What is function overloading and overriding in php?

Overloading is defining functions that have similar signatures, yet have different parameters. Overriding is only pertinent to derived classes, where the parent class has defined a method and the derived class wishes to override that method.

In PHP, you can only overload methods using the magic method __call.

An example of overriding:

<?php

class Foo {
   function myFoo() {
      return "Foo";
   }
}

class Bar extends Foo {
   function myFoo() {
      return "Bar";
   }
}

$foo = new Foo;
$bar = new Bar;
echo($foo->myFoo()); //"Foo"
echo($bar->myFoo()); //"Bar"
?>

Function overloading occurs when you define the same function name twice (or more) using different set of parameters. For example:

class Addition {
  function compute($first, $second) {
    return $first+$second;
  }

  function compute($first, $second, $third) {
    return $first+$second+$third;
  }
}

In the example above, the function compute is overloaded with two different parameter signatures. *This is not yet supported in PHP. An alternative is to use optional arguments:

class Addition {
  function compute($first, $second, $third = 0) {
    return $first+$second+$third;
  }
}

---

Function overriding occurs when you extend a class and rewrite a function which existed in the parent class:

class Substraction extends Addition {
  function compute($first, $second, $third = 0) {
    return $first-$second-$third;
  }
}

For example, compute overrides the behavior set forth in Addition.

Strictly speaking, there's no difference, since you cannot do either :)

Function overriding could have been done with a PHP extension like APD, but it's deprecated and afaik last version was unusable.

Function overloading in PHP cannot be done due to dynamic typing, ie, in PHP you don't "define" variables to be a particular type. Example:

$a=1;
$a='1';
$a=true;
$a=doSomething();

Each variable is of a different type, yet you can know the type before execution (see the 4th one). As a comparison, other languages use:

int a=1;
String s="1";
bool a=true;
something a=doSomething();

In the last example, you must forcefully set the variable's type (as an example, I used data type "something").

---

Another "issue" why function overloading is not possible in PHP: PHP has a function called func_get_args(), which returns an array of current arguments, now consider the following code:

function hello($a){
  print_r(func_get_args());
}

function hello($a,$a){
  print_r(func_get_args());
}

hello('a');
hello('a','b');

Considering both functions accept any amount of arguments, which one should the compiler choose?

---

Finally, I'd like to point out why the above replies are partially wrong; function overloading/overriding is NOT equal to method overloading/overriding.

Where a method is like a function but specific to a class, in which case, PHP does allow overriding in classes, but again no overloading, due to language semantics.

To conclude, languages like Javascript allow overriding (but again, no overloading), however they may also show the difference between overriding a user function and a method:

/// Function Overriding ///

function a(){
   alert('a');
}
a=function(){
   alert('b');
}

a(); // shows popup with 'b'


/// Method Overriding ///

var a={
  "a":function(){
    alert('a');
  }
}
a.a=function(){
   alert('b');
}

a.a(); // shows popup with 'b'

Overloading Example

class overload {
    public $name;
    public function __construct($agr) {
        $this->name = $agr;
    }
    public function __call($methodname, $agrument) {
         if($methodname == 'sum2') {

          if(count($agrument) == 2) {
              $this->sum($agrument[0], $agrument[1]);
          }
          if(count($agrument) == 3) {

              echo $this->sum1($agrument[0], $agrument[1], $agrument[2]);
          }
        }
    }
    public function sum($a, $b) {
        return $a + $b;
    }
    public function sum1($a,$b,$c) {

        return $a + $b + $c;
    }
}
$object = new overload('Sum');
echo $object->sum2(1,2,3);