The role of access specifiers on typedefs and forward declaration

Question

Consider the following sample code for a linked list type class. I wish to declare a method which returns an Iterator, which is a typedef for a Node*. However, Node is a private nested class so in order to make the typdef, i am required to let the compiler know about Node with a forward declaration.

Naively, I thought having both defaulted to private would work; something like this:

class List
{
    class Node;
    typedef Node* Iterator;
public:
   List() : head_(NULL), tail_(NULL) {} 
   Iterator begin() {return head_;}

private:
    class Node
    {
        private:
        int data_;
    };

    Node* head_;
    Node* tail_;
};

int main()
{
    List list;
    List::Iterator = list.begin();
    return 0;
}

which results in a compile time error at line 4:

'typedef class List::Node* List::Iterator' is private
compilation terminated due to -Wfatal-errors.

It isn't difficut to see why the forward declaration class Node; belongs in the private section, but what about typedef Iterator Node*;? This probably owes to my lack of understanding for the typedef keyword, but why does it matter which access specifier is applied? I thought private makes more sense because of the visibility of the Node class.

Does this have something to do with typedefs being part of the public interface? Do all typedefs have to be declared with public visibility?

Edit: Allow me to clarify, I understand that I can stop my compiler from complaining by moving the typedef to public. What I don't understand is why this is necessary.

If I add a semi-colon at the end of the class definitions, I don't get any error. Is this your real code? — Joseph Mansfield, Feb 27 '14 at 14:34
Ah, so it does.. It's just a simplified version. I'll see if i can recreate the error — StickyCube, Feb 27 '14 at 14:38
@StickyCube - maybe you were trying to use `Iterator` type from a non-friend class/function? — Kiril Kirov, Feb 27 '14 at 14:39
Ok, i added a little implementation. I get the error again now — StickyCube, Feb 27 '14 at 14:45

score 1 · Answer 1 · answered Feb 27 '14 at 14:40

1

You can simply make typedef name a public name. Try the following

class List
{
public:
    typedef class Node* Iterator;

Also take into account that for both classes you forgot to place semicolons after closing braces.:)

answered Feb 27 '14 at 14:40

Vlad from Moscow

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score 1 · Answer 2 · answered Feb 27 '14 at 14:40

List::Iterator is private, so the name cannot be used, but the type can be deduced

List l;
List::Iterator it = l.begin(); // Illegal
auto it = l.begin(); // legal

or

template <typename IT>
void foo(IT it);

foo(l.begin()); // legal

BTW, you may have List::Iterator public, and keep List::Node private.

The role of access specifiers on typedefs and forward declaration

2 Answers2