CODEWITHSUNDEEP
c++multithreadingc++11condition-variable
Julian
Julian

Reputation: 4780

How to use a condition_variable to really wait_for no longer than a certain duration

As it turns out, condition_variable::wait_for should really be called condition_variable::wait_for_or_possibly_indefinitely_longer_than, because it needs to reacquire the lock before really timing out and returning.

See this program for a demonstration.

Is there a way to express, "Look, I really only have two seconds. If myPredicate() is still false at that time and/or the lock is still locked, I don't care, just carry on regardless and give me a way to detect that."

Something like:

bool myPredicate();
auto sec = std::chrono::seconds(1);

bool pred;
std::condition_variable::cv_status timedOut;

std::tie( pred, timedOut ) =
    cv.really_wait_for_no_longer_than( lck, 2*sec, myPredicate );

if( lck.owns_lock() ) {
    // Can use mutexed resource.
    // ...
    lck.unlock();
} else {
    // Cannot use mutexed resource. Deal with it.
};

Upvotes: 22

Views: 33311

Answers (3)

iammilind
iammilind

Reputation: 69988

Is there a way to express, "Look, I really only have two seconds. If myPredicate() is still false at that time and/or the lock is still locked, I don't care, just carry on regardless ..."

Yes, there is a way, but unfortunately in case of wait_for it has to be manual. The wait_for waits indefinitely because of Spurious Wakeup. Imagine your loop like this:

while(!myPredicate())
  cv.wait_for(lock, std::chrono::duration::seconds(2);

The spurious wakeup can happen anytime in an arbitrary platform. Imagine that in your case it happens within 200 ms. Due to this, without any external notification wait_for() will wakeup and check for myPredicate() in the loop condition.

As expected, the condition will be false, hence the loop will be true and again it will execute cv.wait_for(..), with fresh 2 seconds. This is how it will run infinitely.

Either you control that updation duration by yourself or use wait_until() which is ultimately called in wait_for().

Upvotes: 0

Carlos Junior
Carlos Junior

Reputation: 11

Actually, the condition_variable::wait_for does exactly what you want. The problem with your example is that you locked a 2-second sleep along with the ready = true assignment, making it impossible for the condition variable to even evaluate the predicate before reaching the time limit.

Put that std::this_thread::sleep_for(2*sec); line outside the lock and see for yourself.

Upvotes: 1

Anton
Anton

Reputation: 6537

I think that you misuse the condition_variable's lock. It's for protecting condition only, not for protecting a time-consuming work.

Your example can be fixed easily by splitting the mutex into two - one is for critical section, another is for protecting modifications of ready condition. Here is the modified fragment:

typedef std::unique_lock<std::mutex> lock_type;
auto sec = std::chrono::seconds(1);
std::mutex mtx_work;
std::mutex mtx_ready;
std::condition_variable cv;
bool ready = false;

void task1() {
    log("Starting task 1. Waiting on cv for 2 secs.");
    lock_type lck(mtx_ready);
    bool done = cv.wait_for(lck, 2*sec, []{log("Checking condition..."); return ready;});
    std::stringstream ss;
    ss << "Task 1 finished, done==" << (done?"true":"false") << ", " << (lck.owns_lock()?"lock owned":"lock not owned");
    log(ss.str());
}

void task2() {
    // Allow task1 to go first
    std::this_thread::sleep_for(1*sec);
    log("Starting task 2. Locking and sleeping 2 secs.");
    lock_type lck1(mtx_work);
    std::this_thread::sleep_for(2*sec);
    lock_type lck2(mtx_ready);
    ready = true; // This happens around 3s into the program
    log("OK, task 2 unlocking...");
    lck2.unlock();
    cv.notify_one();
}

It's output:

@2 ms: Starting task 1. Waiting on cv for 2 secs.
@2 ms: Checking condition...
@1002 ms: Starting task 2. Locking and sleeping 2 secs.
@2002 ms: Checking condition...
@2002 ms: Task 1 finished, done==false, lock owned
@3002 ms: OK, task 2 unlocking...

Upvotes: 12

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