Re: Mutex question
"Vincent Fatica" <vince@blackholespam.net> wrote in message
news:46ae8e21$1@news.vefatica.net...
An app loads my DLL and calls an initialization routine therein. That
routine
contains (essentially)
if ( !OpenMutex(0, FALSE, "MyMutex") )
{
// do some stuff
CreateMutex(NULL, FALSE, "MyMutex");
}
The app unloads my DLL. [The open handle to "MyMutex" remains, according
to
SysInternals's ProcessExplorer.]
If the same instance of the app later loads my DLL (again) and calls its
initialization routine, OpenMutex() fails and "some stuff" is done again
(which
is what I'm trying to avoid).
Why does OpenMutex() fail the second time around?
I see Doug has already pointed out that GetLastError() is your friend. I'd
like to add, though, in case you are wondering why the mutex still exists,
that handles are rundown when a process exits. The unload of the DLL is
beside the point.
Is there a (better) way for my DLL to "leave a mark" so that after
it has been unloaded, an instance loaded in the future will know it
had been loaded/unloaded previously?
Will the DLL_PROCESS_DETACH notification do?
Regards,
Will
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