PHP assign global value the reference of another global value inside a function

Question

While playing around with globals and references in PHP I came across a problem. I wanted to set a variable to the reference of another variable inside a function. To my surprise, the global variable lost its reference after the function call.

In the code below you can see that inside the function $a gets the value 5, but afterwards it has its old value back (1). $x on the other hand has kept the value assigned inside the function.

<?php

$a = 1;
$x = 2;
function test() {
    global $a;
    global $x;

    $a = &$x;
    $x = 5;

    echo PHP_EOL;
    echo $a . PHP_EOL;
    echo $x . PHP_EOL;
}

test();

echo PHP_EOL;
echo $a . PHP_EOL; // $a is 1 here instead of 5
echo $x . PHP_EOL;

$a = &$x;

echo PHP_EOL;
echo $a . PHP_EOL;
echo $x . PHP_EOL;

Outputs:

Why does $a lose its reference after the function is done?

I'm not proficient in globals, as these are evil. I think you need to define global on first declaration of $a, ie before function definition. — PeterM
– PeterM, Commented Mar 31, 2020 at 14:45
As I know global vars can be changed via $GLOBAL array, I mean instead of $a you need to use $GLOBALS[$a] if you want to change it — Banzay
– Banzay, Commented Mar 31, 2020 at 14:51

0stone0 · Accepted Answer · 2020-03-31 15:00:02Z

1

As @Banzay noticed, I believe $a = &$x; only changes the function-scoped variable. You should use $GLOBALS to change the value in a function;

function test() {
    global $a;
    global $x;

    $GLOBALS['a'] = &$x;
    $x = 5;

    echo PHP_EOL;
    echo $a . PHP_EOL;
    echo $x . PHP_EOL;
}

Try online!

edited Mar 31, 2020 at 15:00

answered Mar 31, 2020 at 14:58

0stone0

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1 Comment

ApplePearPerson Over a year ago

That seems to work, although I still find it weird that $x can be changed with a static variable but $a can't with a reference.

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