c++ - Code with multiple inheritance and too much public access -

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i have following chunk of code defining functor composition...

#pragma once #include <tuple>  template<typename args,std::size_t a,typename...f> class _compose{};  template<typename...args,std::size_t a,typename f,typename...g> class _compose<std::tuple<args...>,a,f,g...> :private _compose<std::tuple<args...>,a-1,g...>{         const f f; public:    _compose(f _f,g...g):_compose<std::tuple<args...>,a-1,g...>(g...),f(_f){};     constexpr auto operator()(args...args) const{         return f(_compose<std::tuple<args...>,a-1,g...>::operator()(args...));     } };  template<typename...args,typename f> class _compose<std::tuple<args...>,1,f>{     const f f;   public:     _compose(f _f):f(_f){};     constexpr auto operator()(args...args) const{     return f(args...);     } };  template<typename...args> struct compose{     template<typename...fs>     constexpr auto operator()(fs...f){         return _compose<std::tuple<args...>,sizeof...(fs),fs...>(f...);     }        };

using functor definitions such as:

template<typename t> struct func1{     constexpr auto operator()(t x){return x+1;} };  template<int i,typename t> class func2{     const int i,j; public:     func2(int _j):i(i),j(_j){};     constexpr auto operator()(t x){return x+i+j;} };

i can compose compose<int>()(func1<int>(),func2<1,int>(1))(1) return 4.

now question: weak implementation? 1. went route of multiple inheritance each _compose<i> inherits "downwards" _compose<i-1> , on until _compose<1>. felt elegant solution not sure of caveats. know multiple inheritance frowned upon. 2. unable make _compose constructor more hidden outside world. tried make compose function friend none of attempts compiled. note _compose<...>::operator() intended public compose function returns _compose object.

any suggestions improve above code appreciated. note compile under g++ -std=c++1y.

thanks,


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