Using argc and argv in Caesar cipher in C - c

Question

So, I am trying to use argc and argv in Caesars cipher in order to execute the program with just [./ key ;string] (e.g. ./ 13 Caesar). I have tried in lots of ways, although I must admit I am really lost here. I was thinking I should just use main(void) and ask the input with fgets, but I still have some curiosity in: How could I make it work with “int main(int argc, char *argv[])?”. Any clues you can give me?
Thank you for your help. Here is the code with the current outputs:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int main(int argc, char *argv[])
{
int key;
int result;
char str[60];
int k = 0;
printf("argc =%d\n", argc);
printf("argv =%d\n", argv);
printf("key =%d\n", key);
if (argc != 2)
{
printf("You didn't enter a key.\n");
return 1;
}
else
{
int k = key % 26;
printf("k =%d\n", k);
if (k == 0)
{
printf("Invalid key.\n");
return 1;
}
}
}
Output:
$ ./ceasar 13
argc =2
argv =-13216
key =0
k =0
Invalid key.
Edit: Tentative Answer
int main(int argc, char* argv[])
{
int i;
int key = atoi(argv[1]);
int result;
char str[60];
for (i = 0; i < argc; i++)
{
printf("argv[%d] = %s\n", i, argv[i]);
}
if (argc != 2)
{
printf("You didn't enter a key.\n");
return 1;
}
else
{
if (key == 0)
{
printf("Invalid key.\n");
return 1;
}
}
}

Answer 1

Printing argv with the "%d" specifier is undefined behavior, read printf()'s manual and use "%p".
On the other hand, if you want to print the string you should access the appropriate element with array notation. For example
printf("First Argument: %s\n", argv[1]);
this applies to all arguments, noting that argv[0] is the name of the executable as invoked in the command line.
You should also be careful before accessing argv[1] to check that argc > 2, and always check the current argument + !.