Generic way to convert a string into a numerical type?

Question

I have this class:

template<typename T> class Parser
{
    public:
        Parser() : count(0) {}
        virtual void parse(const string&);
        void get_token(void);
    private:
        T result;
        char token;
        string expression;
        int count;
};

now had the class not been generic, had the result been say, a double, I would have used this method to detect numbers.

while((strchr("1234567890.",token))
{
     /* add token to a "temp" string */
     /* etc. etc. */
}

result = atof(temp.c_str());

But since result is generic, I can't use any method like atof and atoi etc.

What do I do?

¤ You can use a `std::istringstream`. Or you can use a `boost::lexical_cast`, which uses the stream internally. Cheers & hth., — Cheers and hth. - Alf, Jan 07 '12 at 10:47
@Lightness: Yeah, after rereading, I noticed it missed the "generic" part. — Xeo, Jan 07 '12 at 10:53

score 6 · Accepted Answer · edited May 23 '17 at 11:55

6

Boost has this functionality built-in:

 #include <boost/lexical_cast.hpp>

 void Parser<T>::get_token() {
     std::string token = ...;
     result = boost::lexical_cast<T>(token);
 }

Add exception handling as required.

Or, perhaps you don't want to use Boost for some reason:

void Parser<T>::get_token() {
     std::string token = ...;

     std::stringstream ss;
     ss << token;
     ss >> result;
}

Check the error state of ss as required.

More expansive answers may be found on this related question, though it discusses only int specifically.

edited May 23 '17 at 11:55

Community

1
1

answered Jan 07 '12 at 10:46

Lightness Races in Orbit

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+1, Thanks. I'd rather not add boost to my project, so I'll go with `stringstream` – ApprenticeHacker Jan 07 '12 at 10:51
@IntermediateHacker: You only need a header. That's it. The `boost` solution is _far_ more robust. Please consider it again. – Lightness Races in Orbit Jan 07 '12 at 10:52
(If you're writing a parser, I'm baffled that you're not using Boost already!) – Lightness Races in Orbit Jan 07 '12 at 10:52
1

@IntermediateHacker: When writing a parser, I'd resort to a parser generator. Boost.Spirit is pretty awesome in that aspect. – Xeo Jan 07 '12 at 10:58
3

@Xeo Yes, I would have done that if I was like, building a parser for serious work, like a professional programming language / script etc. But I'm actually trying to learn how recursive descent parsers etc. are implemented. Therefore I think, writing from scratch would teach me more. – ApprenticeHacker Jan 07 '12 at 11:03

score 1 · Answer 2 · edited Jan 07 '12 at 14:57

Another generic template based Numeric To String converter. It takes ints and doubles.

#include <sstream>
#include <iostream>
#include <string>
using namespace std;

template <class T>
inline std::string Numeric_To_String (const T& t)
{
    std::stringstream ss;
    ss << t;
return ss.str();
}


int main(int argc, char *argv[])
{
   int i = 9;
   double d = 1.2345;
   string s;

  cout <<"Generic Numeric_To_String( anyDatatype ) \n\n";

  s = Numeric_To_String( i );
  cout <<"int i to string    : "<< s <<"   "<< endl; 

  s = Numeric_To_String( d );
  cout <<"double d to string : "<< s <<"   "<< endl;
  cout <<" \n";   

  return 0;
}

score 0 · Answer 3 · answered Aug 14 '19 at 07:02

With C++17 you can use the templated std::from_chars. https://en.cppreference.com/w/cpp/utility/from_chars

#include <charconv>
#include <iostream>

template <typename Number>
auto stringTo(std::string_view str)
{
    Number number;
    std::from_chars(str.data(), str.data() + str.size(), number);
    return number;
}

int main()
{
    const auto str = std::string("42");
    std::cout << stringTo<long>(str) << '\n';
    std::cout << stringTo<double>(str) << '\n'; 
}

Check the return value of std::from_chars to detect errors.

const auto result = std::from_chars(...);
if (result.ec == std::errc::invalid_argument || result.ec == std::errc::result_out_of_range)
{
   std::cout << "string to number error" << '\n';
}

More info and examples: https://www.bfilipek.com/2018/12/fromchars.html

GCC and clang don't yet support the floating point version of std::from_chars (August 2019).

score 0 · Answer 4 · answered Jan 07 '12 at 14:55

0

If you only have a hand full of types you want to parse, you can use template specialization:

template<>
void Parser<int>::parse(const string&)
{
    result = atoi(string.c_str());
}

template<>
void Parser<float>::parse(const string&)
{
    result = atof(string.c_str());
}

... But this only works if you implement every convertion you need, of course.

answered Jan 07 '12 at 14:55

v01pe

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Regarding `atoi` and `atof` - http://stackoverflow.com/questions/2892951/list-of-deprecated-c-functions. Also you need to use _named_ parameters; `string.c_str()` isn't going to work. – Lightness Races in Orbit Jan 07 '12 at 14:58

Generic way to convert a string into a numerical type?

4 Answers4