How to stringify an expression in C

Question

Is there a way to evaluate an expression before stringification in c?

example:

#define stringify(x)  #x
...
const char * thestring = stringify( 10 *          


        

Answer 1

As the other responses have said, this cannot be done with the C preprocessor. This is one of the many shortcomings of C that are solved by C++, This is the sort of thing that can be accomplished in a very elegant manner using Template Metaprogramming.

To calculate an arithmetic expression at compile time:

#include 
namespace mpl = boost::mpl;
int main(int argc, char *argv[]) {
  const int n = mpl::multiplies, mpl::int_<50> >::value;
  return 0;
}

Here's a string formatting metafunction I found on the boost mailing list archives. This version will convert an int (like the one calculated above) into a string in a base of your choosing:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
namespace mpl = boost::mpl;
struct itoa_ct
{
  // radix for _itoa() goes up to 36, but only bother with 16 here
  typedef mpl::vector_c radix_t;
  template 
    struct radix_convert
    {
      typedef typename mpl::push_back<
        typename radix_convert::type
        , mpl::char_::type::value>
        >::type type;
    };
  template 
    struct radix_convert
    {
      typedef mpl::string<> type;
    };
  template 
    struct apply
    {
      // All bases != 10 consider I as unsigned
      typedef typename radix_convert<
        Radix, static_cast((Radix == 10 && I < 0) ? -I : I)
        >::type converted_t;
      // Prefix with '-' if negative and base 10
      typedef typename mpl::if_<
        mpl::bool_<(Radix == 10 && I < 0)>
        , mpl::push_front >
        , mpl::identity
        >::type::type type;
    };
};

Putting the two together your expression becomes:

const char *thestring = mpl::c_str, mpl::int_<50> >::value>::type>::value;

... and this all gets turned into nothing more than a constant string "500" at compile time :-)