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From: Peter Dimov (pdimov_at_[hidden])
Date: 2001-08-13 13:55:29


From: <duncan_at_[hidden]>

> pdimov_at_[hidden] (Peter Dimov) wrote:
>
> > synchronized_queue<int> q;
> > event e;
> >
> > void threadfunc()
> > {
> > for(;;)
> > {
> > waitFor(e);
> >
> > while(!q.empty())
> > {
> > int m = q.pop();
> > if(m & 1) do_something_a();
> > if(m & 2) do_something_b();
> > if(m & 4) do_something_c();
> > }
> > }
> > }
> >
> > void notifyThread(bool a, bool b, bool c)
> > {
> > q.push(a + 2 * b + 4 * c);
> > e.raise();
> > }
> >
> > Did I utterly miss the point? :-)
>
>
> Surely notifyThread() can happen between end of while loop and before
> the waitFor(e) is entered again. In which case you have a lost wakeup.

Could you elaborate on that? It is possible to 'lose' a wakeup since
notifyThread() can be called several times while the thread is in the while
loop, but what's the problem? The event will remain signalled.

--
Peter Dimov
Multi Media Ltd.

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