Re: a bit of reflection (doesn't work)

From:
=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Wed, 23 Jun 2010 08:07:25 CST
Message-ID:
<ba868232-1c92-4582-96ea-13593aaab1c1@u26g2000yqu.googlegroups.com>
On 23 Jun., 02:11, Frank Bergemann <FBergem...@web.de> wrote:

[..]

i tried this one here to get information, if a class X inherits from a Feature class:
------------------------- snip ----------------------------------------

#include <iostream>
using namespace std;

struct NO
{
        static const int VALUE = 0;
};

struct YES
{
        static const int VALUE = 1;
};

namespace Feature
{

        template <class Base>


I assume, "Base" should better be renamed to
"Derived", because you are supposed to provider
the sub-class type, right?

        class Class
        {
        public:
                Class() { };
                ~Class() { };
        };

        // template for indicator of Feature - default NO
        template <class QueryBase>
        class Indicator
        {
        public:
                typedef NO VALUE;
        };

        // explicit instantiation of indicator of Feature<Base> - YES
        template<class QueryBase>
        class Indicator <Class<QueryBase> >
        {
        public:
                typedef YES VALUE;
        };

}; // namespace Feature

class TestA
{
public:
        TestA() { };
        ~TestA() { };
};

class TestB : public Feature::Class<TestB>
{
public:
        TestB() { };
        ~TestB() { };
};


OK, TestB *derives* from Feature::Class<TestB>,
but that doesn't mean that a template type deduction
will evaluate that TestB is the same as Feature::Class<TestB>

int main(int, char**)
{

        TestB b;
        cout << "Indicator<TestA> = " << Feature::Indicator<TestA>::VALUE::VALUE << endl;
        cout << "Indicator<TestB> = " << Feature::Indicator<TestB>::VALUE::VALUE << endl;
        cout << "Indicator<Feature<TestA> > = " << Feature::Indicator<Feature::Class<TestA> >::VALUE::VALUE << endl;
        return 0;
}

------------------------- snap ----------------------------------------

But it doesn't work:

frank@frank-desktop-2:~$ g++ -o test test.cc
frank@frank-desktop-2:~$ ./test
Indicator<TestA> = 0
Indicator<TestB> = 0
Indicator<Feature<TestA> > = 1

Is there a way to get this information also for Feature::Indicator<TestB> ?


You may want to use boost::is_base_of<Feature::Class<TestB>, TestB>
or - if you use a compiler with C++0x feastures -
std::is_base_of<Feature::Class<TestB>, TestB> to check whether
TestB is a super-class of Feature::Class<TestB>. The is_base_of
traits also works for same types. Given this trait class you could
define a single primary template for Indicator, e.g.

template <class Derived>
struct Indicator {
  typedef typename boost::mpl::if_c<boost::is_base_of<Class<Derived>,
    Derived>::value, YES, NO>::type VALUE;
};

or

template <class Derived>
struct Indicator {
  typedef typename std::conditional<std::is_base_of<Class<Derived>,
    Derived>::value, YES, NO>::type VALUE;
};

HTH & Greetings from Bremen,

Daniel Kr?gler

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