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dispatch_semaphore_t reuse - What am I missing here?

由 余生颓废 提交于 2019-12-05 03:46:54

Yes, this is the expected behavior. If you time out waiting for a signal, when the signal comes it, it will be caught by the next call to dispatch_semaphore_wait for that particular semaphore. Consider the following example:

For example:

dispatch_semaphore_t semaphore = dispatch_semaphore_create(0);
dispatch_time_t timeout;

// in 5 seconds, issue signal

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
    sleep(5);
    NSLog(@"Signal 1");
    dispatch_semaphore_signal(semaphore);
});

// wait four seconds for signal (i.e. we're going to time out before the signal)

NSLog(@"Waiting 1");
timeout = dispatch_time(DISPATCH_TIME_NOW, (int64_t)(4.0 * NSEC_PER_SEC));
if (dispatch_semaphore_wait(semaphore, timeout))
    NSLog(@"Waiting for 1: timed out");
else
    NSLog(@"Waiting for 1: caught signal");

// now, let's issue a second signal in another five seconds

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
    sleep(5);
    NSLog(@"Signal 2");
    dispatch_semaphore_signal(semaphore);
});

// wait another four seconds for signal

// this time we're not going to time out waiting for the second signal, 
// because we'll actually catch that first signal, "signal 1")

NSLog(@"Waiting 2");
timeout = dispatch_time(DISPATCH_TIME_NOW, (int64_t)(4.0 * NSEC_PER_SEC));
if (dispatch_semaphore_wait(semaphore, timeout))
    NSLog(@"Waiting for 2: timed out");
else
    NSLog(@"Waiting for 2: caught signal");

// note, "signal 2" is still forthcoming and the above code's 
// signals and waits are unbalanced

So, when you use the class instance variable, your getSomethingDoneUseMemberSemaphore behaves like above, where the second call to dispatch_semaphore_wait will catch the first signal issued because (a) it's the same semaphore; and (b) if the first call to dispatch_semaphore_signal timed out.

But if you use unique semaphores each time, then the second call to dispatch_semaphore_wait will not respond to the dispatch_semaphore_signal of the first semaphore.

When you call dispatch_semaphore_wait with a timeout, and the thread is still blocked at the timeout, what happens is almost the same as if dispatch_semaphore_signal had been called. One difference is that dispatch_semaphore_signal would wake up any thread, but the timeout wakes up this particular thread. The other difference is that dispatch_semaphore_wait will return a non-zero value instead of 0.

Here's the problem: Whoever was going to call dispatch_semaphore_signal is still going to call it, and then we have one signal too many. This may be hard to avoid; if you have a 10 second timeout then dispatch_semaphore_signal could be called after 10.000000001 seconds. So if you are reusing the semaphore, you have a problem at your hand.

On the other hand, if you are not reusing the semaphore then the worst that happens is that the semaphore count goes to 1. But that's no problem.

Summary: Don't reuse a semaphore if you wait for it with a timeout.

I was able to code up something akin to what I think you're looking for and it appears to work the way you want it to (but again, I'm not 100% sure I understand what you're looking for.):

ArduousTaskDoer.m

@implementation ArduousTaskDoer
{
    dispatch_semaphore_t mSemaphore;
    BOOL mWorkInProgress;
}

- (id)init
{
    if (self = [super init])
    {
        mSemaphore = dispatch_semaphore_create(0);
    }
    return self;
}

- (void)dealloc
{
    mSemaphore = nil;
}

- (void)doWork
{
    @synchronized(self)
    {
        mWorkInProgress = YES;
    }

    // Do stuff
    sleep(10);

    @synchronized(self)
    {
        mWorkInProgress = NO;
    }

    dispatch_semaphore_signal(mSemaphore);
}

- (BOOL)workIsDone
{

    @synchronized(self)
    {
        if (!mWorkInProgress)
        {
            mWorkInProgress = YES;
            dispatch_async(dispatch_get_global_queue(0, 0), ^{
                [self doWork];
            });
        }
    }


    if (dispatch_semaphore_wait(mSemaphore, dispatch_time(DISPATCH_TIME_NOW, (int64_t)2.5 * NSEC_PER_SEC)))
    {
        return NO;
    }

    return YES;
}

@end

...and then the calling code:

ArduousTaskDoer* task = [[ArduousTaskDoer alloc] init];
BOOL isDone = NO;
while(!(isDone = [task workIsDone]))
{
    NSLog(@"Work not done");
}

NSLog(@"Work is done");

// Do it again... Semaphore is being reused
while(!(isDone = [task workIsDone]))
{
    NSLog(@"Work not done");
}

NSLog(@"Work is done");

Hope this helps.

标签
objective-c
grand-central-dispatch
semaphore
nsthread
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