boost :: condition_variable和lock [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);
}
}

对比

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);
}

如果我在调用cond.notify_one ？

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

推荐答案

C ++ 11标准没有规定 notify_one 和 notify_all ;所以没有持有锁，当你发出一个condition_variable是罚款。然而，通常需要信令线程保持锁定，直到它设置由等待线程在它被唤醒之后检查的条件。如果没有，程序可能包含种族。有关示例，请参见此SO问题：升级同步。

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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