Is php function pass by reference?

You can pass a variable by reference to a function so the function can modify the variable. The syntax is as follows:

function foo(&$var)
{
    $var++;
}$a=5;
foo($a);
// $a is 6 here
?>

Note: There is no reference sign on a function call - only on function definitions. Function definitions alone are enough to correctly pass the argument by reference.

The following things can be passed by reference:

• Variables, i.e. foo($a)

• References returned from functions, i.e.:

  function foo(&$var)
{
    $var++;
}
function &bar()
{
    $a = 5;
    return $a;
}
foo(bar());
?>

  See more about returning by reference.

No other expressions should be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid:

function foo(&$var)
{
    $var++;
}
function bar() // Note the missing &
{
    $a = 5;
    return $a;
}
foo(bar()); // Produces a noticefoo($a = 5); // Expression, not variable
foo(5); // Produces fatal errorclass Foobar
{
}foo(new Foobar()) // Produces a notice as of PHP 7.0.7
                  // Notice: Only variables should be passed by reference
?>

tnestved at yahoo dot com ¶

7 years ago

By removing the ability to include the reference sign on function calls where pass-by-reference is incurred (I.e., function definition uses &), the readability of the code suffers, as one has to look at the function definition to know if the variable being passed is by-ref or not (I.e., potential to be modified).  If both function calls and function definitions require the reference sign (I.e., &), readability is improved, and it also lessens the potential of an inadvertent error in the code itself.  Going full on fatal error in 5.4.0 now forces everyone to have less readable code.  That is, does a function merely use the variable, or potentially modify it...now we have to find the function definition and physically look at it to know, whereas before we would know the intent immediately.

ccb_bc at hotmail dot com ¶

3 years ago

// PHP >= 5.6

// Here we use the 'use' operator to create a variable within the scope of the function. Although it may seem that the newly created variable has something to do with '$x' that is outside the function, we are actually creating a '$x' variable within the function that has nothing to do with the '$x' variable outside the function. We are talking about the same names but different content locations in memory.

mike at eastghost dot com ¶

7 years ago

beware unset()  destroys references

$x = 'x';
change( $x );
echo $x; // outputs "x" not "q23"  ---- remove the unset() and output is "q23" not "x"

function change( & $x )
{
    unset( $x );
    $x = 'q23';
    return true;
}

tianyiw at vip dot qq dot com ¶

1 year ago

I designed a class that can easily pass references.

#### Problem 1
function problem(&$value)
{
}
problem(1); // cannot be passed by reference

#### Problem 2

yiangforwork at gmail dot com ¶

2 years ago

function set(&$arr) {
    $test = 10;
    $arr = &$test;
}

$arr = [1, 2, 3];
set( $arr );

var_dump( $arr );    // [1, 2, 3]

If you changed a reference variable with a new `Address`, the variable it originally pointed to won't change.

nickshanks at nickshanks dot com ¶

5 years ago

For anyone wondering, the copy-on-write behaviour just does the Right Thing™ when an array is passed to a function not by-ref which then passes it through to another function by-ref without writing to it. For example:

function do_sort(array $array) : array {
    usort($array, function ($a, $b) {
        return strnatcasecmp($a['name'], $b['name']);
    });

    return

Jason Steelman ¶

2 years ago

Within a class, passing array elements by reference which don't exist are added to the array as null. Compared to a normal function, this changes the behavior of the function from throwing an error to creating a new (null) entry in the referenced array with a new key.

class foo {
    public $arr = ['a' => 'apple', 'b' => 'banana'];
    public function normalFunction($key) {
        return $this->arr[$key];
    }
    public function &referenceReturningFunction($key) {
        return $this->arr[$key];
    }
}$bar = new foo();
$var = $bar->normalFunction('beer'); //Notice Error. Undefined index beer
$var = &$bar->referenceReturningFunction('beer'); // No error. The value of $bar is now null
var_dump($bar->arr);
/**
[
  "a" => "apple",
  "b" => "banana",
  "beer" => null,
],
*/?>
This is in no way a "bug" - the framework is performing as designed, but it took careful thought to figure out what was going on. PHP7.3

rob at librobert dot net ¶

9 months ago

A reference remains a reference, even if it's an element of an array that is not passed by reference.

An example:

$c = 3;
$array = array(
        'a' => 1,
        'b' => 2,
        'c' => &$c
);
var_dump($array);
function func($array) {
        $array['a'] = 4;
        $array['b'] = 5;
        $array['c'] = 6;
        var_dump($array);
}
func($array);
var_dump($array);
var_dump($c);
?>

This will output the following:

array(3) {
  ["a"]=>
  int(1)
  ["b"]=>
  int(2)
  ["c"]=>
  &int(3)
}
array(3) {
  ["a"]=>
  int(4)
  ["b"]=>
  int(5)
  ["c"]=>
  &int(6)
}
array(3) {
  ["a"]=>
  int(1)
  ["b"]=>
  int(2)
  ["c"]=>
  &int(6)
}
int(6)

You could use this to allow a function to have read-write access to part of the array, while limiting it to read-only access for the rest of the array:

$config = [
        'ro' => [
                // ...
        ],
        'rw' => [
                // ...
        ]
];
$copy = $config;
$copy['rw'] = &$config['rw'];
func($copy);
?>

phpnet at holodyn dot com ¶

8 years ago

The notes indicate that a function variable reference will receive a deprecated warning in the 5.3 series, however when calling the function via call_user_func the operation aborts without fatal error.

This is not a "bug" since it is not likely worth resolving, however should be noted in this documentation.

diabolos @t gmail dot com ¶

10 years ago

/*

  This function internally swaps the contents between
  two simple variables using 'passing by reference'.

  Some programming languages have such a swap function
  built in, but PHP seems to lack such a function.  So,
  one was created to fill the need.  It only handles
  simple, single variables, not arrays, but it is
  still a very handy tool to have.

  No value is actually returned by this function, but
  the contents of the indicated variables will be
  exchanged (swapped) after the call.
*/

// ------------------------------------------
// Demo call of the swap(...) function below.

Define:\na = $a\nb = '$b'

After swap(a,b):\na = '$a'\nb = $b

fdelizy at unfreeze dot net ¶

16 years ago

Some have noticed that reference parameters can not be assigned a default value. It's actually wrong, they can be assigned a value as the other variables, but can't have a "default reference value", for instance this code won't compile :

function use_reference( $someParam, &$param =& $POST )
{
...
}
?>

But this one will work :

function use_reference( $someParam, &$param = null )
?>

So here is a workaround to have a default value for reference parameters :

$array1 = array ( 'test', 'test2' );

function

fladnag at zerezo dot com ¶

5 years ago

Beware of using references with anonymous function and "use" keyword :

If you have a PHP version between 5.3 and < 5.3.10, "use" keyword break the reference :

function withRef(&$arg) {
  echo 'withRef - BEGIN - '.$arg."\n"; // 1
  $func = function() use($arg) { /* do nothing, just declare using $arg */ };
  $arg = 2;
  echo 'withRef - END - '.$arg."\n"; // 2
}$arg = 1;
echo 'withRef - BEFORE - '.$arg."\n"; // 1
withRef($arg);
// in PHP 5.3 < 5.3.10 : display 1
// in PHP 5.3 >= 5.3.10 : display 2
echo 'withRef - AFTER - '.$arg."\n";
?>

A workaround is to use a copy of the reference variable in "use" keyword :
  ...
  $arg2 = $arg;
  $func = function() use($arg2) { /* do nothing, just declare using $arg2 */ };

no at spam dot please ¶

7 years ago

agreed : this change produces less readable code.

additionally, it breaks many existing perfectly working codes which are not portable anymore and in some cases will require complex modifications

another issue regards the fatal error that is produced : how the hell am i supposed to do if i want to allow the user to use a value that is not even in a variable, or the return or a function call, or use call_user_func... this produces many occasions for a code to even break at run time

pillepop2003 at yahoo dot de ¶

17 years ago

PHP has a strange behavior when passing a part of an array by reference, that does not yet exist.

    function func(&$a)
    {
        // void();
    }$a['one'] =1;
    func($a['two']);
?>

var_dump($a) returns

    array(2) {
        ["one"]=>
        int(1)
        ["two"]=>
        NULL
    }

...which seems to be not intentional!

Sergio Santana: ssantana at tlaloc dot imta dot mx ¶

18 years ago

Sometimes we need functions for building or modifying arrays whose elements are to be references to other variables (arrays or objects for instance). In this example, I wrote two functions 'tst' and 'tst1' that perform this task. Note how the functions are written, and how they are used.

function tst(&$arr, $r) {
  // The argument '$arr' is declared to be passed by reference,
  // but '$r' is not;
  // however, in the function's body, we use a reference to
  // the '$r' argument array_push($arr, &$r);
  // Alternatively, this also could be $arr[] = &$r (in this case)
} $arr0 = array();          // an empty array
$arr1 = array(1,2,3);   // the array to be referenced in $arr0

// Note how we call the function:

obscvresovl at NOSPAM dot hotmail dot com ¶

17 years ago

Just a simple note...

$num

pallsopp at gmail dot com ¶

5 years ago

The comment by tnestved at yahoo dot com is incorrect as it is based purely on perception and not architecture. The method passing the object should not care whether it is by ref or by val, and nor should the reader.

If we are talking about readability and perception, then the receiving method needs to show that the object coming in is a reference, not an object instance, otherwise the reader is perplexed why the object is not returned.

Good functional headers alleviate all issues in this case.