C program doesnt handle non numeric key - c

Question

Good evening people hope yall are doing well, i come to you all because i need some help regarding some C code i wrote, keep in mind that i just started coding in C so sorry if this question sounds dumb.
Basically I am doing the CS50 and we are writing a program that encripts messages so we first require the user for a command line argument which will be the key we will use to transform the plain text into cipher text. so basically if the user runs the command with lets say a 2 i.e: ./caesar 2 all the words he will type later will be "run" two sides.
my program works how is supposed and if the user types a letter instead of a number the program wont prompt for the users message. however, if the user runs the command line like this for example, ./caesar 8x the program will run eventhough the user typed a letter so any ideas on how to iterate thru the user command argument and if there is a letter the program should run? thanks!
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
#include <cs50.h>
int main(int argc, string argv[])
{
if (argc == 2 && isdigit(*argv[1]))
{
int key = atoi(argv[1]);
string plaintext = get_string("plaintext: ");
printf("ciphertext: ");
for (int i = 0, n = strlen(plaintext); i < n; i++)
{
if (isupper(plaintext[i]))
{
printf("%c", (((plaintext[i] + key) - 65) % 26) + 65);
}
else if (islower(plaintext[i]))
{
printf("%c", (((plaintext[i] + key) - 97) % 26) + 97);
}
else
{
printf("%c", plaintext[i]);
}
}
printf("\n");
return 0;
}
else if (argc == 1)
{
printf("NO KEY\n");
return 1;
}
else if (argc >= 3 || argv[1] == (string) argv[1])
{
printf ("Usage: ./caesar key\n");
return 1;
}
}
EDIT. SOLVED
this is how the code ended up looking, thanks #bruno for the help btw
#include <stdio.h>
#include <ctype.h>
#include <string.h>
#include <stdlib.h>
#include <cs50.h>
int main(int argc, string argv[])
{
if (argc == 2)
{
char *endp;
long lkey;
int errno = 0;
lkey = strtol(argv[1], &endp, 10);
if ((errno != 0) || (*argv[1] == 0) || (*endp != 0) || (lkey < 0) || (((int) lkey) != lkey) || argc >= 3)
// this statement check to see the characters on argv[1] are all digits
{
printf("Usage: ./caesar key\n");
}
else // if the argv[1] is all digits then prompts the user for a plaintext
{
int key = atoi(argv[1]); // converts the key into an integer
string plaintext = get_string("plaintext: ");
printf("ciphertext: ");
for (int i = 0, n = strlen(plaintext); i < n; i++)
// goes thru each of the chars in plaintext and determines if is uppercase, lowercase or none.
{
if (isupper(plaintext[i]))
{
printf("%c", (((plaintext[i] + key) - 65) % 26) + 65);
// it takes 65 and then sums it back to convert the character from the ASCII uppercase index and back
}
else if (islower(plaintext[i]))
{
printf("%c", (((plaintext[i] + key) - 97) % 26) + 97);
// if the case is lower it takes 97 and then adds 97 back just to maintain the ASCII index.
}
else // if it not a lower case nor an uppercase, which means is a symbol then leave it like that.
{
printf("%c", plaintext[i]);
}
}
printf("\n");
return 0;
}
}
else if (argc == 1) // if the user doesnt prompt a key print no key to user
{
printf("NO KEY\n");
return 1;
}
else if (argc >= 3) // if user prompts 3 or more keys into argv then prints error message regargind the usage
{
printf("Usage: ./caesar key\n");
return 1;
}
}

Answer 1

if the user runs the command line like this for example, ./caesar 8x the program will run eventhough the user typed a letter
you get the number from the program argument using atoi, it stops when a non-digit is reached, so the result is the same for 8 and 8x.
If the argument is not compatible with a number since the beginning, atoi will silently return 0, this is why atoi is dangerous. Anyway here you cannot be in that case thank to your test isdigit(*argv[1]) before.
any ideas on how to iterate thru the user command argument and if there is a letter the program should run
Probably you wanted to say the program should not run.
To check if all the argument is compatible with a number you can use strtol to do the conversion :
if (argc == 2) {
char * endp;
long lkey;
errno = 0;
lkey = strtol(argv[1], &endp, 10);
if ((errno != 0) || (*argv[1] == 0) || (*endp != 0) ||
(lkey < 0) || (((int) lkey) != lkey)) {
printf("invalid argument %s\n", argv[1]);
}
else {
int key = (int) lkey;
...
Because you currently use an int for the key I check if the value is compatible with an int, I also check it is positive because this is compatible with your use and in your version you verify the first character is a digit

Answer 2

Good evening,
I don't know if I understood your question right, but I will give it a try.
The problem is that you are checking if the args counter is 2, and the 2nd argument is a digit. if so, do the following code you have written. but if not?
You are not handling the case of argc==2 and argv[1] is not a digit.