PHP Question: Variable Reference

17th April 2011 - 2 minutes read time


What does the following code snipped print?

$a = 5;
$b = 'a';
print $$b;


This will print "5" because $$b tells PHP to use the value of the $b variable as a variable name, so as the value of $b is 'a' we are really looking at the value of $a; the $$b above could be replaced by $a and have the same output. This variable variable feature is a little obscure but it is useful to use in certain circumstances.

This can be taken a step further by using multiple levels of reference to access a variable by adding more dollar signs to the front of the variable name. All of the following will print out 5 because all of the print statements will point back to the variable $a.

$a = 5;
$b = 'a';
$c = 'b';
$d = 'c';
$e = 'd';
$f = 'e';

print $$b;
print $$$c;
print $$$$d;
print $$$$d;
print $$$$$e;
print $$$$$$f;



It prints 5. That's because $$b is actually '$' + 'a' so the result is print $a

Mihai Baboi (Mon, 04/18/2011 - 09:44)

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