PHP “Assign by reference” oddity

Question

I came across a code snippet which included $a = & $b; but hadn\'t tested whether $b actually existed (if (isset($b))). I wasn\'t sure how PHP

Answer 1

Explanation is as simple as this

If you assign, pass, or return an undefined variable by reference, it will get created.

(emphasis mine)

That's what you're doing; Assigning an undefined index by reference so it gets created.

Example #1 Using references with undefined variables

d);
var_dump(property_exists($c, 'd')); // bool(true)
?> 

Example from PHP Manual

Then you have another question:

Meanwhile at the same time, $b[11] shows a value NULL and isset()=FALSE even though its referent (apparently) does exist (!)

That is also explained clearly on the manual

isset — Determine if a variable is set and is not NULL

isset() will return FALSE if testing a variable that has been set to NULL

Since it is NULL, isset() returns FALSE.