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Php passed by value or by reference

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php

As my understanding , when i passed array by value , a copy of array is created. i.e in below program $y & $z should need same memory as of $x. however memory utilization hardly increases. Obviousy my understanding is wrong , Can anyone explain the reason.

for($i=0;$i<1000000;$i++)

        $x[] = $i; // memory usage : 76519792


echo memory_get_usage(); 

function abc($y){

    $y[1] = 1; //memory usage  : 76519948 
    $z[]= $y;   //memory usage : 76520308

}
like image 310
chicharito Avatar asked Oct 22 '22 13:10

chicharito


1 Answers

I heard that php uses copy-on-write: http://en.wikipedia.org/wiki/Copy-on-write

as an example:

<?
for($i=0;$i<100000;$i++)
    $x[] = $i;

// we output the memory use:
echo memory_get_usage().'<br/>';  // outputs 14521040

// here we equate $y to $x, but instead of creating a copy, 
// php engine just creates a pointer to the same memory space
$y = $x;

echo memory_get_usage().'<br/>';  // outputs 14521128

// here we change something in y, now php engine 
// "creates a seperate copy" for y and makes the change
$y[1]=8;

echo memory_get_usage().'<br/>';  // outputs 23569904

?>

and similar behaviour with the function calls:

<?
for($i=0;$i<100000;$i++)
    $x[] = $i;

echo memory_get_usage().'<br/>'; /* 14524968 */

function abc($y){
    echo memory_get_usage().'<br/>'; /* 14524968 */
    $y[1] = 1;
    echo memory_get_usage().'<br/>'; /* 23573752 */
    $z[]= $y;  
    echo memory_get_usage().'<br/>'; /* 23574040 */

}
abc($x);
echo memory_get_usage().'<br/>'; /* 14524968 */
?>

PS: I am testing this on windows, maybe it is different on linux

like image 187
daghan Avatar answered Nov 03 '22 02:11

daghan