Passing Argument by Reference in PHP-CodePudding

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

CodePudding user response：

strtoupper() returns the result of its action, so you need to put that result back into the variable regardless of wether that variable is passed by reference or not.

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

CodePudding user response：

The issue here is not with passing an argument to reference, but how strtoupper itself works.

If instead of strtoupper you were doing something else, such as this

<?php

function increment(&$num)
{
    $num  ;
}

$n = 1;
increment($n);

echo $n."\n";

you would see that the $n variable outside is changed by the function.

But strtoupper doesn't modify the variable in place: it returns the uppercased value instead.

So that's why you need to reassign the value returned by strtoupper to the variable:

<?php
    $str = "ciao";
    increment($str);
    
    function increment(&$str){
        $str = strtoupper($str);
    }
    echo $str."\n";