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Home/ Questions/Q 8743899
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T11:45:28+00:00 2026-06-13T11:45:28+00:00

Currently I have two functions : template<typename Type> bool f(Type* x); template<typename… List> bool

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Currently I have two functions :

template<typename Type> bool f(Type* x);
template<typename... List> bool f(std::tuple<List...>* x);

Is there any way to merge these two functions with an extra template parameter that indicates whether the passed type is a tuple ?

template<typename Type, bool IsTuple = /* SOMETHING */> bool f(Type* x);
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-13T11:45:29+00:00Added an answer on June 13, 2026 at 11:45 am

    Sure, using is_specialization_of (link taken and fixed from here):

    template<typename Type, bool IsTuple = is_specialization_of<Type, std::tuple>::value>
    bool f(Type* x);
    

    The question is, however, do you really want that? Normally, if you need to know if a type is a tuple, you need special handling for tuples, and that usually has to do with its template arguments. As such, you might want to stick to your overloaded version.

    Edit: Since you mentioned you only need a small portion specialized, I recommend overloading but only for the small special part:

    template<class T>
    bool f(T* x){
      // common parts...
      f_special_part(x);
      // common parts...
    }
    

    with

    template<class T>
    void f_special_part(T* x){ /* general case */ }
    
    template<class... Args>
    void f_special_part(std::tuple<Args...>* x){ /* special tuple case */ }
    
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