I don't understand the line "if (index > 63)" and "bit_value = (n >> index) & 1;"

Question

#include "main.h"
/**
 * get_bit - returns the value of a bit at a given index
 * @n: the number
 * @index: the index of the bit we want to print
 * Return: value of the bit
 */
int get_bit(unsigned long int n, unsigned int index)
{
  int bit_value;

  if (index > 63)
    return (-1);

  bit_value = (n >> index) & 1;

  return (bit_value);
}

Answer 1

On your system unsigned long is a 64-bit value. The bit numbers (indexes) range from 0 - 63. If you are looking for an index greater-than 63 that is an error and -1 is returned. Otherwise, you shift the bits to the right (bits 0 - index - 1 fall off the right side (are shifted off)) leaving bits 63 - index. ANDing with 1 simply results in 1 or 0 depending on whether the bit at index is 1 or 0.

An Example

Let's take an example of the number 12 where bits at index 2 (the 4's bit) and 3 (the 8's bit) are both one (bits at indexes 4 - 63 are all 0, represented by .. below). You want to determine whether the bit at index 2 is 1 or 0 (on or off).

The binary representation of the number 12 is:

  ..1100

The index is 2. So shifting the value to the right by 2 (e.g. n >> index) results in the number 3, e.g.:

  ..0011

When you AND a number with 1 you are ANDing all bits with ..00000001 (showing only the final byte, all other bits are 0). The process of ANDing 3 and 1 is, and will result in:

  ..0011  AND
  ..0001
  ------
  ..0001

So in the number 12 the bit at index == 2 is 1. (if you try again with index == 1 or index == 4 the result is 0)

Answer 2

It returns -1 for all positions greater than 64, that's the first if() check.

Then it right shifts the number by given index, and checks whether the bit is set or not.

shifted_number = bit >> index;
check = shifted_number & 1;