C++ convert integer to string at compile time

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First of all, if usually you know the number at run time, you can as easily build the same string. That is, if you have 12 in your program, you can have also "12".

Preprocessor macros can add also quotes to arguments, so you can write:

#define STRINGIFICATOR(X) #X

This, whenever you write STRINGIFICATOR(2), it will produce “2”.

However, it actually can be done without macros (using compile-time metaprogramming). It is not straightforward, so I cannot give the exact code, but I can give you ideas on how to do it:

  1. Write a recursive template using the number to be converted. The template will recurse till the base case, that is, the number is less than 10.
  2. In each iteration, you can have the N%10 digit to be converted into a character as T.E.D. suggests, and using mpl::string to build the compile-time string that appends that character.
  3. You’ll end up building a mpl::string, that has a static value() string.

I took the time to implement it as a personal exercise. Not bad at the end:

#include <iostream>
#include <boost/mpl/string.hpp>

using namespace boost;

// Recursive case
template <bool b, unsigned N>
struct int_to_string2
{
        typedef typename mpl::push_back<
                typename int_to_string2< N < 10, N/10>::type
                                         , mpl::char_<'0' + N%10>
                                         >::type type;
};

// Base case
template <>
struct int_to_string2<true,0>
{
        typedef mpl::string<> type;
};


template <unsigned N>
struct int_to_string
{
        typedef typename mpl::c_str<typename int_to_string2< N < 10 , N>::type>::type type;
};

int
main (void)
{
        std::cout << int_to_string<1099>::type::value << std::endl;
        return 0;
}

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