This is the exercise that I’m doing:

Assignment name : print_bits

Expected files : print_bits.c
Allowed functions: write

Write a function that takes a byte, and prints it in binary WITHOUT A NEWLINE
AT THE END.
Your function must be declared as follows:

void print_bits(unsigned char octet);

Example, if you pass 2 to print_bits, it will print "00000010"

/////////////////////////////////////////////////////////////////////////////////////////////

This is a solution:

#include <unistd.h>

void    print_bits(unsigned char octet)
{
        int             i;
        unsigned char   bits;

        i = 8;
        while (i--)
        {
                bits = (octet >> i & 1) + '0';
                write(1, &bits, 1);
        }
}

I understand the solution, except for one part. If I right shift the char in its binary form, why do I need to perform a Bitwise AND with 1?? What would be the difference? I supposedly should get the same solution, since I’m performing the operation always with the 1 at the final bit of 1. If the bit from the unsigned char is 0, I’ll get 0. If it’s 1, I’ll get 1. But if I try to run the program with an example without the Bitwise AND operation, I get way beyond 0 and 1. Why?

>Solution :

The right-shift operation, denoted with >>, does not just take one bit from the left operand and shift it. It shifts all the bits of the left operand. For example, given 11011001₂ shifted right four bits, the result is 1101₂. If you do not AND that with 1, the result is 13 (1101₂). If you do AND it with 1, the result is 1.

In C implementations using ASCII, the character “0” has code 48, and, if you add 13 to '0', you will get 61, the code for “=”.