views:

368

answers:

2

It works just fine, for plain vanilla functions. The code below works just fine. It prints just what is should:

int __cdecl(int, char)
2
int,char

#include <boost/type_traits.hpp>
#include <boost/function.hpp>
#include <boost/typeof/std/utility.hpp>

#include <iostream>

using std::cout;
using std::endl;

int foo(int, char) {
 return 0;
}
int main() {
    typedef BOOST_TYPEOF(foo) foo_type;;
    typedef boost::function_traits<foo_type> function_traits;

    cout << typeid(foo_type).name() << endl;
    cout << function_traits::arity << endl;
    cout << typeid(function_traits::arg1_type).name() << ",";
    cout << typeid(function_traits::arg2_type).name() << endl;

    return 0;
}

So, the question is, how can one do this if foo is a member function of class bar?

struct bar {
    int foo(int, char) { return 0; }
};

I have tried countless combinations of these constructs: BOOST_TYPEOF_INCREMENT_REGISTRATION_GROUP() BOOST_TYPEOF_REGISTER_TYPE() boost::ref boost::remove_pointer boost::bind boost::mem_fn

etc., etc... No joy.

+1  A: 

Boost Function Types would probably be the natural solution:

#include <boost/function_types/function_type.hpp>
#include <boost/function_types/parameter_types.hpp>
#include <boost/function_types/function_arity.hpp>
#include <boost/typeof/std/utility.hpp>
#include <iostream>

struct bar {
    int foo(int, char) { return 0; }
};

int main() {

    typedef BOOST_TYPEOF(&bar::foo) foo_type;

    std::cout << typeid(foo_type).name() << std::endl;
    std::cout << boost::function_types::function_arity<foo_type>::value << std::endl;
    std::cout << typeid(boost::mpl::at_c<boost::function_types::parameter_types<foo_type>,1>::type).name() << ",";
    std::cout << typeid(boost::mpl::at_c<boost::function_types::parameter_types<foo_type>,2>::type).name() << ",";

    return 0;
}
Kornel Kisielewicz
Jive Dadson
@Jive: sigh, do you people have to get everything served on a plate? :/
Kornel Kisielewicz
Most people do. I've been a professional programmer since 1971, and I've learned that it cannot hurt to spell things out. Thanks for your help.
Jive Dadson
@Jive, no problem, I don't get anything out of it anyway -- although this gave me an inclination to try to do that full C++ reflection library thing again ;>
Kornel Kisielewicz
@Kornel You've got to admit, it's pretty obscure. My dream is a ++C++ that's what C++ would be if Bjarne knew then what he knows now. Without the requirement that C++ be a superset of C, a requirement that has since been dropped, all options would be open. For one thing, compile-time introspection would not have to be done through hacks with template constructs. I've got some ideas, but whom am I kidding? That's one more thing I'll never do.
Jive Dadson
@Jive, yeah I'll admit it. And yes, I'd dream that C++ finally dropped backwards compatibility to C++ -_-. If you want a language that is ++C++ then try D.
Kornel Kisielewicz
@Jive -- BTW, accept the answer if you're satisfied with it ;)
Kornel Kisielewicz
A: 

Kornel Kisielewicz nailed it. Here it is with the solution separated from the test-messages.

#include <boost/function_types/function_type.hpp>
#include <boost/function_types/parameter_types.hpp>
#include <boost/function_types/function_arity.hpp>
#include <boost/typeof/std/utility.hpp>
#include <iostream>

struct bar {
    int foo(int, char) { return 0; }
};

int main() {

    typedef BOOST_TYPEOF(&bar::foo) 
        foo_type;
    int arity = boost::function_types::function_arity<foo_type>::value;
    typedef boost::mpl::at_c<boost::function_types::parameter_types<foo_type>,1>::type
        arg1;
    typedef boost::mpl::at_c<boost::function_types::parameter_types<foo_type>,2>::type
        arg2;

    std::cout << typeid(foo_type).name() << std::endl;
    std::cout << arity << std::endl;
    std::cout << typeid(arg1).name() << ",";
    std::cout << typeid(arg2).name() << std::endl;
    return 0;
}
Jive Dadson