Derived template-class access to base-class member-data
You can use this-> to make clear that you are referring to a member of the class: void Bar<T>::BarFunc () { std::cout << this->_foo_arg << std::endl; } Alternatively you can also use “using” in the method: void Bar<T>::BarFunc () { using Bar<T>::_foo_arg; // Might not work in g++, IIRC std::cout << _foo_arg << std::endl; } … Read more