PHP - Static Local Variables

Question

So I recently ran across a code segment like this:

private static function FOO() {
        static $c = null;
        if ($c == null) {
            $c = new ParentClass_InnerClass();
        }
        return $c;
    }

So what is up with this code? How is this different from:

private static $C;

//other methods and fields    

private static function FOO() {
    if(!$self::c) {
        $self::c = new ParentClass_InnerClass();
    }
    return $self::c;
}

Or are they even the same concept?

The second example is technically not correct. It should be self::$c or static::$c. — ComFreek
– ComFreek, Commented Oct 24, 2013 at 15:17
Read more at PHP: Variable scope - Manual about static local variables — Simon Forsberg
– Simon Forsberg, Commented Oct 24, 2013 at 15:17
$c == null and !$c is not entirely the same thing, is it? — Simon Forsberg
– Simon Forsberg, Commented Oct 24, 2013 at 15:19
@SimonAndréForsberg How do you mean in this case? It's not like $c is ever going to be a boolean false. You are really nitpicking at a piece of code that is irrelevant to the question. — thatidiotguy
– thatidiotguy, Commented Oct 24, 2013 at 15:22

Elias Van Ootegem · Accepted Answer · 2013-10-24 15:23:54Z

9

They are, essentially, the same concept, though the scope is different:

class Foobar
{
    private static $c = null;
    public static function FOO()
    {
        if (self::$c === null)
        {
            self::$c = new stdClass;
        }
        return self::$c;
    }
    public static function checkC()
    {
        if (self::$c === null)
        {
            return false;
        }
        return self::$c;
    }
}

Foobar::checkC();//returns false, as $c is null
//function checkC has access to $c
$returned = Foobar::FOO();//returns object
if ($returned === Foobar::checkC())
{//will be true, both reference the same variable
    echo 'The same';
}

Whereas, if we were to change the code to:

class Foobar
{
    public static function FOO()
    {
        static $c = null;
        if ($c === null)
        {
            $c = new stdClass;
        }
        return $c;
    }
    public static function checkC()
    {
        if ($c === null)
        {
            return false;
        }
        return $c;
    }
}

We will get a notice when calling checkC: undefined variable. the static variable $c is only accessible from within the FOO scope. A private property is scoped within the entire class. That's it really.
But really, do yourself a favour: only use statics if you need them (they're shared between instances, too). And in PHP, it's very rare you'll actually need a static.

answered Oct 24, 2013 at 15:23

Elias Van Ootegem

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3 Comments

sale108 Over a year ago

Hi! I get the difference, but I have to ask how come it doesn't reset $c = null on every call to FOO()? Thanks in advance

Elias Van Ootegem Over a year ago

@sale108 Haven't used PHP in 5 years, but a variable in a function declared as static creates a zval in memory with whatever value you assign to it in the declaration/initialisation statement. Whatever happens to that zval next updated the same object in memory. Essentially, think of static $c = null; as something that is executed once, and then skipped in subsequent calls

sale108 Over a year ago

Thanks! That's enough for me to investigate on my own :)

BenMorel · Accepted Answer · 2013-10-24 15:31:26Z

7

They're the same concept, except that in the first code block, the static variable is local to the function, and such a static variable could be used in any function, even outside the context of a class:

function test() {
    static $counter = 0;
    return $counter++;
}

echo test(); // 0
echo test(); // 1

edited Oct 24, 2013 at 15:31

answered Oct 24, 2013 at 15:16

BenMorel

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3 Comments

thatidiotguy Over a year ago

How? Its a private function. You mean another public function could pass the variable out?

BenMorel Over a year ago

No, this kind of variable is local to the function and cannot be used outside of it. I mean that you could use a static variable in any function, even if it's not a class method.

BenMorel Over a year ago

I updated my answer with an example in a simple function, outside of a class.

Joren · Accepted Answer · 2013-10-24 15:16:26Z

1

In the top example you can access $c only from the local scope (FOO()) and in the bottom example you can access $c from the entire class, so not only form FOO().

answered Oct 24, 2013 at 15:16

Joren

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2 Comments

thatidiotguy Over a year ago

So its forcing the singleton pattern for a private variable, but since its forcing you to go through FOO() you are guaranteeing it is initialized?

Elias Van Ootegem Over a year ago

@thatidiotguy: yes, but forcing a class to be initialized can be done using the Factory pattern, too, aswel as by using a constructor, and the new keyword Don't use singletons

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