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
(
    [0] => 6
    [2] => 8
    [4] => 10
    [6] => 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

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