Are PHP Variables passed by value or by reference?

03/26/2020 23:30:02

Are PHP variables passed by value or by reference?

Verified Answer (303 Votes)

08/04/2008 05:06:56

It's by value according to the PHP Documentation.

By default, function arguments are passed by value (so that if the value of the argument within the function is changed, it does not get changed outside of the function). To allow a function to modify its arguments, they must be passed by reference.

To have an argument to a function always passed by reference, prepend an ampersand (&) to the argument name in the function definition.

<?php
function add_some_extra(&$string)
{
    $string .= 'and something extra.';
}

$str = 'This is a string, ';
add_some_extra($str);
echo $str;    // outputs 'This is a string, and something extra.'
?>
303

Answer #2 (62 Votes)

03/14/2012 11:09:47

In PHP, By default objects are passed as reference copy to a new Object.

See this example.............

class X {
  var $abc = 10; 
}

class Y {

  var $abc = 20; 
  function changeValue($obj)
  {
   $obj->abc = 30;
  }
}

$x = new X();
$y = new Y();

echo $x->abc; //outputs 10
$y->changeValue($x);
echo $x->abc; //outputs 30

Now see this..............

class X {
  var $abc = 10; 
}

class Y {

  var $abc = 20; 
  function changeValue($obj)
  {
    $obj = new Y();
  }
}

$x = new X();
$y = new Y();

echo $x->abc; //outputs 10
$y->changeValue($x);
echo $x->abc; //outputs 10 not 20 same as java does.

Now see this ..............

class X {
  var $abc = 10; 
}

class Y {

  var $abc = 20; 
  function changeValue(&$obj)
  {
    $obj = new Y();
  }
}

$x = new X();
$y = new Y();

echo $x->abc; //outputs 10
$y->changeValue($x);
echo $x->abc; //outputs 20 not possible in java.

i hope you can understand this.

62

Answer #3 (60 Votes)

08/11/2008 08:09:30

It seems a lot of people get confused by the way objects are passed to functions and what pass by reference means. Object variables are still passed by value, its just the value that is passed in PHP5 is a reference handle. As proof:

<?php
class Holder {
    private $value;

    public function __construct($value) {
        $this->value = $value;
    }

    public function getValue() {
        return $this->value;
    }
}

function swap($x, $y) {
    $tmp = $x;
    $x = $y;
    $y = $tmp;
}

$a = new Holder('a');
$b = new Holder('b');
swap($a, $b);

echo $a->getValue() . ", " . $b->getValue() . "\n";

Outputs:

a, b

To pass by reference means we can modify the variables that are seen by the caller. Which clearly the code above does not do. We need to change the swap function to:

<?php
function swap(&$x, &$y) {
    $tmp = $x;
    $x = $y;
    $y = $tmp;
}

$a = new Holder('a');
$b = new Holder('b');
swap($a, $b);

echo $a->getValue() . ", " . $b->getValue() . "\n";

Outputs:

b, a

in order to pass by reference.

60
3
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