[Resolved] Q: Passing references around, or ref to ref?

mflanagan

I have an object function that returns a reference to an array. I store that in a variable (with reference syntax). If I then use that var in an assignment, must I also use ref syntax? For example:

Class A {
    var $_v;

function A($val) { $this->_v = array($val => 0); }

function &GetV() {return $this->_v; }
}

$obj = & new A(50);

$ref = &$obj->GetV();

$subsequentRef = $ref;  // or: $subsequentRef = &$ref

Does the last statement (the assignment to $subsequentRef) need to use the '&'? I'm assuming (uh-oh!) that it does not. I guess I don't care if I get a copy of a reference. If I did use the '&', would I be getting a reference to a reference?

(BTW, I'm not proposing my code sample as an example of good programming practice, just as a sample to explicate my question. Returning a reference to an object variable can be dangerous. One can access those object vars from 'outside' the object, anyway, but still...)

Thanks.

swr

There is no such thing as a "reference to a reference" in PHP.

Analogy if you are familiar with C:

All variables in PHP are like pointers in C. When you say this in PHP:
$a = $b;
It is equivalent to saying this in C:
a = b;
When you say this in PHP:
$a =& $b;
It is equivalent to saying this in C:
a = b;

In non-C terms:

All variables in PHP point to a memory location. When you say "$a = $b;" the contents of $b's memory are copied to $a's memory. When you say "$a =& $b;" you cause $a to point to the same memory location that $b currently points to.

There is no way to cause $a to point to $b using references. You can say "$a = 'b'; echo $$a;" but that uses "variable variables", not references.

mflanagan

Thanks. If I understand you, then, I should say, in the example I gave:

$subsequentRef = $ref;

yes? Both vars would be references to the object's _v variable, correct?

As a follow-on, if my code created a reference to the array in the constructor, thusly:

Class A {
    var $_v;

function A($var) {
    $this->_v = &array($var => 0);
}

function GetV() {
    return $this->_v;
}
}

$obj = & new A(50);

$ref = $obj->GetV();
$subsequentRef = $ref

then GetV wouldn't be defined as returning a reference, and $ref, $subsequentRef, and $obj->_v all point to the array created in the constructor, yes?

ahundiak

I am not sure is swr answered your question or not. swr's c example is correct but a bit misleading. But as swr says, there is no such thing as a reference to a reference in php.

Just to clarify,
$subsequentRef = $ref;
Will in fact create a copy of $ref. So changes made to subsequentRef will not be reflected in ref.

One thing to understand is that a copy is not actually made until the variable actually get's changed. So it's ok to pass around read only copies of objects and arrays without worrying about consuming memory unnecessarily.

ahundiak

References can be confusing. Here is an example of what my explanation was trying to say.

<?php
$a = array(1 => 'OneA', 2=> 'TwoA', 3=> 'ThreeA');
$b =& $a; // Reference
$c =  $b; // Copy

$a[1] = 'OneAA';
$b[2] = 'TwoAB';
$c[3] = 'ThreeAC';

print_r($a);
print_r($b);
print_r($c);
?>

Array
(
[1] => OneAA
[2] => TwoAB
[3] => ThreeA
)
Array
(
[1] => OneAA
[2] => TwoAB
[3] => ThreeA
)
Array
(
[1] => OneA
[2] => TwoA
[3] => ThreeAC
)

mflanagan

I think I understand what you're saying.

Now, what if, in your example, I want $c to refer to $a, but the only thing I have available is $b (which is a reference to $a)?

What you're saying is that I cannot store a reference to $a into $c, unless I say '$c = &$a).

To say it another way, you're saying there is no way to create a reference to a variable unless you have the actual (original) variable available, yes?

So, in my previous post, the last statement in the post (that $ref and $subsequentRef both refer to the object's _v) is incorrect, yes?

ahundiak

$c =& $b;

swr

Originally posted by mflanagan
To say it another way, you're saying there is no way to create a reference to a variable unless you have the actual (original) variable available, yes?

You're thinking in terms of "reference to a variable" which is not how PHP references actually work.

$a = "foo"; // $a points to memory that contains a string
$b =& $a; // $b and $a are now both pointing to the same memory
$c =& $b; // $a, $b, and $c are all pointing to the same memory
$a = "bar!"; // The memory that $a/$b/$c point to now contains a different string
echo $c; // "bar!"

$d = "baz?";
$a =& $d; // $a now points to the same memory as $d (the string "baz?")
$b =& $d; // $a, $b, and $d are all pointing to the same memory
echo $c; // still says "bar!" even though both $a and $b have moved on

If you think $c is a reference to $b, you'd expect the second echo to say "baz?" because of the "$c =& $b;" followed by "$b =& $d;" but that's not how it works. The "$c =& $b;" does not cause $c to reference $b, but rather to reference the same memory that $b references. $b is later changed to reference different memory. $c is not dependent on $b, only an area of memory. $a, $b, and $c all shared the same area of memory but $a and $b went on to share $d's memory, leaving $c as the sole owner of the memory that $a first claimed.