:( handles lack of key or argv[1] failed to execute program due to segmentation fault

Question

    #include <stdio.h>
    #include <cs50.h>
    #include <string.h>
    #include <ctype.h>
    #include <math.h>
    #include <stdlib.h>

    int main(int argc, string argv[])
    {
        string inpt = argv[1];
        int value = atoi(inpt);
        printf("%c\n", argc);
        if (argc != 2 || isalpha(inpt[0]))
        {
            printf("Usage: ./caesar key\n");
            return 1;
        }
        if (isalpha(inpt[1]))
        {
            printf("Usage: ./caesar key\n");
            return 1;
        }
        //printf("%s\n", inpt);
        string plaintext = get_string("plaintext: ");
        //printf("%c\n%i\n", inpt[1], value + 3);
        printf("ciphertext: ");
        for (int i = 0, n = strlen(plaintext); i < n; i++)
        {

            char x = plaintext[i];
            if (isalpha(x))
            {
                if (isupper(x))
               {
                    if ((int)x + value > 90)
                    {
                        int x1 = (int)x - 65;
                        int c = (x1 + value) % 26;
                        printf("%c", c + 65);
                    }
                    else
                    {
                        printf("%c", x + value);
                    }
                }
                 else if (islower(x))
                {
                    if ((int)x + value > 122)
                    {
                        int x2 = (int)x - 97;
                        int f = (x2 + value) % 26;
                        printf("%c", f + 97);
                    }
                    else
                    {
                        printf("%c", x + value);
                    }
                }
            }
            else
            {
                printf("%c", x);
            }
   
    }
        printf("\n");
    }

Answer 1

Does the code check to see that there is exactly one input parameter string BEFORE trying to do anything with argv[1]? Doesn't look like it.

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