Re: [Boost-Users] Why not "if (px != rpx) " ?

From: Peter Dimov (pdimov_at_[hidden])
Date: 2001-12-15 10:36:32

From: "Louis Lavery" <Louis_at_[hidden]>
> From: Peter Dimov <pdimov_at_[hidden]>
> > From: "Louis Lavery" <Louis_at_[hidden]>
> > >
> > >    void share(T* rpx, long* rpn) {
> > >       if (pn != rpn) { // Q: why not px != rpx? A: fails when both ==
0
> > >          ++*rpn; // done before dispose() in case rpn transitively
> > >                  // dependent on *this (bug reported by Ken Johnson)
> > >          dispose();
> > >          px = rpx;
> > >          pn = rpn;
> > >       }
> > >    } // share
> > >
> > > ...I curious to know what it means by "fails when both == 0".
> >
> > It's possible that px == rpx == 0, yet pn != rpn.
>
>     Yes, I see that, but I can't see what's wrong with using the
> test "if ( px != rpx )".
>
> Or, to put it another way, if px == rpx == 0 who cares whether
> pn equals rpn or not?

Two shared_ptr's are either copies of each other (pn == rpn) or they are not
(pn != rpn).

It is possible for two shared_ptr's to be both zero, yet to have different
counts:

shared_ptr<int> a;
shared_ptr<int> b; // b.use_count() == 1
b = a; // b.use_count() == 2

It's also possible to have two NULL shared_ptr's that have the same count:

shared_ptr<int> a;
shared_ptr<int> b(a); // b.use_count() == 2
b = a; // b.use_count() == 2

If you change the test to px != rpx, the first example will leave b with the
same count. This is not correct since after the assignment a and b are
guaranteed to share the same count (and hence use_count() must be at least
2.)

I realize that this is a bit confusing; think about it this way. The test is
an "optimization." The results of share() must be the same even without the
test.

--
Peter Dimov
Multi Media Ltd.