c++ - Need scope resolution operator to call member function template when class template exists with same name -

i have class template happens have same name member function template of classes. function template gets instantiated 1 of classes member function template in question. call member function template within function template need use template keyword, understand , have no problem that. however, need use scope resolution operator (i found out that's that's called) :: specify mean class' member function template , not class template , don't understand why.

that's lot of templated things let me give example:

    //class same name member function below.     //must class template or error doesn't show up.     //also no error if function template instead of class     template <class t>     struct f     {     };      struct base     {         //function same name struct above,         //no error if not templated         template <int n>         void f(){}     };      //template function instantiated t=base.     //no error if not templated     template <class t>     void g(t t)     {         //i understand why template keyword needed here,         //but not why t:: needed         t.t::template f<0>();         //t.template f<0>(); gives error.     } 

the error in question (from g++-4.7)

    error: type/value mismatch @ argument 1 in template parameter list ‘template<class t> struct f’     error:   expected type, got ‘0’ 

it seems compiler parses f<0>() on commented out line (without scope resolution operator) trying instantiate object of type struct f<0>. don't know why doing this, thought should able see t.template i'm trying access member function template.

i'd understand going on here, why t:: needed in case, other placate compiler?

it seems there no problem under msvc , clang, seems g++-specific issue.

i think has two phase lookup. steve jessops notes in answer are important.