Difference between "=" and "&="

[1234homie]

Like in topic title.. What's the difference between using "=" and "&="?

Example:

foreach ($ar as $k => $v)
{
$row &= $rowset[$k];
[...]
}

foreach ($ar as $k => $v)
{
$row = $rowset[$k];
[...]
}

Any help will be appreciated ;d

[Brf]

&= is "assign by reference".

In that case, $row is actually pointing to the same location in memory as $rowset[$k] instead of making a copy of it.

[imkingdavid]

Brf is correct. Here's a little more indepth explanation.

The way I understand it, every variable points to a location within the memory. Let's say you have the following script:

Code: Select all

<?php

$v1 = 'foo';
$v2 = $v1;
$v1 = 'bar';

echo $v2;
// Output: foo
?>

What that did was assign 'foo' to $v1 and then copied the value of $v1 into $v2. However, $v2 points to a separate location in memory. When you change $v1, $v2 remains the same.

Now look at this:

Code: Select all

<?php

$v1 = 'foo';
$v2 &= $v1;
$v1 = 'bar';

echo $v2; // Output: bar
?>

This time, instead of making a new memory location for $v2, we assigned $v2 to point to the SAME location as $v1. In other words, $v1 and $v2 are the same thing, just with different names. Anything done to one is done to the other.

[1234homie]

ok, now situation is being more clearly. thanks for help.

btw. something like $row =& $rowset[$k] will work in the same way?

[Oleg]

References in php are tricky to get right and are best avoided unless you are very sure you need them.

[AmigoJack]

Sorry, you all mixed it up.

1. $a &= $b is the shortcut to $a= $a& $b and as such a bitwise operator.
2. $a =& $b is confusing and better written as $a= &$b, as these are two operators - the whole operation as such is called reference assignment.

If you look up /viewtopic.php you'll only find $row [color=#00F]=&[/color] $rowset[$post_list[$i]];, not $row [color=#00F]&=[/color] $rowset[$post_list[$i]];.

[Brf]

Aha. You are correct. Putting an operator before the "=" is the shortcut syntax.

$a &= $b;
is the same as
$a = ($a & $b);

I was confusing it with the =& when I said it was the ref operator.

[AmigoJack]

the ref operator

...does not exist. While in these cases:

Code: Select all

$a+= 1;  // Increment by one
$b&= 0xFF;  // Only affect rightmost 8bit
$c/= 3;  // Divide by three    

...we have one operator per line, the expression =& is a combination of the assignment operator and a reference treatment. References have nothing to do with operators - it's as faulty as talking about "if-loop"s.

[imkingdavid]

Sorry, you all mixed it up.

Not the first time, either. Thanks for pointing it out.

[Brf]

pot-a-to pot-ah-to