CODEWITHSUNDEEP
php
Kovo
Kovo

Reputation: 1717

PHP Passing by reference

Lets assume the following:

private $array = array(/*really big multi-dimensional array*/);

public function &func1($specific_large_sub_array_key)
{
     return $this->array[$specific_large_sub_array_key]
}

public function func2()
{
    $specificArray = &$this->func1(1);

    $this->func3($specificArray);
}


public function func3($specificArray)
{
    /* do stuff here*/
}

My question is this:

If func3 does not specify that $specificArray is not passed by reference to it, does PHP make a copy of $specificArray when it calls func3 inside of func2? Or does PHP keep the reference and propagate it automatically?

i.e. Will this...

public function func3($specificArray)
{
    unset($specificArray[234]);
}

...affect $array?

Thank you

Note, this example is extremely simplified.

Upvotes: 2

Views: 110

Answers (2)

drew010
drew010

Reputation: 69937

PHP is pretty intelligent as to how it deals with variables and copies.

Take the following example:

// Allocate one variable with content 'Hello'
$var     = 'Hello';

At this point, the Zend Engine has a representation of your string variable with the content, Hello.

Now if you do this:

$varCopy = $var;

You have 2 independent variables ($var and $varCopy), but since their contents are the same, the content only exists in one place in memory (basically a true copy hasn't been made yet). At this point, the two variables reference the same value (Hello) in a symbol table. It will only copy the contents once one of the two variables is modified. This same logic works for 2 copies to any number of copies.

Put simply, PHP is smart enough not to copy the value of the variable or array when it isn't necessary to make a copy.

You can learn more about this on the Reference Counting Basics page on the PHP manual. They even give an example specific to arrays towards the end.

A useful function is memory_get_usage which can show you how much memory PHP is using. You can use this to track the fact that the memory usage will change very little as you pass multiple copies of your array around. This can help prove the point outlined in the reference counting basics section of the manual.

You don't need to know all the details about how it works, but do be aware that PHP is smart in how it creates and manages references.

EDIT:

To answer your actual question directly, no, in func3 PHP will not make a copy of the array even if you don't pass it by reference. It will use references as illustrated in the reference counting basics section, so you can pass it by value without any concern.

If you call unset however, the value you unset will only be removed from the local copy of the array, so it ultimately isn't removed from the source array unless you pass it by reference to the function. But passing it by value does not create a whole new copy of the entire gigantic array. Even removing one value from the copy doesn't create a whole new copy minus the entry you removed (you just have a second array with all identical references to the first, but it is missing the one reference to the removed entry).

Upvotes: 5

hakre
hakre

Reputation: 197624

Can't do multiline comments, so as an answer:

return &$this->array[$specific_large_sub_array_key]
       ^

But to also give you an answer to your question:

i.e. Will this... [...] ...affect $array?

Plain and simple: No. Reason: It's a different variable, not an alias (reference).

Upvotes: 1

Related Questions