Redefining a PHP variable by referencing the variable in a function?

Question

Gee Wiz finding the vocab to express the question can be pretty tricky, but hopefully you can tell me what's wrong with referencing a variable in a function in order to change it.

Example:

    <?php
$a = 5;
$something = 5;

function changeIt($it){
    global $a;
    $it = $it * $a;
};

changeIt($something);

echo $something;
    ?>

This returns 5, so the global variable $something hasn't been redefined.

... not calling changeIt for one, not using reference variables for another. php.net/manual/en/language.references.php . Use &$it in your function declaration instead. — Ry-
– Ry- ♦, Commented May 26, 2011 at 23:42
Your not calling changeIt(). Furthermore, what do you want to change? — Jason McCreary
– Jason McCreary, Commented May 26, 2011 at 23:42
As both minitech and Jason noted you are doing a number of thing incorrectly. It would be very helpful if you told us what exactly you are trying to achieve, forget about the code. — yannis
– yannis, Commented May 26, 2011 at 23:44
Be more specific in what you are trying to do. Are you trying to return the value of $it? make sure you return $it . Are you trying to modify the value of the variable that $it references in a pass by reference type of way? Make sure you specify that you are passing by reference function changeIt(&$it) . — Zak
– Zak, Commented May 26, 2011 at 23:49

Ry- · Accepted Answer · 2011-05-26 23:50:14Z

5

You need to pass $it by reference, and also call changeIt:

<?php
    function changeIt(&$it){
       global $a;
       $it = $it * $a;
    }

    $a = 5;
    $something = 5;

    changeIt($something);
?>

http://ideone.com/9RQQW

edited May 26, 2011 at 23:50

answered May 26, 2011 at 23:45

Ry-♦

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5 Comments

mpen Over a year ago

Just guessing that he wants to do changeIt($something), since he's used $a as a global, but $something is unused.

yannis Over a year ago

+1 for the correct answer and also for including the totally unused $something in it...

3Dom Over a year ago

Hey please check my code above and please tell me why you need that ampersand?

Ry- Over a year ago

@3Dom: You need the ampersand to pass the variable by reference. See the link I posted as a comment to your question.

3Dom Over a year ago

Passing by reference. Now there's another obscure operation I wouldn't have thought necessary, but hey, super duper spot on mate. You have made me a happy man.

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