Is it possible to access outer local variable in PHP?

Question

Is it possible to access outer local varialbe in a PHP sub-function?

In below code, I want to access variable $l in inner function bar. Declaring $l as global $l in bar doesn't work.

function foo()
{
    $l = "xyz";

    function bar()
    {
        echo $l;
    }
    bar();
}
foo();

Answer 1

I don't think this is possible with PHP 5.2...

With PHP 5.3, though, you could probably use a Closure, to do just that...

Edit : took some time to remember the syntax, but here's what it would look like :

function foo()
{
    $l = "xyz";
    $bar = function () use ($l)
    {
        var_dump($l);
    };
    $bar();
}
foo();

And, running the script, you'd get :

$ php temp.php
string(3) "xyz"

A couple of note :

• You must put a ; after the function's declaration !
• You could use the variable by reference, with a & before it's name : use (& $l)

For more informations, as a reference, you can take a look at this page in the manual : Anonymous functions

Answer 2

In PHP 5.3 you must use use keyword.

$bar = function() use(&$l) {
};
$bar();

In PHP 5.2 and earlier this won't work. The syntax you've got isn't a closure, but definition of a global function.

function foo() { function bar() { } }

works the same as:

function foo() { include "file_with_function_bar.php"; }

If you execute function foo twice, PHP will complain that you've tried to re-define (global) function bar.

Answer 3

Just passing it along as an argument to your inner function isn't a solution?