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Home/ Questions/Q 718449
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T05:31:41+00:00 2026-05-14T05:31:41+00:00

In a comparison operator: template<class R1, class R2> bool operator==(Manager<R1> m1, Manager<R2> m2) {

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In a comparison operator:

template<class R1, class R2>
bool operator==(Manager<R1> m1, Manager<R2> m2) {
    return m1.internal_field == m2.internal_field;
}

Is there any way I could enforce that R1 and R2 must have a supertype or subtype relation? That is, I’d like to allow either R1 to be derived from R2, or R2 to be derived from R1, but disallow the comparison if R1 and R2 are unrelated types.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T05:31:41+00:00Added an answer on May 14, 2026 at 5:31 am

    A trait you want might look like this:

    template <typename B, typename D>
    struct is_base_of // check if B is a base of D
    {
        typedef char yes[1];
        typedef char no[2];
    
        static yes& test(B*);
        static no& test(...);
    
        static D* get(void);
    
        static const bool value = sizeof(test(get()) == sizeof(yes);
    };
    

    Then you just need a static assert of some sort:

    // really basic
    template <bool>
    struct static_assert;
    
    template <>
    struct static_assert<true> {}; // only true is defined
    
    #define STATIC_ASSERT(x) static_assert<(x)>()
    

    Then put the two together:

    template<class R1, class R2>
    bool operator==(Manager<R1> m1, Manager<R2> m2)
    {
        STATIC_ASSERT(is_base_of<R1, R2>::value || is_base_of<R2, R1>::value);
    
        return p1.internal_field == p2.internal_field;
    }
    

    If one does not derive from the other, the function will not compile. (Your error will be similar to “static_assert<false> not defined”, and it will point to that line.)

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