How to convert a command-line argument to int?

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Since this answer was somehow accepted and thus will appear at the top, although it’s not the best, I’ve improved it based on the other answers and the comments.

The C way; simplest, but will treat any invalid number as 0:

#include <cstdlib>

int x = atoi(argv[1]);

The C way with input checking:

#include <cstdlib>

errno = 0;
char *endptr;
long int x = strtol(argv[1], &endptr, 10);
if (endptr == argv[1]) {
  std::cerr << "Invalid number: " << argv[1] << '\n';
} else if (*endptr) {
  std::cerr << "Trailing characters after number: " << argv[1] << '\n';
} else if (errno == ERANGE) {
  std::cerr << "Number out of range: " << argv[1] << '\n';
}

The C++ iostreams way with input checking:

#include <sstream>

std::istringstream ss(argv[1]);
int x;
if (!(ss >> x)) {
  std::cerr << "Invalid number: " << argv[1] << '\n';
} else if (!ss.eof()) {
  std::cerr << "Trailing characters after number: " << argv[1] << '\n';
}

Alternative C++ way since C++11:

#include <stdexcept>
#include <string>

std::string arg = argv[1];
try {
  std::size_t pos;
  int x = std::stoi(arg, &pos);
  if (pos < arg.size()) {
    std::cerr << "Trailing characters after number: " << arg << '\n';
  }
} catch (std::invalid_argument const &ex) {
  std::cerr << "Invalid number: " << arg << '\n';
} catch (std::out_of_range const &ex) {
  std::cerr << "Number out of range: " << arg << '\n';
}

All four variants assume that argc >= 2. All accept leading whitespace; check isspace(argv[1][0]) if you don’t want that. All except atoi reject trailing whitespace.

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