C ++ 11:如果不为std :: thread调用join()会发生什么

问题描述

下面给出:

void test()
{
  std::chrono::seconds dura( 20 );
  std::this_thread::sleep_for( dura );
}

int main()
{
  std::thread th1(test);
  std::chrono::seconds dura( 5 );
  std::this_thread::sleep_for( dura );
  return 0;
}

main将在5秒后退出，仍在执行的th1会怎样?

main will exit after 5 seconds, what will happen to th1 that's still executing?

即使您在main中定义的th1线程对象超出范围并被销毁，它是否仍继续执行直到完成?

Does it continue executing until completion even if the th1 thread object you defined in main goes out of scope and gets destroyed?

th1是在执行完毕后还是坐在那里还是在程序终止时以某种方式清理掉?

Does th1 simply sits there after it's finished executing or somehow gets cleaned up when the program terminates?

如果线程是在函数中创建的，而不是在main中创建的，该线程会一直存在直到程序终止或函数超出范围时该怎么办?

What if the thread was created in a function, not main - does the thread stays around until the program terminates or when the function goes out of scope?

如果您想要某种类型的超时行为，简单地不为线程调用join是否安全?

Is it safe to simply not call join for a thread if you want some type of timeout behavior on the thread?

推荐答案

如果在调用析构函数时尚未分离或加入线程，则它将调用std::terminate，我们可以通过转到草稿C ++ 11标准我们看到，该部分30.3.1.3 线程析构函数说:

If you have not detached or joined a thread when the destructor is called it will call std::terminate, we can see this by going to the draft C++11 standard we see that section 30.3.1.3 thread destructor says:

关于此行为的基本原理，我们可以在(不)使用std :: thread

as for a rationale for this behavior we can find a good summary in (Not) using std::thread

，然后是一个示例，并说:

and an example follows and also says:

本文引用了 N2802:重新考虑线程对象的销毁分离，这与先前的建议相抵触，该建议是在可以连接时在销毁时分离的，并且它指出，两个替代方案之一将是联接，这可能导致死锁，另一个替代方案是今天，如果可以连接，则销毁状态为std::terminate.

The article references N2802: A plea to reconsider detach-on-destruction for thread objects which is argument against the previous proposal which was detach on destruction if joinable and it notes that one of the two alternatives would be to join which could lead to deadlocks the other alternative is what we have today which is std::terminate on destruction if joinable.

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