Using * in main() parameters for command-line calculator - c

Question

I'm doing a homework problem which is to create a calculator using the argc and argv[] parameters in the main function.
What we were supposed to do, and I did, was to create a char pointer array containing the operators I want to use.
this is what I have:
const int OTabMax = 10;
const char *OTab[] = {
"+",
"-",
"*",
"/",
"quad",
"wurz",
"sinR",
"sinG",
"cosR",
"cosG"
};
int OTabCheck(char S[]) // char *S
{
int n;
for (n = 0; n < OTabMax; n++){
if (strcmp(S, OTab[n]) == 0) {break;}
}
return n;
}
It returns a number depending on which operator was used, which goes into a switch case where the calculation is done.
looks like this:
switch(key){
case 0:
break;
case 1:
break;
case 2:
x = atof(argv[1]);
y = atof(argv[3]);
z = 0;
z = x*y;
printf("%f * %f = %f\n",x,y,z);
break;
case 3-8;
break;
case 9:
break;
default: printf("Wrong input!\n"); exit(0); break;
}
The case in question is case 2: which whenever I do a calculation like calling the program through the terminal with ./"program_name" 2 * 2
it printf's 2.000000 * 0.000000 = 2.000000
here is the MRE:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
const int OTabMax = 10;
const char *OTab[] = {
"+",
"-",
"*",
"/",
"quad",
"wurz",
"sinR",
"sinG",
"cosR",
"cosG" };
int OTabCheck(char S[]) // char *S
{
int n;
for (n = 0; n < OTabMax; n++){
if (strcmp(S, OTab[n]) == 0) {break;}
}
return n;
}
int main(int argc,char *argv[]){
float x = 0;
float y = 0;
float z = 0;
int key = 0;
if(argc < 3){
exit(0);
}
if(argc == 4){
key = OTabCheck(argv[2]);
}else if(argc == 3){
key = OTabCheck(argv[1]);
}
switch(key){
case 2:
x = atof(argv[1]);
y = atof(argv[3]);
z = 0;
z = x*y;
printf("%f * %f = %f\n",x,y,z);
exit(0);
break;
default: printf("Wrong input!\n"); exit(0); break;
}
}