/**
 * A very trivial example of a bit mask in use
 * Bit masks can make otherwise complex comparisons
 * over multiple values can be quite easy to implement
 * 
 * The core idea is that when you & the value together with the 
 * mask, you old get a non-zero number iff the value has a 
 * one in the same position as in the mask.
 * 
 * @author gtowell
 * Created: April 2021
 * **/

#include <stdio.h>

#define OPTA 1
#define OPTB 2
#define OPTC 4
#define OPTD 8

/**
 * Get input from user.
 * All use input is stored in bits within the single char 
 * that is retured.  Hence, the char should never be read
 * as a char
 * @return a char containing bits set according to use input
 * **/
char getUserSel() {
    char selectedOpts = 0;
    char c[100];
    while (1) {
        printf("Enter a letter: ");
        fgets(c, 99, stdin);
        switch (c[0]) {
            case 'a': case 'A':
                selectedOpts = selectedOpts | OPTA;
                break;
            case 'b': case 'B':
                selectedOpts |= OPTB;
                break;
            case 'c': case 'C':
                selectedOpts |= OPTC;
                break;
            case 'd': case 'D':
                selectedOpts |= OPTD;
                break;
            default:
                return selectedOpts;
            }
    }
}
int main(int argc, char const *argv[])
{
    char userSel = getUserSel();
    //printf("%d  %d   %d\n", userSel, (OPTA | OPTB), (userSel & (OPTA | OPTB)));
    // Two different bitmasks
    if (userSel & (OPTA | OPTB))
        printf("user selected A or B\n");
    if (userSel & (OPTA | OPTC | OPTD))
        printf("user selected A or C or D\n");
    return 0;
}