# callback function return return(\$var & 1)?

I have read the PHP Manuel about array_filter

``````<?php
function odd(\$var)
{
// returns whether the input integer is odd
return(\$var & 1);
}

function even(\$var)
{
// returns whether the input integer is even
return(!(\$var & 1));
}

\$array1 = array("a"=>1, "b"=>2, "c"=>3, "d"=>4, "e"=>5);
\$array2 = array(6, 7, 8, 9, 10, 11, 12);

echo "Odd :\n";
print_r(array_filter(\$array1, "odd"));
echo "Even:\n";
print_r(array_filter(\$array2, "even"));
?>
``````

Even I see the result here :

``````Odd :
Array
(
[a] => 1
[c] => 3
[e] => 5
)
Even:
Array
(
 => 6
 => 8
 => 10
 => 12
)
``````

But I did not understand about this line: `return(\$var & 1);` Could anyone explain me about this?

## Here is Solutions:

We have many solutions to this problem, But we recommend you to use the first solution because it is tested & true solution that will 100% work for you.

### Solution 1

You know `&&` is `AND`, but what you probably don’t know is `&` is a bit-wise `AND`.

The `&` operator works at a bit level, it is bit-wise. You need to think in terms of the binary representations of the operands.

e.g.

`710 & 210 = 1112 & 0102 = 0102 = 210`

For instance, the expression `\$var & 1` is used to test if the least significant bit is `1` or `0`, odd or even respectively.

`\$var & 1`

`010 & 110 = 0002 & 0012 = 0002 = 010 = false (even)`

`110 & 110 = 0012 & 0012 = 0012 = 110 = true  (odd)`

`210 & 110 = 0102 & 0012 = 0002 = 010 = false (even)`

`310 & 110 = 0112 & 0012 = 0012 = 110 = true  (odd)`

`410 & 210 = 1002 & 0012 = 0002 = 010 = false (even)`

`and so on...`

### Solution 2

``````&
``````

it’s the bitwise operator. It does the AND with the corrispondent bit of `\$var` and `1`

Basically it test the last bit of \$var to see if the number is even or odd

Example with \$var binary being 000110 and 1

``````000110 &
1
------
0
``````

0 (false) in this case is returned so the number is even, and your function returns false accordingly

### Solution 3

`\$var & 1` – is bitwise AND
it checks if `\$var` is ODD value

``````0 & 0 = 0,
0 & 1 = 0,
1 & 0 = 0,
1 & 1 = 1
``````

so, first callback function returns TRUE only if \$var is ODD, and second – vise versa (! – is logical NOT).

### Solution 4

It is performing a bitwise AND with \$var and 1. Since 1 only has the last bit set, `\$var & 1` will only be true if the last bit is set in \$var. And since even numbers never have the last bit set, if the AND is true the number must be odd.

### Solution 5

`&` is bitwise “and” operator. With 1, 3, 5 (and other odd numbers) `\$var & 1` will result in “1”, with 0, 2, 4 (and other even numbers) – in “0”.

### Solution 6

An odd number has its zeroth (least significant) bit set to `1`:

``````           v
0 = 00000000b
1 = 00000001b
2 = 00000010b
3 = 00000011b
^
``````

The expression `\$var & 1` performs a bitwise AND operation between \$var and 1 (`1 = 00000001b`). So
the expression will return:

• 1 when `\$var` has its zeroth bit set to 1 (odd number)
• 0 when `\$var` has its zeroth bit set to 0 (even number)

### Solution 7

& is a bitwise AND on \$var.

If \$var is a decimal 4, it’s a binary 100. 100 & 1 is 100, because the right most digit is a 0 in \$var – and 0 & 1 is 0, thus, 4 is even.

### Solution 8

it returns 0 or 1, depending on your \$var

if \$var is odd number, ex. (1, 3, 5 …) it \$var & 1 returns 1, otherwise (2, 4, 6) \$var & 1 returns 0

Note: Use and implement solution 1 because this method fully tested our system.
Thank you 🙂

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0