[RESOLVED] About "function returns an alias"

simongcc

Hello! I am new here.
I have only little experience in programming. Please bear with me if I ask some stupid question.

For instance, I am reading some beginning guides with the following difficulty in understanding.
I cannot see the difference because I cannot understand the concept of values and alias. I can just see both of the result are same when output.
I would like to know more about the behind-scene meaning of the explanation below.

Thanks a lot!
Simon

guide wrote:
By default, values are copied out of the function. A function declared with an & before its name returns a reference (alias) to its return
value:

$names = array("Fred", "Barney", "Wilma", "Betty");
function & find_one($n) {
global $names;
return $names[$n];
}
$person =& find_one(1); // Barney
$person = "Barnetta"; // changes $names[1]

guide wrote:
In this code, the find_one( ) function returns an alias for $names[1], instead of a copy of its value. Because we assign by reference, $person is an alias for $names[1], and the second assignment changes the value in $names[1].

This technique is sometimes used to return large string or array values efficiently from a function. However, PHP's copy-on-write/shallow-copy mechanism usually means that returning a reference from a function is not necessary. There is no point in returning a reference to some large piece of data unless you know you are likely to change that data. The drawback of returning the
reference is that it is slower than returning the value and relying on the shallow-copy mechanism to ensure that a copy of that data is not
made unless it is changed.

laserlight

I cannot see the difference because I cannot understand the concept of values and alias.

A value is the contents of something. An alias is another name for something. So if you return by value, you return a copy of the contents. As such, changing the return value only changes your copy of the contents and thus does not affect the original. If you return by reference, the caller's variable becomes another name for what you returned. Thus changing the return value changes the value of the variable that existed in the function.

This is why you should not return local variables by reference: the caller would have another name for something that no longer exists since the lifetime of a local variable is until the end of the function.

simongcc

laserlight wrote:
This is why you should not return local variables by reference: the caller would have another name for something that no longer exists since the lifetime of a local variable is until the end of the function.

Thanks Lasterlight!
Now I understand more and have a picture of what is going on.
Although I am not sure when to apply this reference stuff. Maybe I experience more and writing more would help.

Simon