对类的指针使用重载的<<运算符

Question

I'm defining a LinkedList class, which makes use of a separately-defined Node class. Each Node has a pointer to the next node.

In the Node class, I have friend ostream& operator<<(ostream& ostr, const Node<T>& m);

In the LinkedList class, I have head, which is a pointer to the first Node.

How can I define a print function for LinkedList that will iteratively call the overloaded << operator for each node in the list? SO far I have

void LinkedList::print() {
  Node* search = head;
  while(search) {
    cout << //print stuff here
    search = search->getNext();
  }
  cout << endl;
}

Answer 1

Fisrtly I would like to refer you to a developed question:

Operator overloading on class templates

Secondly, here some stuff I came up with:

By writing:

friend ostream& operator<<(ostream& ostr, const LinkedList<T>& m);

in class declaration you informed the compiler that you are going to use overloading of operator << to do something with the class(print it). Now you have to define the function.

Please note that it is a global function, not a class method.

Also I "fixed" two bugs in your code. One in the function declaration, in which the LinkedList should be an argument. Intuitive => Because you want to print the list not the node.

Second was in first line in which you have to take the m's head member's adress. In your "C'ish" implementation it is necessary.

ostream& operator<<(ostream& ostr, const LinkedList<T>& m) {
Node* search = &(m.head); // You have to take the adrress of the head
while(search) {
   ostr << //print stuff here
   search = search->getNext();
}
ostr << endl;
return ostr;
}

This might seem a quite odd to return the stream but it actually makes sense in statement like:

LinkedList<Banana> bananas;
LinkedList<Apple> apples;

cout << bananas << apples;

Because bananas takes cout , uses it and then return s it to be used by apples .

Hope I didn't obfuscated it too much or made any mistakes.