Why php does not complain when referencing a non existing variable?

Question

I was wondering why php does not complain when we reference a non existing variable (being it a plain variable or array), is this just the way it is, or there is something e

Answer 1

throws no warning about a non existing variable.

This is how references work. $a = &$b; creates $b if it does not exist yet, "for future reference", if you will. It's the same as in parameters that are passed by reference. Maybe this looks familiar to you:

preg_match($pattern, $string, $matches);

Here the third parameter is a reference parameter, thus $matches does not have to exist at time of the method invocation, but it will be created.

that &NULL is something I didn't really expected, I thought I would get a plain NULL.

Why didn't you expect it? $r['er'] is a reference "to/from" $t. Note that references are not pointers, $r['er'] and $t are equal references to the same value, there is no direction from one to the other (hence the quotation marks in the last sentence).