c++ : meaning of keyword "typename" in this function -

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• where , why have put “template” , “typename” keywords? 5 answers

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 template <typename... ts>     typename std::tuple_element<0, std::tuple<ts...> >::type // or decltype(auto)          callfunction(ts&&... ts)     {         using type = typename std::tuple_element<0, std::tuple<ts...> >::type;         auto = multicache.find(typeid(type));         assert(it != multicache.end());         auto&& fn = boost::any_cast<const std::function<type(ts...)>&>(it->second);         return fn(std::forward<ts>(ts)...);     } 

the meaning of typename std::tuple_element<0, std::tuple<ts...> >::type returned type same of first element of first element in ts..., right?

no, typename keyword, practical purposes, means same thing class.

the difference semantic. in context of template function or class, typename can pod, rather formally declared class. literally means "any type, pod or class", loosely speaking.

that's keyword typename means in generally. in case, has specific purpose:

typename std::tuple_element<0, std::tuple<ts...> >::type 

this tells compiler expect "type" in std::tuple_element<0, std::tuple<ts...> > going class (or typedef), opposed class member.

when have looks like:

classname::identifier 

this can refer either type or class member:

class x {  public:      typedef int y;      int z; }; 

here, x::y refers type, , x::z refers class member. when comes parsing templates, c++ compiler assume default "a::b" going refer class member, unless stick typename in front of it, in case it'll parsed type.