Question

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

int main(int argc, char **argv)
{
        if(argc != 2)
                return 1;
        if(!atoi(argv[1]))
                printf("Error.");
        else printf("Success.");
        return 0;
}

My code works when I enter an argument that is either below or above the value of zero.

[griffin@localhost programming]$ ./testx 1
Success.
[griffin@localhost programming]$ ./testx -1
Success.
[griffin@localhost programming]$ ./testx 0
Error.

Why doesn’t it work?

Answer 1

It’s very simple, atoi returns the number converted which in your case is exactly 0 (as expected).

There is no standard method of checking whether the conversion actually succeeded or not when using atoi.

Since you are writing c++ you could get the same result with better error checking by using a std::istringstream, std::stoi (C++11) or strtol (which is a better interface when dealing with arbitrary numbers).

---

std::istringstream example

#include <sstream>

  ...

std::istringstream iss (argv[1]);
int res;

if (!(iss >> res))
  std::cerr << "error";

---

std::strtol example

#include <cstdlib>
#include <cstring>

  ...

char * end_ptr;

std::strtol (argv[1], &end_ptr, 10);

if ((end_ptr - argv[1]) != std::strlen (argv[1]))
  std::cerr << "error";

---

std::stoi (C++11)

#include <string>

  ...

int res;

try {
  res = std::stoi (argv[1]);

} catch (std::exception& e) {
  std::cerr << "error";
}