A simple problem and a solution Consider this C++ code: #include <iostream> template < typename T> struct Base { void f() { std::cerr << "Base<T>::f

" ; } }; template < typename T> struct Derived : Base<T> { void g() { std::cerr << "Derived<T>::g

" ; f(); } }; The intention of Derived<T>::g is to call Base<T>::f , but what the compiler does instead is produce this error: : In member function ‘void Derived<T>::g()’: :18:10: error: there are no arguments to ‘f’ that depend on a template parameter, so a declaration of ‘f’ must be available :18:10: note: (if you use ‘-fpermissive’, G++ will accept your code, but allowing the use of an undeclared name is deprecated) First, let's see how to fix this. It's easy. All you have to do is to make the compiler understand that the call f depends on the template parameter T . A couple of ways to do this are replacing f() with Base<T>::f() , or with this->f() (since this is implicitly dependent on T ). For example: #include <iostream> template < typename T> struct Base { void f() { std::cerr << "Base<T>::f

" ; } }; template < typename T> struct Derived : Base<T> { void g() { std::cerr << "Derived<T>::g

" ; this ->f(); } }; int main() { Derived< float > df; df.g(); Derived< int > di; di.g(); return 0 ; } main instantiates two Derived objects, parametrized for different types, for reasons that will soon become apparent. This code compiles without errors and prints: Derived<T>::g Base<T>::f Derived<T>::g Base<T>::f Problem fixed. Now, let's understand what's going on. Why does the compiler need an explicit specification for which f to call? Can't it figure out on its own that we want it to call Base<T>::f ? It turns out it can't, because this isn't correct in the general case. Suppose that a specialization of the Base class is later created for int , and it also defines f : template <> struct Base< int > { void f() { std::cerr << "Base<int>::f

" ; } }; With this specialization in place, the main from the sample above would actually print: Derived<T>::g Base<T>::f Derived<T>::g Base<int>::f This is the correct behavior. The Base template has been specialized for int , so it should be used for inheritance when Derived<int> is required. But how does the compiler manage to figure it out? After all, Base<int> was defined after Derived !

Two-phase name lookup To make this work, the C++ standard defines a "two-phase name lookup" rule for names in templates. Names inside templates are divided to two types: Dependent - names that depend on the template parameters but aren't declared within the template.

Non-dependent - names that don't depend on the template parameters, plus the name of the template itself and names declared within it. When the compiler tries to resolve some name in the code, it first decides whether the name is dependent or not, and the resolution process stems from this distinction. While non-dependent names are resolved "normally" - when the template is defined, the resolution for dependent names happens at the point of the template's instantiation. This is what ensures that a specialization can be noticed correctly in the example above. Now, back to our original problem. Why doesn't the compiler look f up in the base class? First, notice that in the call to f() in the first code snippet, f is a non-dependent name. So it must be resolved at the point of the template's definition. At that point, the compiler still doesn't know what Base<T>::f is, because it can be specialized later. So it doesn't look names up in the base class, but only in the enclosing scope. Since there's no f in the enclosing scope, the compiler complains. On the other hand, when we explicitly make the lookup of f dependent by calling it through this-> , the lookup rule changes. Now f is resolved at the point of the template's instantiation, where the compiler has full understanding of the base class and can resolve the name correctly.

Disambiguating dependent type names I've mentioned above that to fix the problem and make the lookup of f dependent, we can either say this->f() or Base<T>::f() . While this works for identifiers like member names, it doesn't work with types. Consider this code snippet: #include <iostream> template < typename T> struct Base { typedef int MyType; }; template < typename T> struct Derived : Base<T> { void g() { // A. error: ‘MyType’ was not declared in this scope // MyType k = 2; // B. error: need ‘typename’ before ‘Base<T>::MyType’ because // ‘Base<T>’ is a dependent scope // Base<T>::MyType k = 2; // C. works! typename Base<T>::MyType k = 2 ; std::cerr << "Derived<T>::g --> " << k << "

" ; } }; int main() { Derived< float > df; df.g(); return 0 ; } Three attempts are shown to declare a local variable k of type MyType . The first two are commented out because they result in compile errors. (A) should be obvious by now - since MyType is non-dependent, it can't be found in the base class - same problem as before. But why doesn't (B) work? Well, because Base<T> can be specialized, so the compiler can't be sure whether MyType is a type or not. A specialization can easily declare a method called MyType instead of it being a type. And neither can the compiler delay this decision until the instantiation point, because whether MyType is a type or not affects how the rest of the definition is parsed. So we must tell the compiler explicitly, at the point of definition, whether MyType is a type or not. It turns out that the default is "not a type", and we must precede the name with typename to tell the compiler it is a type. This is stated in the C++ standard, section 14.6: A name used in a template declaration or definition and that is dependent on a template-parameter is assumed not to name a type unless the applicable name lookup finds a type name or the name is qualified by the keyword typename.

Disambiguating dependent template names While we're at it, the following is yet another example of explicit disambiguation that is sometimes required to guide the compiler when templates and specializations are involved: struct Foo { template < typename U> static void foo_method() { } }; template < typename T> void func(T* p) { // A. error: expected primary-expression before ‘>’ token // T::foo_method<T>(); // B. works! T:: template foo_method<T>(); } The first attempt to call T::foo_method fails - the compiler can't parse the code. As explained before, when a dependent name is encountered, it is assumed to be some sort of identifier (such as a function or variable name). Previously we've seen how to use the typename keyword to explicitly tell the compiler that it deals with a type. So in declaration (A) above can't be parsed, because the compiler assumes foo_method is just a member function and interprets the < and > symbols as comparison operators. But foo_method is a template, so we have to notify the compiler about it. As declaration (B) demonstrates, this can be done by using the keyword template .