Determining even/odd using a array_filter

Question

I am digging deeper into the array_filter() of php. I've understand the basic idea of it but fall into new problem.

$array = ['a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5];

$result = array_filter($array, function ($var){
              return $var & 1;
          });

If I dump the $result it looks like:

array:3 [
  "a" => 1
  "c" => 3
  "e" => 5
]

I want to know how return $var & 1; works behind the scene in the callback function of array_filter() .

Not sure what the actual question is here. Sounds like you are simply not aware what the & operator does? php.net/manual/en/language.operators.bitwise.php — C3roe
– C3roe, Commented May 5, 2020 at 12:44

Robo Robok · Accepted Answer · 2020-05-05 12:53:01Z

4

array_filter() keeps the values that produce truthy result for the callback and removes the values that don't.

This expression actually checks if the number is odd:

$number & 1

Why? because it performs bitwise AND operation with 1. So, odd numbers have 1 as their last digit in binary representation and even numbers have 0.

When you perform bitwise AND, every corresponding digit computes to 1 when both digits are 1 and 0 otherwise. So:

1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
// etc.

Now, you can apply AND operation:

    0001 = 1
AND 0001
  = 0001 = 1 (odd)

    0010 = 2
AND 0001
  = 0000 = 0 (even)

    1001 = 9
AND 0001
  = 0001 = 1 (odd)

I hope you get the idea.

edited May 5, 2020 at 12:53

answered May 5, 2020 at 12:42

Robo Robok

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5 Comments

Brain Foo Long Over a year ago

Hm... That answers the title but not the acutal question.

Robo Robok Over a year ago

I thought that % 1 was OP's trial to determine the odd.

Brain Foo Long Over a year ago

You answer "Piece of cake. You need a modulo operator:". The question was " I want to know how return $var & 1; works behind the scene in the callback function of array_filter() ." :)

Brain Foo Long Over a year ago

No problem. I just was nitpicking

Brain Foo Long Over a year ago

Awesome. Couldn't be better.

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