Reputation: 23
I'm trying to check argument count, which should return 1
if there is no argument or more than 2 arguments in command line:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
int main(int argc, char *argv[])
{
char *key = argv[1]; //taking the key from CL
int n = strlen(key); //length of CL arg string
if (argc != 2) //checks the amount of CL args
{
printf("Usage: ./substitution key\n");
return 1;
}
for (int i = 0; i < n; i++) //validating the key array
{
if (n != 26)
{
printf("Key must contain 26 characters.\n");
return 1;
}
else if (!isalpha(key[i]))
{
printf("Key must contain only alphabetic character.\n");
return 1;
}
for (int j = i + 1; j < n; j++)
{
if (key[i] == key[j])
{
printf("Key must not contain repeated characters.\n");
return 1;
}
}
}
}
When i'm running without an argument it throws an error:
runtime error: null pointer passed as argument 1, which is declared to never be null Segmentation fault (core dumped).
If i'm commenting key validating part, it runs. I'm new in C programming and googled it a lot. Please, help me figure out why?
Upvotes: 1
Views: 1427
Reputation: 35560
You first access argv[1]
, which may not exist, then you check argc
to see if it exists. This is the wrong order. First check argc
, then access argv[1]
:
int main(int argc, char *argv[])
{
if (argc != 2) //checks the amount of CL args
{
printf("Usage: ./substitution key\n");
return 1;
}
char *key = argv[1]; //taking the key from CL
int n = strlen(key); //length of CL arg string
// remaining code...
Upvotes: 2