Reference to static method in PHP?

Question

In PHP, I am able to use a normal function as a variable without problem, but I haven't figured out how to use a static method. Am I just missing the right syntax, or is this not possible?

(EDIT: the first suggested answer does not seem to work. I've extended my example to show the errors returned.)

function foo1($a,$b) { return $a/$b; }

class Bar
{
    static function foo2($a,$b) { return $a/$b; }

    public function UseReferences()
    {
        // WORKS FINE:
        $fn = foo1;
        print $fn(1,1);

        // WORKS FINE:
        print self::foo2(2,1);
        print Bar::foo2(3,1);

        // DOES NOT WORK ... error: Undefined class constant 'foo2'
        //$fn = self::foo2;
        //print $fn(4,1);

        // DOES NOT WORK ... error: Call to undefined function self::foo2()
        //$fn = 'self::foo2';
        //print $fn(5,1);

        // DOES NOT WORK ... error: Call to undefined function Bar::foo2()        
        //$fn = 'Bar::foo2';
        //print $fn(5,1);

     }
}

$x = new Bar();
$x->UseReferences();

(I am using PHP v5.2.6 -- does the answer change depending on version too?)

Peter Bailey · Accepted Answer · 2008-10-10 16:34:18Z

30

PHP handles callbacks as strings, not function pointers. The reason your first test works is because the PHP interpreter assumes foo1 as a string. If you have E_NOTICE level error enabled, you should see proof of that.

"Use of undefined constant foo1 - assumed 'foo1'"

You can't call static methods this way, unfortunately. The scope (class) is relevant so you need to use call_user_func instead.

<?php

function foo1($a,$b) { return $a/$b; }

class Bar
{
    public static function foo2($a,$b) { return $a/$b; }

    public function UseReferences()
    {
        $fn = 'foo1';
        echo $fn(6,3);

        $fn = array( 'self', 'foo2' );
        print call_user_func( $fn, 6, 2 );
     }
}

$b = new Bar;
$b->UseReferences();

edited Oct 10, 2008 at 16:34

answered Oct 10, 2008 at 16:22

Peter Bailey

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2 Comments

Eric Over a year ago

Thanks, call_user_func does the trick. I note that if the function is passed to a method in the other class then, not unnaturally, it needs to be scoped with 'Bar' instead of 'self'.

cjsimon Over a year ago

"PHP handles callbacks as strings, not function pointers." This is exactly what I was looking for

rewbs · Accepted Answer · 2008-10-10 18:32:55Z

10

In php 5.2, you can use a variable as the method name in a static call, but to use a variable as the class name, you'll have to use callbacks as described by BaileyP.

However, from php 5.3, you can use a variable as the class name in a static call. So:

class Bar
{
    public static function foo2($a,$b) { return $a/$b; }

    public function UseReferences()
    {
        $method = 'foo2';
        print Bar::$method(6,2); // works in php 5.2.6

        $class = 'Bar';
        print $class::$method(6,2); // works in php 5.3
     }
}

$b = new Bar;
$b->UseReferences();
?>

answered Oct 10, 2008 at 18:32

rewbs

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Comments

Jiangge Zhang · Accepted Answer · 2015-10-31 12:02:11Z

7

You could use the full name of static method, including the namespace.

<?php
    function foo($method)
    {
        return $method('argument');
    }

    foo('YourClass::staticMethod');
    foo('Namespace\YourClass::staticMethod');

The name array array('YourClass', 'staticMethod') is equal to it. But I think the string may be more clear for reading.

edited Oct 31, 2015 at 12:02

answered Jun 20, 2012 at 6:43

Jiangge Zhang

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1 Comment

Ray Foss Over a year ago

... Strange, it can't resolve it based on the "use" specification. I needed to pass a static method from a controller on the query_builder of a symfony type class entity type.

dwallace · Accepted Answer · 2010-08-20 05:05:38Z

1

In PHP 5.3.0, you could also do the following:

<?php

class Foo {
    static function Bar($a, $b) {
        if ($a == $b)
            return 0;

        return ($a < $b) ? -1 : 1;
    }
    function RBar($a, $b) {
        if ($a == $b)
            return 0;

        return ($a < $b) ? 1 : -1;
    }
}

$vals = array(3,2,6,4,1);
$cmpFunc = array('Foo', 'Bar');
usort($vals, $cmpFunc);

// This would also work:
$fooInstance = new Foo();
$cmpFunc = array('fooInstance', 'RBar');
// Or
// $cmpFunc = array('fooInstance', 'Bar');
usort($vals, $cmpFunc);

?>

answered Aug 20, 2010 at 5:05

dwallace

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Comments

Camilo Martin · Accepted Answer · 2013-12-18 04:02:43Z

1

Coming from a javascript background and being spoiled by it, I just coded this:

function staticFunctionReference($name)
{
    return function() use ($name)
    {
        $className = strstr($name, '::', true);
        if (class_exists(__NAMESPACE__."\\$className")) $name = __NAMESPACE__."\\$name";
        return call_user_func_array($name, func_get_args());
    };
}

To use it:

$foo = staticFunctionReference('Foo::bar');
$foo('some', 'parameters');

It's a function that returns a function that calls the function you wanted to call. Sounds fancy but as you can see in practice it's piece of cake.

Works with namespaces and the returned function should work just like the static method - parameters work the same.

answered Dec 18, 2013 at 4:02

Camilo Martin

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Comments

eddy · Accepted Answer · 2008-10-11 09:25:03Z

0

This seems to work for me:

<?php

class Foo{
    static function Calc($x,$y){
        return $x + $y;
    }

    public function Test(){
        $z = self::Calc(3,4);

        echo("z = ".$z);
    }
}

$foo = new Foo();
$foo->Test();

?>

answered Oct 11, 2008 at 9:25

eddy

2 Comments

Eric Over a year ago

You are just calling Calc(), not storing it in a variable first. Assigning functions to variables and passing them around is useful for writing general purpose algorithms, for example a sort function that accepts different comparison functions for ordering.

Mikko Rantalainen Jan 10 at 15:51

Note that if you expect any inherited class to overwrite the Calc() and you want to call the overridden method from Test() instead of verbatim function in class Foo, then you have to use syntax static::Calc(3,4) instead of self::.... The keyword self is a compile time reference to the class that contains the code, the keyword static refers to the current class (which may or may not be the inherited class).

hek2mgl · Accepted Answer · 2013-01-08 04:01:48Z

0

In addition to what was said you can also use PHP's reflection capabilities:

class Bar {

    public static function foo($foo, $bar) {
        return $foo . ' ' . $bar;
    }


    public function useReferences () {
        $method = new ReflectionMethod($this, 'foo');
        // Note NULL as the first argument for a static call
        $result = $method->invoke(NULL, '123', 'xyz');
    }

}

answered Jan 8, 2013 at 4:01

hek2mgl

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Comments

Joe · Accepted Answer · 2008-10-10 16:51:23Z

-3

"A member or method declared with static can not be accessed with a variable that is an instance of the object and cannot be re-defined in an extending class"

(http://theserverpages.com/php/manual/en/language.oop5.static.php)

answered Oct 10, 2008 at 16:51

Joe

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1 Comment

Eric Over a year ago

This means that you can't invoke the static method using "$this->", it doesn't mean that you can't pass the static method around in a (effectively function-valued) variable.

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