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Home/ Questions/Q 865005
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T09:34:20+00:00 2026-05-15T09:34:20+00:00

I have the following scenario: class my_base { … } class my_derived : public

  • 0

I have the following scenario:

class my_base { ... }

class my_derived : public my_base { ... };


template<typename X>
struct my_traits;

I want to specialize my_traits for all classes derived from my_base including, e.g.:

template<typename Y> // Y is derived form my_base.
struct my_traits { ... };

I have no problems with adding tags, members to my_base to make it simpler. I’ve seen some tricks but I still feel lost.

How can this be done in a simple and short way?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T09:34:20+00:00Added an answer on May 15, 2026 at 9:34 am

    Well, you don’t need to write your own isbaseof. You can use boost’s or c++0x’s.

    #include <boost/utility/enable_if.hpp>
    
    struct base {};
    struct derived : base {};
    
    template < typename T, typename Enable = void >
    struct traits;
    
    template < typename T >
    struct traits< T, typename boost::enable_if<std::is_base_of<base, T>>::type >
    {
      enum { value = 5 };
    };
    
    #include <iostream>
    int main()
    {
      std::cout << traits<derived>::value << std::endl;
    
      std::cin.get();
    }
    

    There are scaling issues but I don’t believe they’re any better or worse than the alternative in the other question.

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