超越PHP - Exceed PHP: call_user_func

Description

mixed call_user_func ( callback function [, mixed parameter [, mixed ...]])

Call a user defined function given by the function parameter. Take the following:


<?php

function barber($type) 

{

    echo "You wanted a $type haircut, no problem";

}

call_user_func('barber', "mushroom");

call_user_func('barber', "shave");

?>

Object methods may also be invoked statically using this function by passing array($objectname, $methodname) to the function parameter.


<?php

class myclass {

    function say_hello() 

    {

        echo "Hello!\n";

    }

}



$classname = "myclass";



call_user_func(array($classname, 'say_hello'));

?>

注: Note that the parameters for call_user_func() are not passed by reference.

<?php function increment(&$var) { $var++; } $a = 0; call_user_func('increment', $a); echo $a; // 0 call_user_func_array('increment', array(&$a)); // You can use this instead echo $a; // 1 ?>

See also: is_callable(), call_user_func_array(), and information about the callback type

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call_user_func

andreyhristov at yahoo dot com
22-Aug-2000 09:09

Let say we construct somehow the name of the function. In some cases we can
use nested  'switch'  but the alternative is to to use for and parse a
string to construct  the  function  name, after than  we also can construct
and the 

parameter(s) for it. It  is like using of eval. I use it in 
one of my projects to construct a name of nested array
like

$ar['1999'['july']['29']['19'] , etc. the problem is in that I'm
parsing a string to construct the name. If I've {1:2:3}  the array name 
will be $ar['1']['2']['3'] but if i've

{1:2:3:4:5} it will  be
$ar['1']['2']['3']['4']['5'].  Eval is the only way (very hard because of
characters escaping is needed)  to do it. So this function is of the type
of EVAL().

USE IT wherever your script is  'self-modified';

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