PHP: Accessing outer static variables

Question

Class A has a member with a function:

class A {
    public function foo() {
        static $x=2;
        function bar() {
            echo "x=$x";
        }
        bar();
    }
}
$a = new A();
$a->foo();

This gives an error in line echo "x=$x";.

How to access $x from class member foo() in function bar()?

I'm using PHP 5.5.9.

No, that results in FATAL ERROR Cannot access self:: when no class scope is active.... $x is no static class member! — WeSee
– WeSee, Commented Oct 3, 2017 at 21:55
$x is out of scope for bar(). There's several options, here's one of them. — ishegg
– ishegg, Commented Oct 3, 2017 at 21:57

Andrew · Accepted Answer · 2017-10-03 22:00:33Z

2

Don't try to declare your static variables within a function.

class A {
    protected static $x = 2;

    public function foo() {
        function bar() {
            echo "x=" . self::$x;
        }

        bar();
    }
}

$a = new A();
$a->foo();

answered Oct 3, 2017 at 22:00

Andrew

20.4k13 gold badges110 silver badges122 bronze badges

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

13 Comments

aknosis Over a year ago

Given context it is hard to say if this is the right answer or not. But I would imagine that this scenario (static class property) is more likely what OP wanted to use.

WeSee Over a year ago

Why not declare static variables inside functions?

WeSee Over a year ago

A solution: php.net/manual/de/language.variables.scope.php#98811 But a bad one!

ishegg Over a year ago

This doesn't work. bar() is in the global scope, it can't see $x: Fatal error: Cannot access self:: when no class scope is active in /Applications/MAMP/htdocs/trello/up.php on line 8

WeSee Over a year ago

I also thought of using the use keyword for function bar(): function bar() use ($x) { .... But use is only available for closures :-( Do we have to wait for a future PHP release?

|

WeSee · Accepted Answer · 2017-10-17 07:53:37Z

Finally found a solution, based on the closure idea of @ishegg:

class A {
    public function foo() {
        static $x=2;
        $bar = function () use (&$x,&$bar) {
            echo "x=$x\n";
            if (++$x<6) $bar();
        };
        $bar();
    }
}
$a = new A();
$a->foo();

This recursion is possible because the closure is passed to itself as reference. And since a reference is just a pointer it can be evaluated before the function is fully defined. So the solution effectively relies on closures passed by reference.

Please note that $x has also be passed to as reference &$x since otherwise we just get a copy of $x and the static variable $x never changes.

Runs like a charm: https://3v4l.org/U293s

Result:

x=2 
x=3 
x=4 
x=5

Collectives™ on Stack Overflow

PHP: Accessing outer static variables

2 Answers 2

13 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

13 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related