

One, which @Guiroux mentions is to sleep_until a non-reachable time:

std::this_thread::sleep_until(std::chrono::system_clock::now() + std::chrono::hours(std::numeric_limits<int>::max()));

或者让它无限期地等待永远不会满足的条件。

Or have it wait indefinitely for a condition that will never be fulfilled.

std::condition_variable cv;
std::mutex m;
std::unique_lock<std::mutex> lock(m);
cv.wait(lock, []{return false;});

但是我看不出原因。

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