How to export error-handling to a function in C using <errno.h>

Question

I am checking if user enters the correct number and type of cmd arguments when calling main.

I thought it would be a great idea to write a function, which prints out some text, so I can reuse it when checking for NULL pointers. I included <errno.h>.

void errcall()
{
  perror("Error printed by perror()");
  exit(EXIT_FAILURE);
}

Then I wrote a function to check if arguments are right and sufficient.

void err_cmd_handle(int argc_input)
{
  if(argc_input==1 || argc_input>2)
    errcall(); 
}

When I call this in main, giving int argc as an argument to err_cmd_handle(), then I get a success, even when I did not give any arguments besides starting the program. Why does the condition fail to check correctly?

int main(int argc,char* argv[])
{
  err_cmd_handle(argc);
  return 0;
}

Answer 1

If you take your complete code with required headers:

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

void errcall()
{
  perror("Error printed by perror()");
  exit(EXIT_FAILURE);
}
void err_cmd_handle(int argc_input)
{
  if(argc_input==1 || argc_input>2)errcall(); 
}

int main(int argc,char* argv[])
{
    err_cmd_handle(argc);
    return 0;
}

and then run it with no arguments, the output is:

Error printed by perror(): Success 

Clearly the validation has worked because errcall() has been called. The text "Success" is simply because the value of errno is zero - because nothing has set it.

Your original code before you changed the question had:

if(argc_input==1 || argc_input>2)errcall();
  else exit(EXIT_SUCCESS); 

So when you stated:

then I get a success, even when I did not give any arguments besides starting the program

It was reasonable to assume that it was terminating via exit(EXIT_SUCCESS); - that is clearly not the case. It was also you I originally assumed that errno had nothing to do with the question because it seemed that errcall() cannot have been called if it returned EXIT_SUCCESS. Hopefully you see why your question caused so much confusion and comment?

Answer 2

Errno hasn't been set.

void errcall()
{
  printf("Value of errno: %d\n", errno);
  printf("Error: %s\n",strerror(errno));
  perror("Error printed by perror()");
  exit(EXIT_FAILURE);
}

int main(int argc,char* argv[])
{
    if(argc!=2) 
    {
       errno=EINVAL;
       errcall();
    }

    return 0;
}