CODEWITHSUNDEEP
phpvariablesclosuresglobal
user2180613
user2180613

Reputation: 739

Using global inside a closure in PHP

Somehow, the following results in NULL. Why is this the case? I find it important to know, because the use statements does not support variable variables.

function a() {
    $a = "a";
    $aa = function() {
        global $a;
        var_dump($a);
    };
    $aa();
}
a();

Upvotes: 0

Views: 192

Answers (5)

newacct
newacct

Reputation: 122489

global $a access a global variable called $a. It is not the local variable in the function a. Since you have never initialized a global variable called $a, it is NULL.

In order for local variables to be usable inside an anonymous function, they must be captured using a use clause. Local variables can either be captured by value or by reference.

If you want to capture by value (which means the value of the local variable $a is copied into the closure at the time it's defined):

function a() {
    $a = "a";
    $aa = function() use ($a) {
        var_dump($a);
    };
    $aa();
}
a();

If you want to capture by reference (which means inside the closure it directly references the variable $a):

function a() {
    $a = "a";
    $aa = function() use (&$a) {
        var_dump($a);
    };
    $aa();
}
a();

In this case both would be the same. But if you modified the variable after the closure was created, and before it is run, it would give a different result (the capture by value would still have the previous value).

Upvotes: 0

Vlad Preda
Vlad Preda

Reputation: 9920

Variable $t is the variable variable you can access with $$a. Note that inside the function a(), $t is never used, so should be what you want.

$t = 3;
$a = 't';
function a()
{
    global $a, $$a;
    $x = $$a;
    $b = function()  use ($x) {
        echo $x;
    };
    $b();
}
a();

The code prints 3.

Upvotes: 0

Leo Zhuang
Leo Zhuang

Reputation: 489

Maybe you can make a reference to the variable variables.

function a() {

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

    $ref = &$$a;

    $aa = function() use ($ref) {
        var_dump($ref);
    };
    $aa();
}
a();

which outputs: string(8) "variable"

Upvotes: 0

Alex Shesterov
Alex Shesterov

Reputation: 27565

The value is NULL because there is no global with name $a.

The following would print "global": 

$a = "global"; // global variable initialization

function a() {
    $a = "a";
    $aa = function() {
        global $a;
        var_dump($a);
    };
    $aa();
}
a();

Upvotes: 1

ahmed
ahmed

Reputation: 360

You can write this : 

function a() {
    $a = "a";
    function b() {
        global $a;
        var_dump($a);
    };
    $aa = "b";
    $aa();
}
a();

Upvotes: 0

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