Arrays as Function Argument

[nunoperalta]

Hi, everyone! :)

I would like to know if an array, when given as function argument, is cloned (occupying more memory), or it is just referenced (but not modified as global variable).

I will explain:
$array = array(2, 3);
$var = $array; // this is reference, if I change $var, I will change $array

-------

And I am asking about this:

function checkArray($ar) {
return ($ar[1] > 0);
}

$array = array(1, 2, 3);
$var = checkArray($arr ay);

In this case, will $array be recreated/cloned to checkArray()?
I just want to know, because I have a function that may *SOMETIMES* need to check two arrays, and I have always to give those arrays as argument, but I don't want to waste memory when those arrays are not needed...

Thanks in advance for your help!

[Dormilich]

$array = array(2, 3);
$var = $array; // this is reference, if I change $var, I will change $array

I doubt that, references are usually made with the & operator

$var = &$array;

[nunoperalta]

I doubt that, references are usually made with the & operator

$var = &$array;

this is deprecated in PHP6. I am sure about what I said in that part. You can test yourself :) Thanks, anyway.

[Dormilich]

this is deprecated in PHP6. I am sure about what I said in that part. You can test yourself :) Thanks, anyway.

I tested i (though PHP 5.2.10) and yours didn't work.

and I'd like to know where this (about deprecation) is stated for data types other than objects (where this indeed is deprecated).

[nunoperalta]

I tested i (though PHP 5.2.10) and yours didn't work.

wow... how come I tested this long ago and it worked before!!!

http://www.nunoperalta.com/rubbish/test.php

It failed, really!

That's ok then, I will use reference operator in the function arguments, then, so the arrays will not be cloned, right? Will it work even if the given arguments are null?

Thanks!

and I'd like to know where this (about deprecation) is stated for data types other than objects (where this indeed is deprecated).

well, I read about $var = &new Object(); , and I thought it was for all references in this way... $var = & *something*;
Sorry :)

[nunoperalta]

Just tested again.
Only Objects are referenced automatically. Maybe I tested with an object, and thought it would be the same for Arrays.

I thought that, because other programming languages, like Python, do that.
Thanks again!