Are arrays in PHP copied as value or as reference to new variables and when passed to functions

Question

1  When an array is passed as an argument to a method or function  is it passed by reference  or by value   2  When assigning an array to a variable  is the new variable a reference to the original array  or is it new copy  What about doing this    a   array 1 2 3    b    a    Is  b a reference to  a

User · Answer

This thread is a bit older but here something I just came across:

Try this code:

$date = new DateTime();
$arr = ['date' => $date];

echo $date->format('Ymd') . '<br>';
mytest($arr);
echo $date->format('Ymd') . '<br>';

function mytest($params = []) {
    if (isset($params['date'])) {
        $params['date']->add(new DateInterval('P1D'));
    }
}

http://codepad.viper-7.com/gwPYMw

Note there is no amp for the $params parameter and still it changes the value of $arr['date']. This doesn't really match with all the other explanations here and what I thought until now.

If I clone the $params['date'] object, the 2nd outputted date stays the same. If I just set it to a string it doesn't effect the output either.

User · Answer

With regards to your first question  the array is passed by reference UNLESS it is modified within the method   function you re calling  If you attempt to modify the array within the method   function  a copy of it is made first  and then only the copy is modified  This makes it seem as if the array is passed by value when in actual fact it isn t   For example  in this first case  even though you aren t defining your function to accept  my array by reference  by using the  amp  character in the parameter definition   it still gets passed by reference  ie  you don t waste memory with an unnecessary copy    function handle array  my array                  read from but do not modify  my array     print r  my array                my array effectively passed by reference since no copy is made     However if you modify the array  a copy of it is made first  which uses more memory but leaves your original array unaffected    function handle array  my array                modify  my array      my array      New value                my array effectively passed by value since requires local copy     FYI - this is known as  lazy copy  or  copy-on-write

User · Answer

For the second part of your question  see the array page of the manual  which states  quoting        Array assignment always involves value   copying  Use the reference operator to   copy an array by reference    And the given example     lt  php  arr1   array 2  3    arr2    arr1   arr2     4      arr2 is changed                   arr1 is still array 2  3    arr3    amp  arr1   arr3     4     now  arr1 and  arr3 are the same   gt     For the first part  the best way to be sure is to try  -   Consider this example of code    function my func  a         a     30      arr   array 10  20   my func  arr   var dump  arr     It ll give this output    array   0   gt  int 10   1   gt  int 20   Which indicates the function has not modified the  outside  array that was passed as a parameter   it s passed as a copy  and not a reference   If you want it passed by reference  you ll have to modify the function  this way    function my func  amp   a         a     30      And the output will become    array   0   gt  int 10   1   gt  int 20   2   gt  int 30   As  this time  the array has been passed  by reference     Don t hesitate to read the References Explained section of the manual   it should answer some of your questions  -

User · Answer

When an array is passed to a method or function in PHP  it is passed by value unless you explicitly pass it by reference  like so   function test  amp  array         array  new      hey       a    array 1 2 3      prints  0  gt 1 1  gt 2 2  gt 3  var dump  a   test  a      prints  0  gt 1 1  gt 2 2  gt 3  new   gt  hey   var dump  a     In your second question   b is not a reference to  a  but a copy of  a     Much like the first example  you can reference  a by doing the following    a   array 1 2 3    b    amp  a     prints  0  gt 1 1  gt 2 2  gt 3  var dump  b    b  new      hey      prints  0  gt 1 1  gt 2 2  gt 3  new   gt  hey   var dump  a

User · Answer

In PHP arrays are passed to functions by value by default  unless you explicitly pass them by reference  as the following snippet shows    foo   array 11  22  33    function hello  fooarg       fooarg 0    99     function world  amp  fooarg       fooarg 0    66     hello  foo   var dump  foo       original array not modified  array passed-by-value  world  foo   var dump  foo       original array modified  array passed-by-reference   Here is the output   array 3       0   gt    int 11     1   gt    int 22     2   gt    int 33    array 3       0   gt    int 66     1   gt    int 22     2   gt    int 33

User · Answer

To extend one of the answers  also subarrays of multidimensional arrays are passed by value unless passed explicitely by reference    lt  php  foo   array  array 1 2 3   22  33    function hello  fooarg       fooarg 0  0    99     function world  amp  fooarg       fooarg 0  0    66     hello  foo   var dump  foo       original array not modified  array passed-by-value  world  foo   var dump  foo       original array modified  array passed-by-reference   The result is   array 3       0   gt    array 3         0   gt      int 1       1   gt      int 2       2   gt      int 3         1   gt    int 22     2   gt    int 33    array 3       0   gt    array 3         0   gt      int 66       1   gt      int 2       2   gt      int 3         1   gt    int 22     2   gt    int 33

User · Answer

By default   Primitives are passed by value   Unlikely to Java  string is primitive in PHP Arrays of primitives are passed by value Objects are passed by reference Arrays of objects are passed by value  the array  but each object is passed by reference    lt  php  obj new stdClass     obj- gt field  world     original array  obj     function example  hello         hello 0 - gt field  mundo      change will be applied in  original      hello 1  new stdClass       change will not be applied in  original          example  original    var dump  original      array 1     0   gt  object stdClass  1  1      field    gt  string 5   mundo          Note   As an optimization  every single value is passed as reference until its modified inside the function   If it s modified and the value was passed by reference then  it s copied and the copy is modified

User · Answer

TL DR  a  the method function only reads the array argument    implicit  internal  reference b  the method function modifies the array argument     value c  the method function array argument is explicitly marked as a reference  with an ampersand     explicit  user-land  reference   Or this  - non-ampersand array param  passed by reference  the writing operations alter a new copy of the array  copy which is created on the first write  - ampersand array param   passed by reference  the writing operations alter the original array     Remember - PHP does a value-copy the moment you write to the non-ampersand array param  That s what copy-on-write means  I d love to show you the C source of this behaviour  but it s scary in there  Better use xdebug debug zval     Pascal MARTIN was right  Kosta Kontos was even more so   Answer  It depends   Long version  I think I m writing this down for myself  I should have a blog or something     Whenever people talk of references  or pointers  for that matter   they usually end up in a logomachy  just look at this thread    PHP being a venerable language  I thought I should add up to the confusion  even though  this a summary of the above answers   Because  although two people can be right at the same time  you re better off just cracking their heads together into one answer   First off  you should know that you re not a pedant if you don t answer in a black-and-white manner  Things are more complicated than  yes no    As you will see  the whole by-value by-reference thing is very much related to what exactly are you doing with that array in your method function scope  reading it or modifying it   What does PHP says   aka  change-wise    The manual says this  emphasis mine         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   amp   to the   argument name in the function definition   As far as I can tell  when big  serious  honest-to-God programmers talk about references  they usually talk about altering the value of that reference  And that s exactly what the manual talks about  hey  if you want to CHANGE the value in a function  consider that PHP s doing  pass-by-value    There s another case that they don t mention  though  what if I don t change anything - just read  What if you pass an array to a method which doesn t explicitly marks a reference  and we don t change that array in the function scope  E g     lt  php function readAndDoStuffWithAnArray  array         return  array 0     array 1     array 2       x   array 1  2  3    echo readAndDoStuffWithAnArray  x     Read on  my fellow traveller   What does PHP actually do   aka  memory-wise    The same big and serious programmers  when they get even more serious  they talk about  memory optimizations  in regards to references  So does PHP  Because PHP is a dynamic  loosely typed language  that uses copy-on-write and reference counting  that s why     It wouldn t be ideal to pass HUGE arrays to various functions  and PHP to make copies of them  that s what  pass-by-value  does  after all     lt  php     filling an array with 10000 elements of int 1    let s say it grabs 3 mb from your RAM  x   array fill 0  10000  1        pass by value  right  RIGHT  function readArray  arr        lt -- a new symbol  variable  gets created here     echo count  arr      let s just read the array    readArray  x     Well now  if this actually was pass-by-value  we d have some 3mb  RAM gone  because there are two copies of that array  right     Wrong  As long as we don t change the  arr variable  that s a reference  memory-wise  You just don t see it  That s why PHP mentions user-land references when talking about  amp  someVar  to distinguish between internal and explicit  with ampersand  ones     Facts  So  when an array is passed as an argument to a method or function is it passed by reference   I came up with three  yeah  three  cases  a  the method function only reads the array argument b  the method function modifies the array argument c  the method function array argument is explicitly marked as a reference  with an ampersand        Firstly  let s see how much memory that array actually eats  run here     lt  php  start memory   memory get usage     x   array fill 0  10000  1   echo memory get usage   -  start memory     1331840   That many bytes  Great     a  the method function only reads the array argument  Now let s make a function which only reads the said array as an argument and we ll see how much memory the reading logic takes    lt  php  function printUsedMemory  arr          start memory   memory get usage         count  arr            read      x    arr 0           read    minor assignment       arr 0  -  arr 1      read      echo memory get usage   -  start memory     let s see the memory used whilst reading     x   array fill 0  10000  1      this is 1331840 bytes printUsedMemory  x     Wanna guess  I get 80  See for yourself  This is the part that the PHP manual omits  If the  arr param was actually passed-by-value  you d see something similar to 1331840 bytes  It seems that  arr behaves like a reference  doesn t it  That s because it is a references - an internal one     b  the method function modifies the array argument  Now  let s write to that param  instead of reading from it    lt  php  function printUsedMemory  arr         start memory   memory get usage          arr 0    1     WRITE       echo memory get usage   -  start memory     let s see the memory used whilst reading     x   array fill 0  10000  1   printUsedMemory  x     Again  see for yourself  but  for me  that s pretty close to being 1331840  So in this case  the array is actually being copied to  arr   c  the method function array argument is explicitly marked as a reference  with an ampersand   Now let s see how much memory a write operation to an explicit reference takes  run here  - note the ampersand in the function signature    lt  php  function printUsedMemory  amp  arr      lt ----- explicit  user-land  pass-by-reference        start memory   memory get usage          arr 0    1     WRITE       echo memory get usage   -  start memory     let s see the memory used whilst reading     x   array fill 0  10000  1   printUsedMemory  x     My bet is that you get 200 max  So this eats approximately as much memory as reading from a non-ampersand param

[php] Are arrays in PHP copied as value or as reference to new variables, and when passed to functions?

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