CS50 Caesar segmentation fault - c

Question

I am new here and working on the second homework Caesar of cs50, it seems most of my review is correct except the last one -- I cannot handle the situation of lacking argv[1], which means if I only type ./caesar, it will return segmentation fault. I am wondering why this code if (argc != 2) cannot return 0 when argc == 1, however it works when argc > 1, I find that is weird. Can anyone help me?? Thanks in advance!
# include <stdio.h>
# include <cs50.h>
# include <string.h>
# include <ctype.h>
# include <math.h>
# include <stdlib.h>
int check_the_key(int argc, string y);
int main(int argc, string argv[])
{
string x = argv[1];
int y = argc;
int k = check_the_key(y, x);
if (k == 0)
{
printf("ERROR!!!!!\n");
return 1;
}
else
{
// printf("The key is %i\n", k);
string text = get_string("Input your text:");
int i;
int n;
printf("ciphertext: ");
for (i = 0, n = strlen(text); i < n; i++)
{
if (islower(text[i]))
{
printf("%c", (text[i] - 97 + k) % 26 + 97 );
}
else if (isupper(text[i]))
{
printf("%c", (text[i] - 65 + k) % 26 + 65);
}
else
{
printf("%c", text[i]);
}
}
printf("\n");
return 0;
}
}
int check_the_key(int argc, string y)
{
int number = argc;
string key = y;
int numberkey = atoi(key);
if (argc != 2)
{
return 0;
}
else
{
if (numberkey > 0)
{
return numberkey;
}
else
{
return 0;
}
}
}

Answer 1

I know what is going on! Because I need to pass some value into atoi(), if I only call ./caesar, there is no value I can pass into atoi(), so it causes segmentation fault. Which means I need to change code order slightly, put int numberkey = atoi(key); inside the else loop. So the code will run if (argc != 2) first, if no, then go to the next step! Here is the code after change.
int check_the_key(int argc, string y)
{
int number = argc;
string key = y;
if (argc != 2)
{
return 0;
}
else
{
int numberkey = atoi(key);
if (numberkey > 0)
{
return numberkey;
}
else
{
return 0;
}
}
}