In C++ classes have the feature of having their name injected into their scope ([class]/2): The class-name is also inserted into the scope of the class itself; this is known as the injected-class-name. For purposes of access checking, the injected-class-name is treated as if it were a public member name. And the code snippet you … Read more
They could. This is the way most early implementations of templates worked, and is still the way the Microsoft compiler worked. It was felt (in the committee) that this was too error prone; it made it too easy to accidentally hijack a name, with the instantiation in one translation unit picking up a local name, … Read more
David Joyner had the history, here is the reason. The problem when compiling B<T> is that its base class A<T> is unknown from the compiler, being a template class, so no way for the compiler to know any members from the base class. Earlier versions did some inference by actually parsing the base template class, … Read more
I believe that this question is duplicate, but I cannot find it now. C++ Standard says that you should fully qualify name according to 14.6.2/3: In the definition of a class template or a member of a class template, if a base class of the class template depends on a template-parameter, the base class scope … Read more
This part explains it: C++ Standard 03 14.8.1.6: [Note: For simple function names, argument dependent lookup (3.4.2) applies even when the function name is not visible within the scope of the call. This is because the call still has the syntactic form of a function call (3.4.1). But when a function template with explicit template … Read more
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
C++ answer (general answer) Consider a template class Derived with a template base class: template <typename T> class Base { public: int d; }; template <typename T> class Derived : public Base<T> { void f () { this->d = 0; } }; this has type Derived<T>, a type which depends on T. So this has … Read more
Koenig Lookup, or Argument Dependent Lookup, describes how unqualified names are looked up by the compiler in C++. The C++11 standard ยง 3.4.2/1 states: When the postfix-expression in a function call (5.2.2) is an unqualified-id, other namespaces not considered during the usual unqualified lookup (3.4.1) may be searched, and in those namespaces, namespace-scope friend function … Read more