C code to count the number of '1' bits in an unsigned char

Question

I need C code to return the number of 1\'s in an unsigned char in C. I need an explanation as to why it works if it\'s not obvious. I\'ve found a lot of code for a 32-bit nu

Answer 1

an unsigned char is a "number" in just the same way that a 32-bit float or integer is a "number", what the compiler deems them to represent is what changes.

if you picture a char as its bits:

01010011 (8 bits);

you can count the set bits by doing the following:

take the value, lets say x, and take x % 2, the remainder will be either 1 or 0. that is, depending on the endianness of the char, the left or right most bit. accumulate the remainder in a separate variable (this will be the resulting number of set bits).

then >> (right shift) 1 bit.

repeat until 8 bits have been shifted.

the c code should be pretty simple to implement from my pseudocode, but basically

public static int CountSetBits(char c)
{
    int x = 0;
    int setBits = 0;
    while (x < 7)
    {
       setBits = setBits + c % 2;
       c = c >> 1;
       x = x + 1;
    }
}