Help deducing the derived class inside a base class

This is a discussion on Help deducing the derived class inside a base class within the C++ Programming forums, part of the General Programming Boards category; Hi! In my program, I've been wanting to determine the the derived type of a base class. So far, this ...

  1. #1
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    Feb 2011
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    6

    Help deducing the derived class inside a base class

    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!

  2. #2
    a guy with long hair Xupicor's Avatar
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    Poland
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    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

  3. #3
    The larch
    Join Date
    May 2006
    Posts
    3,573
    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();
    }
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  4. #4
    C++ Witch laserlight's Avatar
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    Location
    Singapore
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    23,628
    Quote 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.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  5. #5
    The larch
    Join Date
    May 2006
    Posts
    3,573
    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);
    }
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  6. #6
    C++ Witch laserlight's Avatar
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    Oct 2003
    Location
    Singapore
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    23,628
    Maybe I should ask a more over-arching question: why do you want to print typeid(object).name() in the first place?
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  7. #7
    Registered User
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    Feb 2011
    Posts
    6
    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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