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


Paul M. Foster

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