What is the exact purpose of a statement like this?
$myObj =& $existingObject
Is this extending $myObj with the properties and methods of $existingObject?
What does the equals sign (=) and the ampersand (&) married together here do?
Um first of all, &= and =& are two different things. Which is it?
&= is a bitwise and with the righthand side=& is better written as = & (with a space), is assigning something as a reference.$myObj becomes a reference to $existingObject instead of being copied. So any change to $myObj also changes $existingObject. (see this article)
This is a pass by reference. It means anything you do to $existingObject will also be done to $myObj because one is a reference to the other.
Based on the names of the variables, they are objects. Objects are always passed by reference, so = & would be redundant.