do_op

Subject

Assignment name  : do_op
Expected files   : *.c, *.h
Allowed functions: atoi, printf, write
--------------------------------------------------------------------------------

Write a program that takes three strings:
- The first and the third one are representations of base-10 signed integers
  that fit in an int.
- The second one is an arithmetic operator chosen from: + - * / %

The program must display the result of the requested arithmetic operation,
followed by a newline. If the number of parameters is not 3, the program
just displays a newline.

You can assume the string have no mistakes or extraneous characters. Negative
numbers, in input or output, will have one and only one leading '-'. The
result of the operation fits in an int.

Examples:

$> ./do_op "123" "*" 456 | cat -e
56088$
$> ./do_op "9828" "/" 234 | cat -e
42$
$> ./do_op "1" "+" "-43" | cat -e
-42$
$> ./do_op | cat -e
$

Commented solution

do_op

Check this page to understand the switch statement : (todo)

do_op.c

#include <stdio.h>
#include <stdlib.h>

int main(int ac, char *av[])
{
    /* checking the number of arguments
     */
    if (ac == 4)
    {
        /* we then switch over the operator (av[2][0])
         */
        switch(av[2][0])
        {
            /* if the operator is + we simply make an
             * addition and print the result to the screen
             */
            case '+':
                printf("%d", atoi(av[1]) + atoi(av[3]));
                break;
            /* if the operator is - we simply make a
             * substraction and print the result to the screen
             */
            case '-':
                printf("%d", atoi(av[1]) - atoi(av[3]));
                break;
            /* if the operator is * we simply make a
             * multiplication and print the result to the screen
             */
            case '*':
                printf("%d", atoi(av[1]) * atoi(av[3]));
                break;
            /* if the operator is / we simply make a
             * divisions and print the result to the screen
             */            
            case '/':
                printf("%d", atoi(av[1]) / atoi(av[3]));
                break;
            /* if the operator is % we simply make a
             * modulo and print the result to the screen
             */
            case '%':
                printf("%d", atoi(av[1]) % atoi(av[3]));
                break;
        }
    }
    /* finally, we write a newline
     */
    printf("\n");
}