On Mon, Apr 26, 2010 at 12:24 AM, Paul M Foster pa...@quillandmouse.com 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
Paul M Foster wrote:
[snip]
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
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