The program is supposed to multiply two integers and print the result of the multiplication. If the program doesn't receive 2 arguments it should print Error and return 1.

I compiled it with < gcc -Wall -pedantic -Werror -Wextra -std=gnu89 3-mul.c -o mul>

when I run it with <./mul int int > or <./mul int int int > it's outputting correct result but when I run it with just <./mul > it's saying segmentation fault and I expect it to output error. My code is below. Thanks in advance

#include <stdio.h>
#include <stdlib.h>

int main (int argc, char *argv[])
{
  int x = atoi(argv[1]);
  int y = atoi(argv[2]);

if (argc == 3)
   {
       printf("%d\n", x * y);
   }
else
   { 
       printf("Error\n");
       return 1;
    }

return 0;
}

CodePudding user response：

You want this:

int main(int argc, char* argv[]) // char* argv[] here
{
  if (argc != 3)                // test argc *before* dereferencing argv[1] and argv[2]
  {
    printf("Error\n");
    return 1;
  }

  int x = atoi(argv[1]);
  int y = atoi(argv[2]);
  printf("%d\n", x * y);  
  return 0;
}

You are dereferencing argv[1] and argv[2] before testing if argc is 3. Instructions are executed sequencially. If you execute your program without arguments on the command line, argc will be 1 and argv[1] is out of bounds hence the seg fault.

Moreover the signature of main is wrong, it should be int main (int argc, char *argv[]). The second argument of main is an array of pointers to char.