同时调用`sync.Cond`的`Wait（）`方法安全吗？

Question

Is calling the Wait() method of sync.Cond safe when according to the documentation, it performs an Unlock() first?

Assume we are checking for a condition to be met:

func sample() {
    cond = &sync.Cond{L: &sync.Mutex{}} // accessible by other parts of program

    go func() {
        cond.L.Lock()
        for !condition() {
            cond.Wait()
        }
        // do stuff ...
        cond.L.Unlock()
    }()

    go func() {
        cond.L.Lock()
        mutation()
        cond.L.Unlock()

        cond.Signal()
    }()
}

And:

func condition() bool {
    // assuming someSharedState is a more complex state than just a bool
    return someSharedState
}

func mutation() {
    // assuming someSharedState is a more complex state than just a bool
    // (A) state mutation on someSharedState
}

Since Wait() performs an Unlock, should (A) has a locking of it's own? Or being atomic?

Answer 1

Yes it is safe to call Wait even when it calls L.Unlock() first but it is essential that you acquired the lock before calling Wait and before checking your condition since, in this situation, both are not thread safe.

Wait atomically unlocks c.L and suspends execution of the calling goroutine. After later resuming execution, Wait locks c.L before returning.

1. The goroutine that calls Wait acquired the lock, checked the condition and found it to be unsatisfactory.
2. Now it waits but in order to allow for changes of the condition it needs to give the lock back. Wait does that automatically for you and then suspends the goroutine.
3. Now changes of the condition can happen and eventually the goroutine is awoken by Broadcast or Signal. It then acquires the lock to check the condition once again (This must happen one-by-one for each waiting goroutine or else there would be no way telling how many goroutines are running around freely now).