[Freeswitch-dev] Why is _thread_cond_timedwait written the way it is?

Richard Lewis Haggard haggard at msn.com
Wed Apr 11 05:27:39 MSD 2012

I was looking at _thread_cond_timedwait  in thread_code. There is a section
of the code that I do not understand the purpose of. Could someone please
explain to me what the purpose of the other part of the code is?


Before the code enters a loop it serializes access to the passed in cond
structure pointer, increments a counter and makes a local copy of a value
before releasing the cond again.


    generation = cond->generation;


What is the purpose of 'num_waiting' and 'generation'?


The next part looks odd to me. Why does the code unlock the mutex when what
is desired is to lock it?



The code then enters a do { } while (true); loop.


    do {


The first thing it does is to wait for access to a semaphore. This is a
limited time wait and may time out before the event is encountered. The
result of this wait is checked later.


        res = WaitForSingleObject(cond->semaphore, timeout_ms);


The 'cond' structure is going to be modified again and the code needs to
have exclusive access. 



I do not know what the purpose of this is. Can someone explain it to me?
What is the purpose of this particular block?


        if (cond->num_wake) {
            if (cond->generation != generation) {
                rv = APR_SUCCESS;
            } else {
                wake = 1;
        else if (res != WAIT_OBJECT_0) {
            rv = APR_TIMEUP;


If neither 'break' executes then the code releases the critical section. The
code would be less obtuse (and smaller) if the two 'break' functions were
removed. That would save having duplicate LeaveCriticalSection() functions.
In any case, the code releases the critical section.




What is the purpose of the next bock? It only executes if the semaphone
acquisition was successful but the 'cond->generation' has changed while
attempting to determine if it is OK to execute the mutex lock. 


        if (wake) {
            wake = 0;
            ReleaseSemaphore(cond->semaphore, 1, NULL);

Note: If the two 'break' statements were removed from above and the
following lines added. At this point in the code the critical section would
have been released so it would not be necessary to explicitly do so outside
of the forever loop.

        else {


Under what conditions can 'cond->generation' change?


The bottom of the loop is encountered.


    } while (1);


To get here one of the two 'break' statements must have been hit. The
critical section is still locked and must be released. Note: if my suggested
changes were implemented then it would not be necessary to have this line:



And now, finally, the mutex gets locked.





Richard Lewis Haggard 


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