/**
 * Example showing a use of arrays of function pointers
 * The core idea here is that by putting the function pointers into an array
 * you can cleverly avoid sets of cumbersome if statements
 * 
 * @author gtowell
 * Created: April 2021
 * **/
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#ifndef NULL
#define NULL (void *)0
#endif

// typedef for my function pointers
typedef int(*myfuns)(int, int);

/**
 * Apply the operation passed in as a function pointer to the 
 * other params
 * @param f -- a pointer to a function.  The function must return 
 *             an int and it must take two ints as param
 * @param a -- an int
 * @param b -- another int
 * @return the value that result from applying the passed function
 * to the two params
 * **/
int op(int (*f)(int, int), int a,int b) {
    return (*f)(a, b);
}

/**
 * Multiply two integers 
 * @param a an integer
 * @param b another integer
 * @return the product of the two integers
 * **/
int mult(int a, int b) {
    return a * b;
}

/**
 * Divide two integers but go UP if remainder > 0.
 * @param a an integer
 * @param b another integer
 * @return Divide two integers but go UP if remainder > 0.
 * **/
int divU(int a, int b) {
    return  (a / b)  + ((a % b == 0) ? 0 : 1);
}

/** 
 * Normal integer division
 * @param a an integer
 * @param b another integer
 * @return Divide two integers.
 * **/
int divV(int a, int b) {
    return  (a / b);
}

/**
 * get input from the user.  This input is assumed to be of the form
 * OP i i
 * where OP is ether m, d or u or q
 * and i is an integer
 * This program is fragile.
 * **/
int main(int argc, char const *argv[])
{
    myfuns funarr[128];
    funarr['m'] = *mult;
    funarr['*'] = mult;
    funarr['d'] = divV;
    funarr['u'] = divU;

    char lin[256];
    char op;
    int v1;
    int v2;
    while (fgets(lin, 255, stdin)) {
        sscanf(lin, "%c %d %d", &op, &v1, &v2);
        if (op=='q')
            break;
        printf("%d\n", (funarr[op])(v1, v2));
    }
    return 0;
}