Decimal to Binary Conversion in C

Question

Implement the inspect_bits function that checks if given number contains 2 or more consecutive ones in its binary representation. If it does, the function should return 1. Otherwise, it should return 0.

For example, inspect_bits(13) should return 1 as it contains 2 consecutive ones in its binary representation (1101).

The program is working for certain numbers, 13 in this case, and a few others, however, the program is not working for '8','10' and for some other numbers.

#include <stdlib.h>
#include <stdio.h>

int inspect_bits(unsigned int number)
{
    int arr[32];
    int i=0,temp,c=0;;

    while(number > 0)
    {
        arr[i] = number%2;
        number = number/2;
        i++;
    }
    for (int j=i-1; j>=0; j--)
    {
        temp = arr[j];
            for (int k=j+1; k>0; k--)
            {
                if (temp==arr[k])
                c++;
            }
    }
    if (c != 0)
        return 1;
    else
        return 0;
}

#ifndef RunTests
int main ()
{
    printf("%d", inspect_bits(13));
}
#endif

Answer 1

That looks way too complicated, as an algorithm implementation and to debug. This is simpler:

int inspect_bits(unsigned int number)
{
   while (number)
   {
      if ((number & 3) == 3)
          return 1;
      number >>= 1;
   }
   return 0;
}

The loop tests the last two bits of number against 3, which is 00000011 in binary. If this matches, then there were (at least!) two consecutive bits in number . If not, the bits in number are shifted to the right and tested again. The loop can stop when number reaches 0 -- technically even while (number > 2) as only values higher than 2 have more than two consecutive bits.

Answer 2

To determine whether the unsigned integer n has two adjacent bits set, it is enough to test:

(n & (n >> 1)) != 0

There n >> 1 is the number n shifted one bit to the right, which is the same as an integer division by 2. You can visualize this method by numbering the bits:

n            7 6 5 4 3 2 1 0
n >> 1         7 6 5 4 3 2 1

The bitwise and ( & ) is a binary operator. Its result will have a bit set if both operands have the respective bit set. In the expression above, it will have bit 0 set if n has both bit 0 and 1 set; it will have bit 1 set if n has both bit 1 and 2 set and so on. It will have no bits set (and therefore be zero) if no two adjacent bits of n are set.