On Mon, Apr 26, 2010 at 12:24 AM, Paul M Foster <paulf@xxxxxxxxxxxxxxxxx> wrote: > Here is some code: > > $a = new my_object; > $b = $a; > > My understanding of this operation under PHP 5+ is that $b will now be > essentially a "reference" to $a, *not* a *copy* of the $a object. Is > this correct? > > There are cases where I strictly want a *copy* of $a stored in $b. In > cases like this, I supply $a's class with a copy() method, and call it > like this: > > $b = $a->copy(); > > Is this reasonable, or do people have a better/more correct way to do > this? > > Paul > > -- > Paul M. Foster > I've not used it, but isn't that what clone() is for? Andrew -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
