Hey,
put this in thedocs somewhere?
regards,
Derick
On Wed, 9 Jun 2004, Sara Golemon wrote:
> That's true. What I left out of my explanation (in order to keep it simple)
> is that when you "copy" a variable, a new label is created to point to the
> same zval, and the zval's refcount is incrme
In general, if you don't mean for it to be a reference semantically, you
are best off not using references at all. PHP will do the right thing.
Andi
At 10:32 AM 6/9/2004 +0200, Bert Slagter wrote:
Derick Rethans wrote:
On Wed, 9 Jun 2004, Stephan Schmidt wrote:
Yes, that is correct.
No, it is not
Derick Rethans wrote:
On Wed, 9 Jun 2004, Stephan Schmidt wrote:
Yes, that is correct.
No, it is not. 'copying' a variable copies the structure to contain the
variable, but not the data, but does NOT make a reference as that's a
different concept.
You are right that the data associated with a vari
On Wed, 9 Jun 2004, Stephan Schmidt wrote:
> Hi,
>
> > I think, and I could be completely wrong, that copying a
> > variable actually
> > creates a reference. The data is only copied when the
> > variable referenced is
> > modified.
> Yes, that is correct.
No, it is not. 'copying' a variable copi
Hi,
> I think, and I could be completely wrong, that copying a
> variable actually
> creates a reference. The data is only copied when the
> variable referenced is
> modified.
Yes, that is correct.
Stephan
--
PHP Internals - PHP Runtime Development Mailing List
To unsubscribe, visit: http://w
