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Home/ Questions/Q 7415591
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Editorial Team
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Editorial Team
Asked: May 29, 20262026-05-29T07:19:56+00:00 2026-05-29T07:19:56+00:00

boost::condition_variable cond; boost::mutex mut; //thread1 { read_socket() cond.notify_one(); } //thread2 { for(;;) { …

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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);
}
}

versus

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);
}

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

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T07:19:57+00:00Added an answer on May 29, 2026 at 7:19 am

    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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