Help deducing the derived class inside a base class

**DrPiratePhD** · 02-05-2011

Hi!

In my program, I've been wanting to determine the the derived type of a base class. So far, this seems to work fine:

Code:

class BaseClass
{
public:
      template<class derived>
      void PrintType(derived object)
      {
           cout << "type: " << typeid(object).name() << endl;
      }
};

class DerivedClass : public BaseClass
{
protected:
      void Foo()
      {
           PrintType(*this);
      }
};

int main()
{
          DerivedClass test_object;
          test_object.Foo();
          cin.get();
}

However, in the context of what I intend to write, PrintType() is always called within a derived class of BaseClass and the argument is always going to be *this" (with or without * works for me). Not only will the argument always be the same, but it is the only argument for my program to function as I want it. So I was wondering: is there a way to re-write PrintType() such that it has it takes no arguments and can deduce the derived type on its own?

Thanks!

**Xupicor** · 02-05-2011

Either revise your design, or... Try to use a pattern that's already there. Maybe visitor pattern?
Visitor pattern - Wikipedia, the free encyclopedia
Visitor Pattern

**anon** · 02-06-2011

Yes, it could do that if the base class was polymorphic (contained a virtual function), in which case RTTI determines the most derived type of the given value.

Code:

#include <iostream>
#include <typeinfo>
using namespace std;

class BaseClass
{
public:
      virtual ~BaseClass() {}
      void PrintType()
      {
           cout << "type: " << typeid(*this).name() << endl;
      }
};

class DerivedClass : public BaseClass
{
public:
      void Foo()
      {
           PrintType();
      }
};

int main()
{
          DerivedClass test_object;
          test_object.Foo();
}

**laserlight** · 02-06-2011

Originally Posted by anon

Yes, it could do that if the base class was polymorphic (contained a virtual function), in which case RTTI determines the most derived type of the given value.

However, before you reach for this solution, take Xupicor's advice.

**anon** · 02-06-2011

Well, eventually you could just use a free function:

Code:

#include <iostream>
#include <typeinfo>
using namespace std;

class BaseClass
{
};

class DerivedClass : public BaseClass
{
};

template <class T>
void PrintType(const T& value)
{
    cout << "type: " << typeid(value).name() << endl;
}

int main()
{
          DerivedClass test_object;
          PrintType(test_object);
}

**laserlight** · 02-06-2011

Maybe I should ask a more over-arching question: why do you want to print typeid(object).name() in the first place?

**DrPiratePhD** · 02-08-2011

It's not that important really. I was just working on a windows API wrapper class. I wanted a simple base class encapsulating class registration and window creation. Upon inheriting from the base class, the window class name automatically becomes the name of the inherited class. I know there are probably MUCH better ways of doing it, but that is what seemed most obvious to me.

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