Reference assignment to a more complex variable

Question

Let's say I have an example like this one:

$foo = 'Hello ';
$bar = 1;

$abc =& $foo . $bar;

if (true) {
    ++$bar;

    if (true)
    {
        ++$bar;
    }
}

echo $abc;

I am expecting $abc to return Hello 3, but it actually returns Hello only. I'm really confused. Is there something I've gotten wrong with references in PHP?

A reference is a symbolic link to another variable - It can't keep track of multiple variables using a single reference. — nickb
– nickb, Commented Nov 11, 2013 at 21:52
No, $true is not defined, so that entire if block is skipped. — Samuel Cook
– Samuel Cook, Commented Nov 11, 2013 at 21:54

Nick Rolando · Accepted Answer · 2013-11-11 22:01:03Z

2

A reference variable is like an alias to the same object/variable, and it can only reference one variable at a time.

I'm not really sure how to help your situation because I don't know what you're trying to do, but..

$foo = 'Hello ';
$bar = 1;

$abc =& $bar;

++$bar;
++$bar;

echo $foo . $abc;

http://www.php.net/manual/en/language.references.whatdo.php

answered Nov 11, 2013 at 22:01

Nick Rolando

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1 Comment

aborted Over a year ago

This is what I was looking for. Thank you! My situation is much more complex and the suggested solutions wont work, but all I needed to know was how the reference assignment operator works like and you pointed it out.

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