Örnekler
The result for
<?php
class Foo{
public function method(){
echo "call Method no paramater";
}
public function method($par){
echo "call Method has a paramater";
}
}
$foo = new Foo();
echo $foo->method("param");]
?>
can be achieved by
<?php
class Foo{
public function method($par = null){
echo "call Method no paramater";
}
}
$foo = new Foo();
echo $foo->method("param");]
?>
Php will give access to both $foo->method() and $foo->method("param")
Bir PHP sınıfının aşırı yüklenmesi
overload() işlevini kullanan basit bir örnek:
Örnek 1 - Bir PHP sınıfının aşırı yüklenmesi
<?php
class OO {
var $a = 111;
var $elem = array('b' => 9, 'c' => 42);
// Bir özelliği döndürmek için geriçağırım yöntemi
function __get($prop_name, &$prop_value)
{
if (isset($this->elem[$prop_name])) {
$prop_value = $this->elem[$prop_name];
return true;
} else {
return false;
}
}
// Bir özelliği tanımlamak için geriçağırım yöntemi
function __set($prop_name, $prop_value)
{
$this->elem[$prop_name] = $prop_value;
return true;
}
}
// Nesneyi burada aşırı yüklüyoruz
overload('OO');
$o = new OO;
echo "\$o->a: $o->a\n"; // print: $o->a: 111
echo "\$o->b: $o->b\n"; // print: $o->b: 9
echo "\$o->c: $o->c\n"; // print: $o->c: 42
echo "\$o->d: $o->d\n"; // print: $o->d:
// $elem dizisine yeni bir öğe ekleyelim
$o->x = 56;
// stdclass'ı örnekleyelim (PHP 4'te yerleşiktir)
// $val aşırı yüklenmez!
$val = new stdclass;
$val->prop = 555;
// içinde $val nesnesi olar bir diziyi "a"'ya atayalım'
// Fakat __set() bunu $elem dizisine koyacak
$o->a = array($val);
var_dump($o->a[0]->prop);
?>
add a note
User Contributed Notes
Bir PHP sınıfının aşırı yüklenmesi
Anonymous
26-Jun-2011 07:20
author at dereleased dot com
21-Dec-2009 05:30
While the term "overloading" in many other languages specifically means declaring multiple methods with the same name and having the compiler intelligently select which one should be run at any given time, this (as some have pointed out) is not the case in PHP, for a very simple yet important reason:
PHP is a weakly/dynamically typed language; the only exceptions to this rule are constants (which have certain type restrictions, e.g. array) and Type-Hinted method arguments. Because of this, traditional overloading style is simply not possible, as there is no way for the interpreter to know which method to call (reliably).
However, with PHP's ability to accept an indefinite number of undefined arguments via func_get_args() and related functions, as well as simply provide defaults to values to allow you to declare just one method and perform simple tests to see if some/all of the arguments were indeed passed, traditional means of method overloading are not entirely necessary.
It may not be what you're used to, but it works.
scurvysquid at yahoo dot com
03-Jun-2009 01:43
Overloading is defining a method multiple times with different input parameters, then conditionally running whichever method matches the provided input.
It looks like the "overload" function just enables magic methods for a class, which can be used in implementing an extremely janky simulation of overloading.
Also, it seems that magic methods are enabled by default in php5 regardless of whether the overload function has been run on a class.
Anonymous
26-Mar-2009 05:48
You can't redeclare because PHP won't know which one to use. Use a default value and test conditionally instead such as
<?php
public function method($arg=false){
if($arg==false){
//do something
} else {
//do something else
}
}
?>
ly dot sitthykun at yahoo dot com
21-Jan-2009 02:47
<?php
class Foo{
public function method(){
echo "call Method no paramater";
}
public function method($par){
echo "call Method has a paramater";
}
}
$foo = new Foo();
echo $foo->method("param");]
?>
#output:
Fatal error: Cannot redeclare Foo::method() in class_overload.php on line 6
why?
we can not create the same method name.