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Home/ Questions/Q 8264653
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Editorial Team
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Editorial Team
Asked: June 8, 20262026-06-08T04:32:01+00:00 2026-06-08T04:32:01+00:00

I have been looking around QMetaData and QObject calling the className(). I want to

  • 0

I have been looking around QMetaData and QObject calling the className(). I want to replicate this in my own class but I cannot figure out how QT does it. What I mean is, I do the following but it does not output the correct class (ignore any syntax errors):

#include <iostream>

#define CLASS(name)              \
    std::string className() {    \
        return #name;            \     
    }                           

class A {
public:
   CLASS(A) 

   A(){}
   ~A(){}  

   void output() {
       std::cout << className() << std::endl;
   }
};

class B: public A{
public:
   CLASS(B)
   B(){}
   ~B(){}
};

int main() {
   B b;
   b.output();    // This obviously outputs "A" but I would
                  // like it to output "B" from the base class
                  // function
   return 0;
}

How Qt has it is you do not have to add any more code just the Q_OBJECT macro and you can get the className of the derived class from the base class even if the base class has the output function (in Qt I mean the QObject::debugObjectTree()). How does Qt accomplish this effect without adding any extra code to the derived classes (except maybe a macro), where the base class can output the classname of its derived class?

Thanks in advance.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-08T04:32:03+00:00Added an answer on June 8, 2026 at 4:32 am

    You need dynamic dispatch (i.e. virtual functions in c++) for this.

    Just add virtual before std::string className() in your macro:

    #define CLASS(name)              \
    virtual std::string className() {    \
        return #name;            \     
    }  
    

    http://ideone.com/Kr0Sc

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