#include <stdio.h>

template <typename TYPE>
struct Base {
        Base() {
                //this unqualified call works fine within the base class
                voidFunc();
        }

        void voidFunc() {
                printf("voidFunc()!\n");
        }

        //a public variable
        TYPE m_publicVar;
};

template <typename TYPE>
struct Child : public Base<TYPE> {
        Child() {
                //calling fully qualified works fine
                Base<TYPE>::voidFunc();

                //this spits out 
                //"error: there are no arguments to `voidFunc' that depend on a
template parameter, so a declaration of `voidFunc' must be available
                //        (if you use `-fpermissive', G++ will accept your
code, but allowing the use of an undeclared name is deprecated)"
                //and will compile and link with -fpermissive
                voidFunc();

                if (m_publicVar) { //this var access will not compile regardles
of -fpermissive
                        printf("m_publicVar!\n");
                }
        }

};

int main(int argc, const char* argv) {
        Child<int> child;
        return 0;
}


-- 
           Summary: template inheritance loses base class scope
           Product: gcc
           Version: 3.4.6
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
        AssignedTo: unassigned at gcc dot gnu dot org
        ReportedBy: tim at deadlyninja dot com
  GCC host triplet: i686-pc-linux-gnu-3.4.6
GCC target triplet: i686-pc-linux-gnu-3.4.6


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=28464

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