Re: one last boost conversion question: notify_one

From: Mark Sizer (yg-boost-users_at_[hidden])
Date: 2003-05-22 16:11:49

Again, duh on me. Thank you.

I don't want to use the 'while' paradigm, because I want to be able to 
free all the blocked threads on shutdown so they all gracefully 
terminate (destructors called, etc...). Getting a default constructed 
message back is OK as long as it's rare.

I've refactored to this (it seems to work, but I'm posting it anyway):

MSG mqueue::get()
{
   MSG msgResult;
   { boost::mutex::scoped_lock lockQueue( _mutexQueue );
     if ( _queueOut.empty() )
     {
       _conditionMessageReady.wait( lockQueue );
     }

     if ( !_queueOut.empty() )
     {
       msgResult = _queueOut.front();
       _queueOut.pop();
     }
     else
     {
       // this happens after a flush() call (on shutdown)
       // (and the occasional race condition)
     }
   }
   return msgResult;
}

( and changed post back to .notify_one() )
Thanks again,
  - Mark

Christopher Currie wrote:

> Mark,
> 
> The problem is that your "get" waits on the condition before it checks 
> if the condition has already been met. So if you call "get" after the 
> producer has already "post"ed, you'll sit on the condition forever. The 
> canonical form of waiting on a condition looks like this:
> 
> 
>>MSG mqueue::get();
>>{
>>  MSG msgResult;
>>  { boost::mutex::scoped_lock lockQueue( _mutexQueue );
>>    while ( _queueOut.empty() )
>>
>  >       _conditionMessageReady.wait( lockQueue );
> 
>>    
>>
>  >     msgResult = _queueOut.front();
> 
>>    _queueOut.pop(); 
>>  }
>>  return msgResult;
>>}
>>
> 
> Now if you call get and there's something in the queue, there's no need 
> to wait on the condition.
> 
> 
>>P.S. The lockQueue is locked when the condition is done waiting, right?
>>
> 
> Yes, the lock is automatically released when entering wait, and acquired 
> before exiting it.
> 
> Hope this helps!
> 
> Christopher
> 
> 
> 
> 
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