CODEWITHSUNDEEP
phpreferencepass-by-reference
Firsh - justifiedgrid.com
Firsh - justifiedgrid.com

Reputation: 5676

How can a referenced value stay in memory when the original variable that holds it has a new referenced value?

I found this awfully old comment in the PHP docs comments, but can't get my mind around it why it outputs "hihaha" and not "eita" in the first example. $a is changed and I'd assume that "hihaha" is removed for good. If not, then why is it so that if the change is to null or assigning a copy of another variable, then the "hihaha" IS removed for good?

// 1. example
$a = "hihaha";
$b = &$a;
$c = "eita";
$a = &$c; // why doesn't this purge "hihaha" from existence?
echo $b; // shows "hihaha" WHY?

// 2. example
$a = "hihaha";
$b = &$a;
$a = null;
echo $b; // shows nothing (both are set to null)

// 3. example
$a = "hihaha";
$b = &$a;
$c = "eita";
$a = $c;
echo $b; // shows "eita"

Is this a "good way" towards the circular references problem?

Upvotes: 1

Views: 35

Answers (2)

Don't Panic
Don't Panic

Reputation: 41820

Starting with $a = "hihaha";, when you do $b = &$a;, $b is not referencing $a. It is referencing the content of $a. As it says in PHP: What References Do:

$a and $b are completely equal here. $a is not pointing to $b or vice versa. $a and $b are pointing to the same place.

Then after $c = "eita";, when you do $a = &$c;, $a is now referencing the content of $c ("eita").

This does not affect $b at all. $b is still referencing the original content of $a ("hihaha"). Pointing $a at something else does not change that.

In case you have a more mspaint learning style, here is a visual aid representing the first four statements of example 1:

enter image description here

In the second example, $a and $b are still pointing at the same content when $a is set to null, so $b is now referencing null as well. Visually:

enter image description here

Upvotes: 2

Steve
Steve

Reputation: 1963

Think of a variable as pointing to a reference - to break down Example 1...

1

$a = "hihaha";

$a points to the reference for the string hihaha, lets call it R1

2

$b =& $a;

Here we are saying, point $b to the same reference as$a (R1)

3

$c = "eita";

$c points to the reference for the string eita, lets call it R2

4

$a =& $c;

Now we say, point $a to the same reference as $c ($b still points to R1)

At this stage,
$a and $c point to R2,
$b points to R1
- should be easy to guess what happens next!

5

echo $b; // hihaha

We now know that echoing $b will output R1!

Hope that helps!

Have a read of http://php.net/manual/en/language.references.whatdo.php

Upvotes: 0

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