Array pass by reference in php
You can pass a variable by reference to a function so the function can modify the variable. The syntax is as follows:
$a = 5 ;
foo ( $a );
// $a is 6 here
?>
Note: There is no reference sign on a function call — only on function definitions. Function definitions alone are enough to correctly pass the argument by reference.
- Variables, i.e. foo($a)
- References returned from functions, i.e.:
No other expressions should be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid:
foo ( $a = 5 ); // Expression, not variable
foo ( 5 ); // Produces fatal error
foo (new Foobar ()) // Produces a notice as of PHP 7.0.7
// Notice: Only variables should be passed by reference
?>
User Contributed Notes 18 notes
By removing the ability to include the reference sign on function calls where pass-by-reference is incurred (I.e., function definition uses &), the readability of the code suffers, as one has to look at the function definition to know if the variable being passed is by-ref or not (I.e., potential to be modified). If both function calls and function definitions require the reference sign (I.e., &), readability is improved, and it also lessens the potential of an inadvertent error in the code itself. Going full on fatal error in 5.4.0 now forces everyone to have less readable code. That is, does a function merely use the variable, or potentially modify it. now we have to find the function definition and physically look at it to know, whereas before we would know the intent immediately.
// Here we use the ‘use’ operator to create a variable within the scope of the function. Although it may seem that the newly created variable has something to do with ‘$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.
$x = 10 ;
(function() use ( $x ) $x = $x * $x ;
var_dump ( $x ); // 100
>)();
var_dump ( $x ); // 10
beware unset() destroys references
$x = ‘x’;
change( $x );
echo $x; // outputs «x» not «q23» —- remove the unset() and output is «q23» not «x»
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
Parameters passed by references can have default values.
You can find out if a variable was actually passed by using func_num_args():
function refault ( & $ref = ‘Do I have to be calculated?’ ) echo ‘NUM ARGS: ‘ . func_num_args (). «\n» ;
echo «ORI VALUE: < $ref >\n» ;
if( func_num_args () > 0 ) $ref = ‘Yes, expensive to calculate result: ‘ . sleep ( 1 );
else $ref = ‘No.’ ;
echo «NEW VALUE: < $ref >\n» ;
>
$result = ‘Do I have to be calculated?’ ;
refault ( $result );
echo «RESULT: < $result >\n» ;
// NUM ARGS: 1
// ORI VALUE: Do I have to be calculated?
// NEW VALUE: Yes, expensive to calculate result: 0
// RESULT: Yes, expensive to calculate result: 0
refault ();
// NUM ARGS: 0
// ORI VALUE: Do I have to be calculated?
// NEW VALUE: No.
?>
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.
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.
*/
print «
Define:\na = $a \nb = ' $b '
» ;
swap ( $a , $b );
print «
After swap(a,b):\na = ' $a '\nb = $b
» ;
function swap (& $arg1 , & $arg2 )
// Swap contents of indicated variables.
$w = $arg1 ; $arg1 = $arg2 ; $arg2 = $w ;
>
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’ ]);
>);
var_dump ( $data );
do_sort ( $data ); // does not affect value of $data
var_dump ( $data );
$data = do_sort ( $data );
var_dump ( $data );
I designed a class that can easily pass references.
#### Problem 1
function problem (& $value )
>
problem ( 1 ); // cannot be passed by reference
#### Problem 2
class problem2
static function __callStatic ( $name , & $arguments ) // cannot take arguments by reference
>
>
?>
My solution 👇
class Reference
function __construct (public mixed & $data )
>
function test ( $value )
$value = & Reference :: get ( $value ); // values OR reference
$value = «test- $value » ;
return $value ;
>
echo test ( 1 ), PHP_EOL ; // test-1
$val = 2 ;
echo test ( Reference :: create ( $val )), PHP_EOL ; // test-2
echo $val , PHP_EOL ; // test-2
#### Problem solving 2 ####
class TestCall
static function __callStatic ( $name , $arguments )
$value = & Reference :: get ( $arguments [ 0 ]);
$value = » $name — $value » ;
return $value ;
>
>
echo TestCall :: test ( 3 ), PHP_EOL ; // test-3
$val = 4 ;
echo TestCall :: test ( Reference :: create ( $val )), PHP_EOL ; // test-4
echo $val , PHP_EOL ; // test-4
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’ );
AddTo ( «indirect test» , «test» , $array1 );
AddTo ( «indirect POST test» , «test» );
echo «Array 1 » ;
print_r ( $array1 );
echo «_POST » ;
print_r ( $_POST );
?>
And this scripts output is :
Array 1 Array
(
[0] => test
[1] => test2
[indirect test] => test
)
_POST Array
(
[indirect POST test] => test
)
Of course that means you can only assign default reference to globals or super globals variables.
If you changed a reference variable with a new `Address`, the variable it originally pointed to won’t change.
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 :
$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 */ >;
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.
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:
tst ( $arr0 , & $arr1 ); // We are passing a reference to ‘$arr1’ in the call !
print_r ( $arr0 ); // Contains just the reference to $arr1
array_push ( $arr0 , 5 ); // we add another element to $arr0
array_push ( $arr1 , 18 ); // we add another element to $arr1 as well
print_r ( $arr1 );
print_r ( $arr0 ); // Changes in $arr1 are reflected in $arr0
function tst1 (& $arr , & $r ) <
// Both arguments ‘$arr’ and ‘$r» are declared to be passed by
// reference,
// again, in the function’s body, we use a reference to
// the ‘$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:
tst1 ( $arr0 , $arr1 ); // ‘tst1’ understands ‘$r’ is a reference to ‘$arr1’
print_r ( $arr0 ); // Contains just the reference to $arr1
array_push ( $arr0 , 5 ); // we add another element to $arr0
array_push ( $arr1 , 18 );
print_r ( $arr1 );
print_r ( $arr0 ); // Changes in $arr1 are reflected in $arr0
I hope this is somehow useful
php array assign by copying value or by reference? [duplicate]
I heard that PHP can select how to assign arrays, depends on array size. It can assign by copying value (as any scalar type) or by reference. PHP always assign array to variables by copying a value, as it says in manual. Or it can assign by reference. ?
Are you just wanting to know if it’s possible to assign an array by reference? Or do you want to know if there’s some way to set it up to do that automatically?
i want to know, that sometimes PHp engine can pass $a = array(1,2,3); $b = $a; by reference without ‘&’ sign.
3 Answers 3
Assigning arrays by reference is possible when assigning array variables to other variables:
// By value. $a = array(1,2,3); $b = $a; array_push($a, 5); print_r($b); // $b is not a reference to $a Array ( [0] => 1 [1] => 2 [2] => 3 ) // By Reference $a = array(1,2,3); $b = &$a; // Here we assign by reference array_push($a, 5); print_r($b); // $b is a reference to $a and gets the new value (3 => 5) Array ( [0] => 1 [1] => 2 [2] => 3 [3] => 5 ) You can’t assign something by reference unless you are referencing something that already exists. Equally, you can’t copy something that doesn’t exist.
. neither copies or references — it just creates a new array fills it with values, because the values were specified literally.
$x = 1; $y = 2; $z = 3; $a = array($x,$y,$z); . copies the values from $x , $y and $z into an array 1 . The variables used to initialise the array values still exist in their own right, and can be modified or destroyed without affecting the values in the array.
. creates an array of references to $x , $y and $z (notice the & ). If, after running this code, I modify $x — let’s say I give it a value of 4 — it will also modify the first value in the array. So when you use the array, $a[0] will now contain 4 .
See this section of the manual for more information of how reference work in PHP.
1 Depending on the types and values of the variables used as the array members, the copy operation may not happen at the time of the assignment even when assigned by-value. Internally PHP uses copy-on-write in as many situations as possible for reasons of performance and memory efficiency. However, in terms of the behaviour in the context of your code, you can treat it as a simple copy.