[Resolved] references (specifically returned function references)

section31

I understand the benefits of using references for variables every now and then, but sometimes I see code where people return references in their functions. ex.

<?php
function &foo()
{
   /* ...code... */
   return $bar;
}

$foo = &foo();

// output $foo later

?>

Now is this only for efficiency purposes, so lets say the functions doesn't return a copy of the variable and then have to destroy it....Instead it returns a reference? I'm just not understanding why they would do this.

drew010

this may help explain it a bit better.
http://php.net/manual/en/language.references.return.php

section31

Originally posted by drew010
this may help explain it a bit better.
http://php.net/manual/en/language.references.return.php

hehe, i read that before I posted this and no it doesn't help 🙁 The manual basically shows how to implement different types of references, not on how they will be beneficial.

drew010

it is a bit confusing, ill use an example from a comment on the page to try to explain it...

<?
//obscvresovl at NOSPAM dot hotmail dot com

$var = 1;
$num = NULL;

function &blah()
{
   $var =& $GLOBALS["var"]; # the same as global $var;
   $var++;
   return $var;
}

$num = &blah();

echo $num; # 2

blah();

echo $num; # 3

?>

normally when a function 'return's a value, a copy of the value is made and whatever variable is catching the assignment gets a copy of the value.
i.e.

function x() {
  $x = 5;
  return $x;  //return a copy of $x
}

$var = $x; //$var receives the copy

if you use a reference, what it does is not return a copy, but the variable receiving the return will become a reference to the function.
in relation to the first example this means that the memory address of $num, after the first call to blah() will be assigned as a reference to the return of the function. that way, when a value is returned from the reference, a copy is not made, but the value is returned to the memory address of $num.
thats why he can call blah() later with no lvalue needed. return $var is actually taking the memory address of $num and storing the contents there.

sort of hypothetically, returning a reference tells php to actually think about where to put the return value, whereas a regular return just blindly hands off a copy of the return value. when you make a reference php actually remembers where it was placed in the first place, so on subsequent calls it can then give the return value back to the same place (memory address).

i hope it made sense its kinda hard for me to explain, just one of the things you can see in your head but cant put to words easily.

section31

No, that makes perfect sense...somehow I flew by that example thinking that wasn't relating to what i was talking about it.

Thanks