John Davidson

Message:

I started studying PHP and I have a question about passing arguments by reference in a function.

I wrote this code:

<?php

    $str = "ciao";

    increment($str);

    

    function increment(&$str){

        strtoupper($str);

    }

    echo $str."\n";

?>

The result is "ciao" instead of "CIAO".

Why if I pass a variable, like $str, by reference to a function, the original variable dosen't come modify?. I though String $str is immutable in php (like Java) but it's not so. To modify the original value I should write

$str=strtoupper($str);

instead of

strtoupper($str);

So, in general, why if I pass an argument by reference in a function in PHP I have to save in the same varible the modify that i do in the function's body?

I hope to be clear, thanks

Luca

---

My Answer:
In PHP, when you pass a variable by reference to a function, any changes made to the variable within the function will affect the original variable outside of the function. However, in your code, you are not actually modifying the variable $str within the function increment.

The function strtoupper() returns the modified string, but you are not assigning it back to the variable $str within the function. That's why the original variable $str remains unchanged.

To modify the original variable $str within the function, you should assign the result of strtoupper() back to the variable $str like this:

php
function increment(&$str){
$str = strtoupper($str);
}


By doing this, the original variable $str will be modified to "CIAO" as expected.