How do I check if a C++ string is an int?

Question

When I use getline, I would input a bunch of strings or numbers, but I only want the while loop to output the "word" if it is not a number. So is there any way to check if "word" is a number or not? I know I could use atoi() for C-strings but how about for strings of the string class?

int main () {
  stringstream ss (stringstream::in | stringstream::out);
  string word;
  string str;
  getline(cin,str);
  ss<<str;
  while(ss>>word)
    {
      //if(    )
        cout<<word<<endl;
    }
}

Answer 1

You might try boost::lexical_cast. It throws an bad_lexical_cast exception if it fails.

In your case:

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

Answer 2

Ok, the way I see it you have 3 options.

1: If you simply wish to check whether the number is an integer, and don't care about converting it, but simply wish to keep it as a string and don't care about potential overflows, checking whether it matches a regex for an integer would be ideal here.

2: You can use boost::lexical_cast and then catch a potential boost::bad_lexical_cast exception to see if the conversion failed. This would work well if you can use boost and if failing the conversion is an exceptional condition.

3: Roll your own function similar to lexical_cast that checks the conversion and returns true/false depending on whether it's successful or not. This would work in case 1 & 2 doesn't fit your requirements.

Answer 3

You can use boost::lexical_cast, as suggested, but if you have any prior knowledge about the strings (i.e. that if a string contains an integer literal it won't have any leading space, or that integers are never written with exponents), then rolling your own function should be both more efficient, and not particularly difficult.

Answer 4

template <typename T>
const T to(const string& sval)
{
        T val;
        stringstream ss;
        ss << sval;
        ss >> val;
        if(ss.fail())
                throw runtime_error((string)typeid(T).name() + " type wanted: " + sval);
        return val;
}

And then you can use it like that:

double d = to<double>("4.3");

or

int i = to<int>("4123");

Answer 5

If you're just checking if word is a number, that's not too hard:

#include <ctype.h>

...

string word;
bool isNumber = true;
for(string::const_iterator k = word.begin(); k != word.end(); ++k)
    isNumber &&= isdigit(*k);

Optimize as desired.

Answer 6

Another version...

Use strtol, wrapping it inside a simple function to hide its complexity :

inline bool isInteger(const std::string & s)
{
   if(s.empty() || ((!isdigit(s[0])) && (s[0] != '-') && (s[0] != '+'))) return false ;

   char * p ;
   strtol(s.c_str(), &p, 10) ;

   return (*p == 0) ;
}

Why strtol ?

As far as I love C++, sometimes the C API is the best answer as far as I am concerned:

• using exceptions is overkill for a test that is authorized to fail
• the temporary stream object creation by the lexical cast is overkill and over-inefficient when the C standard library has a little known dedicated function that does the job.

How does it work ?

strtol seems quite raw at first glance, so an explanation will make the code simpler to read :

strtol will parse the string, stopping at the first character that cannot be considered part of an integer. If you provide p (as I did above), it sets p right at this first non-integer character.

My reasoning is that if p is not set to the end of the string (the 0 character), then there is a non-integer character in the string s, meaning s is not a correct integer.

The first tests are there to eliminate corner cases (leading spaces, empty string, etc.).

This function should be, of course, customized to your needs (are leading spaces an error? etc.).

Sources :

See the description of strtol at: http://www.cplusplus.com/reference/clibrary/cstdlib/strtol/.

See, too, the description of strtol's sister functions (strtod, strtoul, etc.).