Write a program that can be used at the command prompt as a calculating utility.
The usage of the program is shown below.
C> calc <switch> <n> <m>
Where n and m are two integer operands. switch can be any one of the arithmetic or comparison operators. If the arithmetic operator is supplied, the output should be the result of the operation. If the comparison operator is supplied then the output should be True or False.
Ans:
#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<stdlib.h>
int main(int ac, char *av[])
{
int i, a, b;
a = b = 0;
if (ac != 4)
{
puts("Wrong number of arguements");
printf("\n.Write like this : ");
printf("\"calc\" \"switch\"");
printf("\"number 1\" \"number 2\"\n");
exit(1);
}
for (i = 0; av[2][i] != '\0'; i++)
//Converting av[2][i] string into numeric and saving in "a".
{
a *= 10;
a += av[2][i] - 48;
}
for (i = 0; av[3][i] != '\0'; i++)
//Converting av[3][i] string into numeric and saving in "b".
{
b *= 10;
b += av[3][i] - 48;
}
switch (av[1][0])
{
case '+':
printf("%d %c %d = %d", a, av[1][0], b, a + b);
break;
case '-':
printf("%d %c %d = %d", a, av[1][0], b, a - b);
break;
case '/':
printf("%d %c %d = %.2f", a, av[1][0], b, float(a) / float(b));
break;
case '*':
printf("%d %c %d = %d", a, av[1][0], b, a * b);
break;
case '%':
printf("%d %c %d = %d", a, av[1][0], b, a % b);
break;
case '<':
if (a < b)
printf("%d %c %d is True.", a, av[1][0], b);
else
printf("%d %c %d is False.", a, av[1][0], b);
break;
case '>':
if (a > b)
printf("%d %c %d is True.", a, av[1][0], b);
else
printf("%d %c %d is False.", a, av[1][0], b);
break;
case '=':
if (a == b)
printf("%d %c %d is True.", a, av[1][0], b);
else
printf("%d %c %d is False.", a, av[1][0], b);
break;
default:
printf("Operator '%c' is not found.\n", av[1][0]);
}
_getch();
return 0;
}