通过引用传递与通过值传递

Question

<?php
function sum($y) {
$y = $y + 5; 
}

$x = 5;
sum($x);
echo $x;
?>

So I have this code. The questions are: What does it output? The answer: 5. How do I make it to output 10? The answer: sum(&$x).

The problem is that I don't understand why the answer to the first question is 5. When you make sum($x), shouldn't it replace the function with $x, so $x= 5+5=10? Why the answer is 5? I really don't understand. Someone explaind me something related to pointers and the adress, but I didn't understand. I never understood the concept of pointers, and I googled it and apparently there are no pointers in php, so I'm super confused. My friend said that a variable is formed of a value, and the memory adress of that value. Can someone explain me like I'm 5 years old why the answer is 5 and not 10? Please

Answer 1

Let's pretend $x is a piece of paper with 5 written on it.

function sum($y) {
  $y = $y + 5; 
}

Here $y is the value of what you have written. You add 5 to such value in your mind, but the note is left untouched.

function sum(&$y) {
  $y = $y + 5; 
}

With the reference operator (&$y), you pass the very paper to the function, and it overwrites what's written on it.

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For primitive values like numbers, I wouldn't bother and always return the value you want:

function valuePlusFive($x) {
  return $x + 5;
}

$x = 5;
$x = valuePlusFive($x);