Passing by reference vs. passing by value - php

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.
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);

Answer 2

This is not a very good explanation from theory point of view, but this should help you understand the concept:
when you declare function like this and then call it:
function ($argument) {...}
The argument you pass there is passed by value. This means that inside the scope of a function you will be working with a copy of an argument you passed. You can imagine that before the function is called the copy of argument has been made and inside the function you're working with the copy, while original remains untouched. Once the function is finished the copy is no more
However when you declare it like this:
function (&$argument) {...}
You are passing argument by reference, meaning that you are working directly with a variable you've passed. So in this case no copies are made, you took the argument from one place, put it inside the function and changed it.

Answer 3

In PHP, "The scope of a variable is the context within which it is defined." (according to the docs). So inside your function, $y (a copy of the value you passed in) is being operated on, but it is not returned by the function. So when the function ends, the value is no longer accessible outside the function.
If you want to pass the variable in by reference (similar to a pointer in C) then you can add a & like so:
function sum(&$y) {
$y = $y + 5;
}
Now when you call this code:
$x = 5;
sum($x);
echo $x;
it will output 10. Maybe a better way to do this would be to return a value from your function, and output that value:
function sum($y) {
return $y + 5;
}
$x = 5;
echo sum($x);