CODEWITHSUNDEEP
c++data-structuresoperator-overloading
killometers90
killometers90

Reputation: 25

How does operator++ overloading work to iterate over my LinkedList

This is my LinkedList namespace

namespace LinkedList {     
    template<class T>
    class Node{
    public:
        Node(const T& data)
            :data_(data), next_(nullptr) {}
        T data_;
        Node<T> *next_;

        Node<T> *operator++(){
            return next_;
        }
    };


    

    template<class T>
    class LinkedList{
    public:
        LinkedList()
            : head_(nullptr), tail_(nullptr) {}
        
        ~LinkedList(){
            Node<T> *curr = head_;
            Node<T> *next;
            while(curr != nullptr){
                next = curr->next_;
                delete curr;
                curr = next;
            }
        }
        void Append(const T& data) {
            Node<T> *tmp = new Node<T>(data);
            if(head_ == nullptr) {
                head_ = tmp;
                tail_ = tmp;
            } else if(head_ == tail_){
                head_->next_ = tmp;
                tail_ = tmp;
            } else {
                tail_->next_ = tmp;
                tail_ = tmp;
            }
        }

        void Present(){
            for(Node<T> *curr = head_; curr != nullptr; curr=curr->next_){
                std::cout << curr->data_ << std::endl;
            }
        }

        Node<T> *begin(){
            return head_;
        }

        Node<T> *end(){
            return nullptr;
        }
    private:
        Node<T> *head_;
        Node<T> *tail_;
    };
}

I was reading into Iterators and wanted to make my LinkedList object compatible with range based for loops. This is what I got from reading about range-based for loops

for(auto x: list){ }

is equivalent to

for(; begin != end; ++begin) { //*begin};

I thought I was being cheeky and could skip a couple steps of operator overloading (!=, *) for my LinkedList to work with range based for loops by programming my iterator into my Node class for no other reason besides that I felt like it. This is only my second day of learning C++ so this might be a stupid question but I am still a little confused on why this doesn't work:

LinkedList::LinkedList<int> list;
list.Append(3);
list.Append(5);
for(auto x: list) { //blah blah blah}

I know that my prefix increment is the core issue, when the for loop does ++__begin, what I am hoping it does is ___begin=_begin->next however it does not. Why is this? How the operator overloading works in my mind is is that whatever member variables I reference in the overloading function, it is referencing from the instance I am operating on. And when I return the ptr, it is setting whatever instance I am operating on to this ptr. I know my understanding of this is wrong because it doesn't work, so please someone explain to me how it actually works lol.

Upvotes: 0

Views: 64

Answers (1)

Useless
Useless

Reputation: 67713

You've written a

template<class T>
Node<T>* Node<T>::operator++();

which would get invoked like

Node<int> n;
Node<int> *p = ++n; // calls the pre-increment operator

Note this is quite different to the canonical form

Node<T>& Node<T>::operator++();

which returns a reference to the incremented object that you can use in an expression.

But there's another problem: operator overloading works

When an operator appears in an expression, and at least one of its operands has a class type or an enumeration type, ...

But Node<T>* is not a class type. It's a pointer, and pointers already have their own built-in operators. You can't overload them. There is no Node<T>*::operator++().

Write an iterator. It's not that bad, and they actually work!

NB. Iterators are designed to generalize pointers, and to have a compatible interface. You can use raw pointers if your container is basically an array - because incrementing a pointer is already the right way to iterate over an array.

It's only when we want to provide array-like or pointer-like access to a non-contiguous structure that we have to do this extra work.

Upvotes: 1

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