NSCondition, what if no lock when call signal?

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From this Apple's document about NSCondition, the usage of NSCondition should be:

Thead 1:

[cocoaCondition lock];
while (timeToDoWork <= 0)
    [cocoaCondition wait];
timeToDoWork--;
// Do real work here.
[cocoaCondition unlock];

Thread 2:

[cocoaCondition lock];
timeToDoWork++;
[cocoaCondition signal];
[cocoaCondition unlock];

And in the document of method signal in NSConditon:

You use this method to wake up one thread that is waiting on the condition. You may call this method multiple times to wake up multiple threads. If no threads are waiting on the condition, this method does nothing. To avoid race conditions, you should invoke this method only while the receiver is locked.

My question is:

I don't want the Thread 2 been blocked in any situation, so I removed the lock and unlock call in Thread 2. That is, Thread 2 can put as many work as it wish, Thread 1 will do the work one by one, if no more work, it wait (blocked). This is also a producer-consumer pattern, but the producer never been blocked.

But the way is not correct according to Apple's document So what things could possibly go wrong in this pattern? Thanks.

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Ken Thomases On BEST ANSWER

Failing to lock is a serious problem when multiple threads are accessing shared data. In the example from Apple's code, if Thread 2 doesn't lock the condition object then it can be incrementing timeToDoWork at the same time that Thread 1 is decrementing it. That can result in the results from one of those operations being lost. For example:

Thread 1 reads the current value of timeToDoWork, gets 1
Thread 2 reads the current value of timeToDoWork, gets 1
Thread 2 computes the incremented value (timeToDoWork + 1), gets 2
Thread 1 computes the decremented value (timeToDoWork - 1), gets 0
Thread 2 writes the new value of timeToDoWork, stores 2
Thread 1 writes the new value of timeToDoWork, stores 0

timeToDoWork started at 1, was incremented and decremented, so it should end at 1, but it actually ends at 0. By rearranging the steps, it could end up at 2, instead. Presumably, the value of timeToDoWork represents something real and important. Getting it wrong would probably screw up the program.

If your two threads are doing something as simple as incrementing and decrementing a number, they can do it without locks by using the atomic operation functions, such as OSAtomicIncrement32Barrier() and OSAtomicDecrement32Barrier(). However, if the shared data is any more complicated than that (and it probably is in any non-trivial case), then they really need to use synchronization mechanisms such as condition locks.