c++ - Derive class A from a template class Base<A> so that Base<A> can use A::B? -

template <typename t> class base { private:     typename t::b c; };  class : public base<a> { public:     class b; }; 

is possible? vc++ 2013 says b not member of a.

i go (live example):

template<typename t> struct impl; template<typename t> struct nested;  template <typename t> class base { private:     typename nested<t>::type c; };  struct a;  template<> struct impl<a>   { class b { }; }; template<> struct nested<a> { using type = typename impl<a>::b; };  struct : base<a>, impl<a> {     //... }; 

here, class impl contains part of a not depend on base, is, nested class b. hence a derives both base<a> , impl<a>.

class nested contains alias specifying type of above nested class. base reads type nested , defines data member, of type.

we need declare a before specialize impl , nested it. , need specializations before defining a because @ point base<a> instantiated, , requires nested<a> complete, in turn requires impl<a> complete.

the main difference philip's answer responsibilities more separated:

• base not derive anything, has less chance of being polluted. change type of data member typename nested<t>::type, that's it.

• nested pure type trait. defines alias type, that's it.

• impl implementation class. contains definition of nested class b or possibly else not depend on base.

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by way, stroustrup's 4th edition of the c++ programming language has following code on page 771:

template<typename n> struct node_base : n::balance_type { };  template<typename val, typename balance> struct search_node : node_base<search_node<val, balance> > {     using balance_type = balance; }; 

which has same problem.