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c++templatesusingtypename
Wilson
Wilson

Reputation: 511

Confused about C++ nested dependent type name

Effective C++ told me that I'd better use typename when encountered nested dependent type name.

The following example code is easy to understand:

template <typename ElementType>
class BST {
private:
  class LinkNode {
  public:
    ElementType data;
    LinkNode *left, *right;
    explicit LinkNode() {}
  };

public:
  void some_func();
}

template <typename ElementType>
void BST<ElementType>::some_func() {
  // or `using NodePtr = typename BST<ElementType>::LinkNode *;`
  typedef typename BST<ElementType>::LinkNode * NodePtr;
  ...
}

However, after I added using aliases in the template class BST, it seemed that keyword typename is not neccessary anymore.

Here you can see:

template <typename ElementType>
class BST {
private:
  class LinkNode {
  public:
    ElementType data;
    LinkNode *left, *right;
    explicit LinkNode() {}
  };

  using NodePtr = LinkNode *; // the only difference between these two code blocks

public:
  void some_func();
}

template <typename ElementType>
void BST<ElementType>::some_func() {
  // typename is not neccessary here!
  BST<ElementType>::NodePtr ptr;
  ...
}

Does anyone could figure out that?

Upvotes: 4

Views: 379

Answers (1)

Jans
Jans

Reputation: 11250

That effect is not directly tied to type alias through using, it's a result of name lookup for member of the current instantiation.

Inside BST both BST and BST<ElementType> expression refer to the current instantiation and the members of it can be found without the need of the prefix typename you could do:

template <typename ElementType>
void BST<ElementType>::some_func() {
    BST::NodePtr ptr; // or 
    BST<ElementType>::LinkNode * ptr2;
}

resulting in the same thing. But now let assume that some_func is also a template member function defined as:

template <typename ElementType>
struct BST {
    class LinkNode { /*...*/ };

    using NodePtr = LinkNode *;

    template <typename T>
    void some_func();
};


template <typename ElementType>
template <typename T>
void BST<ElementType>::some_func() {
    BST<T>::NodePtr ptr;     // (1)
    BST<T>::LinkNode * ptr2  // (2)
}

Now neither (1) nor (2) will compile because B<T> is no longer the current instantiation, hence in these cases you need typename.

The relevant part of the standard [temp.res]/7:

Within the definition of a class template or within the definition of a member of a class template following the declarator-id, the keyword typename is not required when referring to the name of a previously declared member of the class template that declares a type or a class template. [...]

Upvotes: 1

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