Question

I´m trying to break a number into an array of numbers (in php) in the way that for example:

• 25 becomes (16, 8, 1)
• 8 becomes (8)
• 11 becomes (8, 2, 1)

I don´t know what the correct term is, but I think the idea is clear.

My solution with a loop is pretty straightforward:

   $number = rand(0, 128);        $number_array_loop = array();     $temp_number = $number;    while ($temp_number > 0) {        $found_number = pow(2, floor(log($temp_number, 2)));        $temp_number -= $found_number;             $number_array_loop[] = $found_number;    } 

I also have a recursive solution but I can´t get that to work without using a global variable (don´t want that), the following comes close but results in arrays in arrays:

   function get_numbers($rest_number) {             $found_number = pow(2, floor(log($rest_number, 2)));         if ($found_number > 0) {            $temp_array[] = get_numbers($rest_number - $found_number);            $temp_array[] = $found_number;        }             return $temp_array;    }     $number_array_recursive = array();    $number_array_recursive = get_numbers($number); 

However, using something like pow(floor(log())) seems a bit much for a simple problem like this.

It seems to me that the problem calls for a recursive solution with some very simple maths, but I just don´t see it.

Answer 1

You can check each bit of the input number with the following (untested) function.

function checkBit($var, $pos) {     return ($var & (1 << $pos)); } 

It checks the bit at position $pos in the variable $var by using a bitwise AND function. I’ll show you with 4-bit numbers for brevity.

• 1 = 0001
• 2 = 0010
• 4 = 0100
• 8 = 1000

If I want to check position 0 (the rightmost bit) of the number 3, I’d call the function like this:

$number = 3; checkBit($number, 0); 

Internally, checkBit is going to shift the constant 1 to the left 0 times because I passed in a 0. It’s then going to bitwise AND (&) the result with the number I passed in, 3. Since 3 = 0011 and 1 = 0001 the result is true, since the 0th bit is set in both arguments to the bitwise AND operator.