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Home/ Questions/Q 8111785
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T02:14:55+00:00 2026-06-06T02:14:55+00:00

Is there anyway to specialize a template like this, making the specialization apply only

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Is there anyway to specialize a template like this, making the specialization apply only if T has a member function hash? (Note this is only an example of what I am trying to do. I know that it would make more sense for each class that has a hash function to check it on its own in the operator== member function, but I just want to know if this kind of thing is possible.)

template <class T>
bool equals(const T &x, const T &y)
{
    return x == y;
}

template <class T> // somehow check if T has a member function 'hash'
bool equals<T>(const T &x, const T &y)
{
    return x.hash() == y.hash() && x == y;
}

I would prefer a pre-C++11 solution if possible.

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  1. Editorial Team
    Editorial Team
    2026-06-06T02:14:56+00:00Added an answer on June 6, 2026 at 2:14 am

    Here’s an example from my own code. As you might guess from one of the structure names this is based on the principle that Substitution Failure is Not an Error. The structure has_member_setOrigin defines two versions of test. The first one cannot be satisfied if U does not have a member setOrigin. Since that is not an error in a template substitution it just acts as if it does not exist. The resolution order for polymorphic functions thus finds test(...) which would otherwise have a lower priority. The value is then determined by the return type of test.

    This is followed by two definitions of callSetOrigin (equivalent to your equals) using the enable_if template. If you examine enable_if you’ll see that if the first template argument is true then enable_if<...>::type is defined, otherwise it is not. This again creates a substitution error in one of the definitions of callSetOrigin such that only one survives.

    template <typename V>
    struct has_member_setOrigin
    {
        template <typename U, void (U::*)(const Location &)> struct SFINAE {};
        template <typename U> static char test(SFINAE<U, &U::setOrigin>*);
        template <typename U> static int test(...);
        static const bool value = sizeof(test<V>(0)) == sizeof(char);
    };
    
    template<typename V>
    void callSetOrigin(typename enable_if <has_member_setOrigin<V>::value, V>::type &p, const Location &loc) const
    {
        p.setOrigin(loc);
    }
    
    template<typename V>
    void callSetOrigin(typename enable_if <!has_member_setOrigin<V>::value, V>::type &p, const Location &loc) const
    {
    }
    

    Forgot I provided a definition of enable_if as well:

    #ifndef __ENABLE_IF_
    #define __ENABLE_IF_
    
    template<bool _Cond, typename _Tp>
    struct enable_if
    { };
    
    template<typename _Tp>
    struct enable_if<true, _Tp>
    { typedef _Tp type; };
    
    #endif /* __ENABLE_IF_ */
    
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