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

Answer 1

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.

Answer 2

"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)

Answer 3

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();
?>

Answer 4

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');
    }

}

Answer 5

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();

Answer 6

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);

?>