PHP function confusion

Question

I have the following scenario (simplified):

function changeFruit($fruit) {
 changeAgain($fruit);
}
function changeAgain($fruit) {
 $fruit = "Orange";
}
MAIN:
$fruit = "Apple";
changeFruit($fruit);
echo $fruit // Will show up as "Apple", How do I get it to show up as "Orange"??

EDIT: FORGOT TO ADD. THE SCENARIO CANNOT USE RETURN STATEMENTS - JUST &$variable

Thanks! Matt Mueller

• Anax: To the person who downvoted the question, you should seriously go back and check your very first code.
• Mez: And also, good practice to leave a comment if you downvote a question, to help improve the question)
• Tom: Anax: It wasn't me who downvoted you, but it seems that you haven't read much of the manual. Try this: php.net/manual/en/functions.arguments.php
• Matt: Hey I don't want to start a flame war here.. the issue DID NOT involve return statements.. I wanted more information on &$variable. I apologize it was late and a poorly written question and description, but my question's intent was not entirely trivial.

Answer 1

When you pass something that is not an object to a function in PHP, php makes a copy of that to use within the function.

To make it not use a copy, you need to tell PHP you are passing a reference.

This is done with the & operator

function changeFruit(&$fruit) {
 changeAgain($fruit);
}
function changeAgain(&$fruit) {
 $fruit = "Orange";
}
$fruit = "Apple";
changeFruit($fruit);
echo $fruit;

It would be more sensible, and better practice, to use return values of the functions (as this makes things easier to read)

function changeFruit($fruit) {
 return changeAgain($fruit);
}
function changeAgain($fruit) {
 // do something more interesting with$fruit here
 $fruit = "Orange";
 return $fruit;
}
$fruit = "Apple";
$fruit = changeFruit($fruit);
echo $fruit

• Matt: Thanks for digging into the question a little more. I forgot to add, that I can't return values!
• Paul Lammertsma: I was about to write exactly this. Entirely correct and very complete. Clean answer and clear examples!
• Mez: Thanks Paul - I tend to only answer if I'm not going to be giving one liners, and there's something I can go into in detail regarding :D

Answer 2

function changeFruit($fruit) {
 return changeAgain($fruit);
}
function changeAgain($fruit) {
 return $fruit = "Orange";
}
MAIN:
$fruit = "Apple";
$fruit = changeFruit($fruit);
echo $fruit;

Hope that helps!

Note: the return from the changeAgain function and overwriting $fruit = changeFruit($fruit);

• Mez: changeFruit also needs to return

Answer 3

You are not returning the values from your functions. Try this:

function changeFruit($fruit) {
 return changeAgain($fruit);
}
function changeAgain($fruit) {
 $fruit = "Orange";
 return $fruit;
}
MAIN:
$fruit = "Apple";
$fruit = changeFruit($fruit);