Re: can't release mutex

[David Sugar]

I think what he really wants to do is use the Common C++ Semaphore with
a count of 1, since semaphore can block with a "timeout", and with a "1"
count in effect acts as a mutex...

The larger question though of what to do with the stuck thread, perhaps
from a broken library, is important also.  He could cancel the thread in
question if it timed out, and put cleanup code in the thread destructor
when he invokes delete to join it, but if the library allocated
resources he cannot determine for them to get cleanly deallocated.

Conrad T. Pino wrote:
> You're on the right track.  Please pardon my rusty recollection syntax.
> Revise:
> 
>       mutex.enterMutex();
> 
> to a form that accepts a timeout.  This example won't compile:
> 
>       const int timeout = 500;
> 
>       while (1) {
>               if ( isPending( pendingInput ) ) {
>                       int result = mutex.enterMutex( timeout );
> 
>                       if ( result == success ) {
>                               ---> protected function call (a function that 
> can't work 
>                               in multithread env) <---       
>                               mutex.leaveMutex();
>                       } else {
>                               // do something else with pending input
>                       }
>               }
>       }
> 
> While the above will implement your timeout idea it does nothing about
> the fundamental problem.  When the library "breaks" can be discoverd
> and the blocked mutex/thread combination can be cancelled/restared
> HOWEVER PROPRIETARY LIBRARY STATE IS INCONSISTENT & LIKELY DANGEROUS.
> 
> If the library is dynamically loaded under your control then unload it
> and reload it to get it back into a consistent state.  If it's auto or
> static linked then you may have an unsolveable problem.
> 
> If a reliable system is your goal then one of these MUST be true:
> 
>               proprietary library NEVER fails
>       OR
>               proprietary library can be forced back to a consistent
>               state because it has a reload or reinitialize method
> 
> Good luck.
> 
>> -----Original Message-----
>> From: address@hidden
>> [mailto:address@hidden Behalf Of Wakan
>> Sent: Saturday, January 27, 2007 07:18
>> To: address@hidden
>> Subject: can't release mutex
>>
>> Hi,
>> I'm using commoncpp library in a project that includes a multithread 
>> server that
>> can accept many tcp connection and each connection generates a thread.
>> In this server I'm using a thirdy party proprietary library (compiled), 
>> that (as the producer says)
>> can't work in multithread environment.
>> As each connection thread needs to call functions of this thirdy party 
>> library, I've used a mutex to protect the piece of
>> code where this external library function is called.
>> In this way I can use this library fine, because all calls are 
>> serialized and the protected function is called by one thread each time.
>> This is an example:
>> ...
>>     while(1) {
>>         if(isPending(pendingInput)) {
>>                 mutex.enterMutex();
>>                 ---> protected function call (a function that can't work 
>> in multithread env) <---       
>>                 mutex.leaveMutex();
>>        }
>>     }
>> ...
>> in 99% of cases it works fine. But it accidentally happens that a thread 
>> can't exit from the mutex
>> maybe because the thirdy party function freezes itself.
>> When this happens, the server can accept other connections, soit still 
>> works, but each thread can't enter in the mutex proteced piece of code.
>> I'm thinking about something, like a timeout, that exits the mutex after 
>> a time, if the thread is not exiting...
>> Can someone help me to resolve this problem? Are there other solutions?
>> Thanks in advance,
>> Regards
>> Carlo
> 
> 
> 
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