how the value incresed without increasing in the this php code

Question

<?php

function doSomething( &$arg ){
    $return = $arg;
    $arg += 1;
    return $return;
}

$a = 3;
$b = doSomething( $a );

echo $a.'  ';
echo $b;
?>

I know the answer a=4 and b=3 I understand how b= 3 but how come value of the a increased

Because you passed the argument by reference? (`&$arg`) – DarkBee Dec 05 '22 at 08:53 — DarkBee, Dec 05 '22 at 08:53

score 3 · Answer 1 · answered Dec 05 '22 at 08:53

The & operator tells PHP not to copy the variable when passing it to the function. Instead, a reference to the variable is passed into the function, thus the function modifies the original variable instead of a copy.

Therefore $arg += 1; will increase $a

how the value incresed without increasing in the this php code

1 Answers1