Question

For clarification purposes I need the program to print the numbers that are input for a and b, not the actual letters a and b.
Okay here’s the revised program per yall’s suggestions:

int main (int argc, char *argv[])   {    int a; /*first number input*/      int b; /*second number input*/       a = atoi(argv[1]); /*assign to a*/      b = atoi(argv[2]); /*assign to b*/       if (a < b)         printf('%s\n', a < b); /* a is less than b*/         else {            printf('%s\n', a >= b); /* a is greater than or equal to b*/         }       if (a == b)         printf('%s\n', a == b);  /* a is equal to b*/         else {            printf('%s\n', a != b); /* a is not equal to b*/         }       return 0;   } /* end function main*/   

lol, now when I run the program I get told

 8 [main] a 2336 _cygtls::handle_exceptions: Error while dumping state    Segmentation fault  

What the heck does that mean? (If you haven’t noticed by now I am pretty hopeless at this stuff lol).

Answer 1

Based on your edit, I think you’re looking for this:

#include <stdio.h>  int main (int argc, char *argv[]) {     int a; /*first number input*/     int b; /*second number input*/      a = atoi(argv[1]); /*assign to a*/     b = atoi(argv[2]); /*assign to b*/      if (a < b)         printf('%d < %d\n', a, b); /* a is less than b*/     else         printf('%d >= %d\n', a, b); /* a is greater than or equal to b*/      if (a == b)         printf('%d == %d\n', a, b);  /* a is equal to b*/     else         printf('%d != %d\n', a, b); /* a is not equal to b*/      return 0; } 

This code:

wfarr@turing:~$ ./foo 1 2 1 < 2 1 != 2