提高:: condition_variable和锁定 [英] boost::condition_variable and lock

问题描述

boost::condition_variable cond;
boost::mutex mut;

//thread1
{
"read_socket()"
cond.notify_one();
}

//thread2
{
for(;;)
{
...
boost::unique_lock<boost::mutex> lock(mut);
cond.wait(lock);
}
}

与

boost::condition_variable cond;
boost::mutex mut;

//thread1
{
"read_socket()"
boost::unique_lock<boost::mutex> lock(mut);
cond.notify_one();
}

//thread2
{
for(;;)
{
...
boost::unique_lock<boost::mutex> lock(mut);
cond.wait(lock);
}

有没有影响，如果我打电话cond.notify_one前省略了锁（）？

Is there an impact if I omit the lock before calling cond.notify_one() ?

推荐答案

在C ++ 11标准没有规定任何要求 notify_one 和 notify_all ;所以当你一个信号是condition_variable罚款不持有锁。然而，它往往是必要的信令线程持有锁，直到它设置它唤醒后等待的线程检查的条件。如果不是这样，该程序可以包含比赛。有关示例，请参阅本SO问题：升压同步

The C++11 standard does not state any requirement for notify_one and notify_all; so not holding the lock when you signal a condition_variable is fine. However, it's often necessary for the signaling thread to hold the lock until it sets the condition checked by the waiting thread after it's woken up. If it does not, the program may contain races. For an example, see this SO question: Boost synchronization.

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