How to check std::string if its indeed an integer?

Question

The following code converts an std::string to int and the problem lies with the fact that it cannot discern from a true integer or just a random string. Is there a systematic method for dealing with such a problem?

#include <cstring>
#include <iostream>
#include <sstream>

int main()
{
    std::string str =  "H";

    int int_value;
    std::istringstream ss(str);
    ss >> int_value;

    std::cout<<int_value<<std::endl;

    return 0;
}

EDIT: This is the solution that I liked because it is very minimal and elegant! It doesn't work for negative numbers but I only needed positive ones anyways.

#include <cstring>
#include <iostream>
#include <sstream>

int main()
{
    std::string str =  "2147483647";

    int int_value;
    std::istringstream ss(str);

    if (ss >> int_value)
        std::cout << "Hooray!" << std::endl;

    std::cout<<int_value<<std::endl;


    str =  "-2147483648";
    std::istringstream negative_ss(str);

    if (ss >> int_value)
        std::cout << "Hooray!" << std::endl;

    std::cout<<int_value<<std::endl;

    return 0;
}

Answer 1

WhozCraig's approach is much nicer and I wanted to expand on it using the approach that the C++ FAQ uses which is as follows:

#include <iostream>
#include <sstream>
#include <string>
#include <stdexcept>

class BadConversion : public std::runtime_error {
public:
  BadConversion(std::string const& s)
    : std::runtime_error(s)
    { }
};



inline int convertToInt(std::string const& s,
                              bool failIfLeftoverChars = true)
{
  std::istringstream i(s);
  int x;
  char c;
  if (!(i >> x) || (failIfLeftoverChars && i.get(c)))
    throw BadConversion("convertToInt(\"" + s + "\")");
  return x;
}


int main()
{
    std::cout << convertToInt( "100" ) << std::endl ;
    std::cout << convertToInt( "-100" ) << std::endl ;
    std::cout << convertToInt( "  -100" ) << std::endl ;
    std::cout << convertToInt( "  -100  ", false ) << std::endl ;

    // The next two will fail
    std::cout << convertToInt( "  -100  ", true ) << std::endl ;
    std::cout << convertToInt( "H" ) << std::endl ;
}

This is robust and will know if the conversion fails, you also can optionally choose to fail on left over characters.

Answer 2

You may try to use Boost lexical_cast, it will throw an exception if the cast failed.

int number;
try
{
     number = boost::lexical_cast<int>(str);
}
catch(boost::bad_lexical_cast& e)
{
    std::cout << str << "isn't an integer number" << std::endl;
}

EDIT Accorinding to @chris, You may also try to use std::stoi since C++11. It will throw std::invalid_argument exception if no conversion could be performed. You may find more information here: std::stoi

Answer 3

/* isdigit example */
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main ()
{
  char str[]="1776ad";
  int year;
  if (isdigit(str[0]))
  {
    year = atoi (str);
    printf ("The year that followed %d was %d.\n",year,year+1);
  }
  return 0;
}