RTTI type traits in C++

RTTI Type Traits Example

// RTTI Type Traits
// Example by Patrick Hebron

template <typename T> struct TypeTraits
{
	static const char* name() { return "TYPE_NAME"; }
};

struct Base
{
	Base() {}
	virtual ~Base() {}
	
	virtual void testPrint() = 0;
	virtual void testPrintTraitName() = 0;
};

template <typename DerivedT>
struct BaseImpl : public Base
{
	BaseImpl() {}
};

#define DECLARE_DERIVED_TYPE(Type,Symbol) \
template<> struct TypeTraits<struct Type> \
{ \
	typedef Type type; \
	static const char* name() { return Symbol; } \
}; \
struct Type : public BaseImpl<Type> \

DECLARE_DERIVED_TYPE( Foo, "FOO_TYPE" )
{
	Foo() {}
	
	void testPrint() { std::cout << "FOO::testPrint()" << std::endl; };
	void testPrintTraitName() { std::cout << TypeTraits<Foo>::name() << std::endl; };
	
	void testFoo() { std::cout << "FOO::testFoo()" << std::endl; };
};

DECLARE_DERIVED_TYPE( Bar, "BAR_TYPE" )
{
	Bar() {}
	
	void testPrint() { std::cout << "BAR::testPrint()" << std::endl; };
	void testPrintTraitName() { std::cout << TypeTraits<Bar>::name() << std::endl; };
	
	void testBar() { std::cout << "BAR::testBar()" << std::endl; };
};

int main(int argc, const char * argv[])
{
	using namespace std;
	
	Foo* aFoo = new Foo();
	Bar* aBar = new Bar();
	
	Base* aFooBaseCast = aFoo;
	Base* aBarBaseCast = aBar;
	
	aFoo->testPrint();
	aFoo->testPrintTraitName();
	aFoo->testFoo();
	
	cout << endl;
	
	aBar->testPrint();
	aBar->testPrintTraitName();
	aBar->testBar();
	
	cout << endl;
	
	aFooBaseCast->testPrint();
	aFooBaseCast->testPrintTraitName();
	
	cout << endl;
	
	aBarBaseCast->testPrint();
	aBarBaseCast->testPrintTraitName();
	
	cout << endl;
	
	delete aFoo;
	delete aBar;
	
    return 0;
}