The typedef declaration in C++ cannot be a template. However, C++11 added an alternative syntax using the using declaration to allow parametrized type aliases: template <typename T> using Bar = typename Foo<T>::Bar; Now you can use: Bar<int> x; // is a Foo<int>::Bar
First of all, this is how you should provide a definition for member functions of a class template: #include “LinkedArrayList.h” template<typename ItemType> void LinkedArrayList<ItemType>::insert (int index, const ItemType& item) {} template<typename ItemType> ItemType LinkedArrayList<ItemType>::remove (int index) {return ItemType();} template<typename ItemType> int LinkedArrayList<ItemType>::find (const ItemType& item) {return -1;} Secondly, those definitions cannot be put in a … Read more
If you slightly change your declaration, you get an entire different story template <class T> void foo(T::type& v); That isn’t unambiguous anymore. It could declare a variable of type void that is initialized by a bit-wise AND expression. The entire declaration would be templated. Of course, this semantically is all nonsense, but it syntactically is … Read more
Short answer: because the Standard says so. Longer answer: prior to Standardization, C++ templates required the class keyword for all template parameters. However, to stress the fact that templates could also be of non-class (i.e. builtin) type, an alternative keyword typename was introduced. However, in C++98, template-template parameters could only be of class-type, and this … Read more
Define the function in the .h file. Works fine for me a.h #include <vector> #include <iostream> using namespace std; class A { public: template< typename T> static double foo( vector<T> arr ); }; template< typename T> double A::foo( vector<T> arr ){ cout << arr[0]; } main.cpp #include “a.h” int main(int argc, char *argv[]) { A … Read more
namedtuple() is a factory function for tuple subclasses. Here, ‘whatsmypurpose’is the type name. When you create a named tuple, a class with this name (whatsmypurpose) gets created internally. You can notice this by using the verbose argument like: Point=namedtuple(‘whatsmypurpose’,[‘x’,’y’], verbose=True) Also you can try type(p) to verify this.
In list<tNode<T>*>::iterator, you have a dependant name, that is, a name that depends on a template parameter. As such, the compiler can’t inspect list<tNode<T>*> (it doesn’t have its definition at this point) and so it doesn’t know whether list<tNode<T>*>::iterator is either a static field or a type. In such a situation, the compiler assumes that … Read more
The typename is required by the standard. Template compilation requires a two step verification. During the first pass the compiler must verify the template syntax without actually supplying the type substitutions. In this step, std::map::iterator is assumed to be a value. If it does denote a type, the typename keyword is required. Why is this … Read more
Short answer: Whenever referring to a nested name that is a dependent name, i.e. nested inside a template instance with unknown parameter. Long answer: There are three tiers of entities in C++: values, types, and templates. All of those can have names, and the name alone doesn’t tell you which tier of entity it is. … Read more
One of the new features in C++20 is Down with typename. In C++17, you had to provide the typename keyword in nearly all†dependent contexts to disambiguate a type from a value. But in C++20, this rule is relaxed a lot. In all contexts where you need to have a type, the typename keyword is … Read more