c++ - Delegate to another object's operator-> -

writing @ alias template deduce return type of type t's operator->, far have this

template <typename t> using arrow = decltype(std::declval<t&>().operator->()); 

which works class types, doesn't work pointers. similar problem exists trying call ->

template <typename t> struct d {     t t;     arrow<t> f() {          // not valid call .operator->() on pointers         return t.operator->();     } }; 

how can make function correct return type declared, , delegate correctly both class types , pointers?

for pointer, type of operator->() own type, , resulting object has same value. level of indirection, helper struct can specialized pointer types

template <typename t> struct arrowhelper {     using type = decltype(std::declval<t&>().operator->());     type operator()(t& t) const {         return t.operator->();     } };  template <typename t> struct arrowhelper<t*> {     using type = t*;     constexpr type operator()(t* t) const noexcept {         return t;     } }; 

to simplify usage, alias template , function can defined easily

template <typename t> using arrow = typename arrowhelper<t>::type;  template <typename t> arrow<t> apply_arrow(t& t) {     return arrowhelper<t>{}(t); } 

the delegating member function becomes

template <typename t> struct d {     t t;     arrow<t> f() { return apply_arrow(t); }     };