Question

This program is supposed to determine how many units are stored in the value of the variable c_val, if each unit is stored as a set bit.
My question is: why did the author write if (c % 2 == 1) count++; then shift c to the right with this statement c = c >> 1;?

#include <stdio.h>
#include <cstdlib>

int main(){
    unsigned char c_val;
    printf("char value = ");
    scanf("%c", &c_val);
    int count = 0;
    unsigned char c = c_val;
    while(c){
        if (c % 2 == 1) count++;
        c = c >> 1;
    }
    printf("%d bits are set", count);
    system("pause");
}

Answer 1

The data size of type char is always one byte – no exceptions. This code, however, calculates the popcount – that is, the number of 1 bits – in c_val.

We can translate the relevant code from

while (c) {
    if (c % 2 == 1) count++;
    c = c >> 1;
}

to

while (c != 0) {
    if (c & 0x1 == 1) count++; /* if the rightmost bit of c is 1, then count++ */
    c = c / 2;
}

The last change I made works because right-shifting an unsigned integral data type (in this case, unsigned char) is equivalent to dividing by 2, with round-toward-zero semantics.

We can think of c as a conveyor belt of bits – zero bits come in from the left, and one bit falls off the right on each loop iteration. If the rightmost bit is a 1, we increase the count by 1, and otherwise the count remains unchanged. So, once c is filled with zero bits, we know that we have counted all the one bits, and exactly the one bits, so count contains the number of one bits in c_val.