Do static and global modifiers for variables implement a non-modifiable reference?

Question

The PHP doc says

PHP implements the static and global modifier for variables in terms of references.

<?php
function test_global_ref() {
    global $obj;
    $new = new stdClass;
    $obj = &$new;
}

function test_global_noref() {
    global $obj;
    $new = new stdClass;
    $obj = $new;
}

test_global_ref();
var_dump($obj);
test_global_noref();
var_dump($obj);
?>

Since the program yields NULL as the first output, is this to say that the implemented reference is non-modifiable(hence the reference to &$new is nullified somehow)? The doc says the implementation results in an unexpected behaviour. Is there a logical explanation to this?

a true global variable imported inside a function scope with the global statement actually creates a reference to the global variable Looking at this, I simply feel the first one said NULL because it was pointing to the address of a local variable which got garbage collected when the function call finished. See onecompiler.com/php/3yxabtfmm — mighty_dev
– mighty_dev, Commented Feb 8, 2023 at 8:36

shingo · Accepted Answer · 2023-02-08 11:20:50Z

1

This is not about global or static, this is about the concept of reference.

Think about the following codes:

$a = "a"; $b = "b";

$r = &$a;
var_dump($a, $b, $r); # a, b, a

$r = &$b;
var_dump($a, $b, $r); # a, b, b

It's easy to understand, but the important thing is the statement $r = &$b; means copy the reference of $b to $r, so both $b and $r refer to the same value.

Next if you do:

$r = $a;
var_dump($a, $b, $r); # a, a, a

The statement $r = $a; means copy the value of $a to $r, so the value of $r changes from "b" to "a". Since both $b and $r refer to the same value, the value of $b also becomes "a".

Finally if you do:

$r = "r";
var_dump($a, $b, $r); # a, r, r

Still only the value of $b to $r is changed, $a keeps its original value.

Back to your question, your first function is almost equivalent to:

function test_global_ref(&$r) {
    $b = "b";
    $r = &$b;
}

$a = "a";
test_global_ref($a);

I changed the variable names and values to those corresponding to the above example, hope this is easier to understand. So the global variable $a is passed to the function as a reference $r, when you copy the reference of $b to $r, the global variable $a won't be influenced.

answered Feb 8, 2023 at 11:20

shingo

29.8k7 gold badges37 silver badges70 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

brittle_spirit Over a year ago

So, changing the reference simply overrides the existing reference without affecting the initial reference?

shingo Over a year ago

Yes, each variable has its own reference, changing one's doesn't affect other's.

Collectives™ on Stack Overflow

Do static and global modifiers for variables implement a non-modifiable reference?

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related