Subj : Re: recursive mutexes
To   : comp.programming.threads
From : Uenal Mutlu
Date : Sun May 15 2005 06:11 pm

"Uenal Mutlu" wrote
> "Peter Dimov" wrote
> > Uenal Mutlu wrote:
> > > "David Schwartz" wrote
> > >>     Yes, it can and must know that it already has the lock. A
> > >> function that operates on X should be defined as being called with X
> > >> locked, or with X unlocked. It could even take a boolean that
> > >> indicates whether the caller has locked the object or not, though
> > >> this is rarely the right thing to do.
> > >
> > > You are unnecessarily complicating things and make the program
> > > slower by doing these testings.
> >
> > No, he's in fact making the program faster.
>
> Are you sure? Let's see:
>
> void X::f()
> {
>    Lock();
>    for (int i = 0; i < 1000; i++)
>      f();
>    Unlock();
> }
>
> void X::g()
> {
>   if (!IsLocked())
>     return;
>   //...do..something...
> }
>
> Are you saying the above version of g() is faster than this one:  ?
> void X::g()
> {
>   //...do..something...
> }

Besides being slower, the first version is also buggy. And I don't know
what he or you want do if the object is not already locked. I guess you will do the following:
void X::g()
{
  bool fILockedItHere = false;
  if (!IsLocked())
     {
       Lock();
       fILockedItHere = true;
     }
  //...do..something...
  if (fILockedItHere)
     Unlock();
}

But, this is not thread safe!!! :-) Do you know why?
The only consequence is: my method is the only right one, believe me! :-)

.