I'm running into an issue with changing a global variable to another reference inside a function.
$one = 1;
$two = 2;
$ref = &$one;
change();
echo $ref;
function change(){
global $ref, $two;
$ref = &$two;
}
The result of the code is "1". I don't really understand why, I would have guessed changing the global variable inside the function it would be persistent after leaving the function scope?
What would be a good workaround (apart from using the GLOBALS array)?
Actually that was a great question. In order to have references returned from functions, use the & before the function name as below.
$one = 1;
$two = 2;
$ref = &$one;
$ref = &change();
echo $ref;
$two = 3;
echo $ref;
function &change(){
global $two;
return $two;
}
Output:
23
In PHP, reference is NOT pointer. It is something like alias of another variable. I will explain what happen with your code:
$one = 1;
$two = 2;
$ref = &$one;
After three commands above, we have:
variables | $one | $ref | $two |
content | 1 | 2 |
As you see, $one and $ref refer to the same content, it is what the term reference meaning. Continue:
global $ref, $two;
According to this document, above command the same as:
$ref =& $GLOBALS['ref'];
$two =& $GLOBALS['two'];
So, we have:
variables | $one (global) | $ref(global) | $ref (local) | $two (global) | $two (local) |
content | 1 | 2 |
Yes, we have 5 variables! Continue:
$ref = &$two;
It is actually:
$ref (local) = &$two (local);
So we have:
variables | $one (global) | $ref(global) | $ref (local) | $two (global) | $two (local) |
content | 1 | 2 |
And, the last command:
echo $ref;
Actually is:
echo $ref (global);
And, 1 is the correct value!
Additional:
change();
echo $two;
function change(){
global $ref, $two;
$ref = &$two;
$ref = 9;
}
The result of this code is 9;
----- EDIT -----
I did not read the question carefully. My answer is for the part The result of the code is "1". I don't really understand why. The answer of Jonathan Gagne is what you are looking for.
9 only because you assign it inside the function. He just has to use & before the function name. If he changes the value of his variable $two it is not referencing to the $ref variable. Look my solution please you will understand what he was asking for.
$one = 1; $two = 2; $ref = &$one; echo $ref; change(); echo $ref; $two = 3; echo $ref; it will output 199 instead 193 as he wants. However, your answer was really instructive and descriptive of the PHP global variable. But did not represent a solution for this question. Thanks for all btw!
$ref = $two;