When a key of 0 is used in my caeser cipher, my if statement returns a 1

Question

In my program I used a if statement to check if key = atoi(argv[1]) == 0, and if key is == 0, then my program returns a 1. However, it wasn't until I tested my program for corner cases that I realised (unrealisticly) what if the user enters a key of 0?

Now, because of the if statement, I am not sure how I can correct this so that a key of 0 can be entered and my program can still check for strings being inputted for the argument.

---

int main(int argc, string argv[]){

if (argc != 2){
    printf("ERROR! Please enter a positive number and try again.\n");
    return 1;
    }

int key = atoi(argv[1]);

if (key == 0){
    printf("ERROR! Please enter a positive number and try again.\n");
    return 1;
    }

Answer 1

I don't believe the assignment indicates that the key cannot be 0. I think it might mention that it would not scramble the message, but I believe your code is still supposed to handle a key of 0 (i.e., outputting a string identical to the input).

Also, sort of an aside, I believe generally when check50 checks to see if you handle an improper amount of arguments, it expects an output explaining the usage, like "Usage: ./caesar.c (what argv[1] should be) (what argv[2] should be) etc.." (without parentheses of course, and using square brackets around optional arguments).

Answer 2

Per the spec:

Incidentally, you can assume that the user will only type integers at the prompt. You don’t have to worry about them typing, say, foo, just to be difficult