Accessing class member pointers - c++

Question

Suppose I have the following definition of List and Node:
template <class T>
class List {
public:
class Iterator;
class ConstIterator;
//Constructors and Destructors.
List() : head(NULL), tail(NULL), size(0) {}
List(const List& list);
~List();
//Methods
Iterator begin();
ConstIterator begin() const;
Iterator end();
ConstIterator end() const;
void insert(const T& data);
void insert(const T& data, const Iterator& iterator);
void remove(const Iterator& iterator);
int getSize() const;
Iterator find(const T& item);
ConstIterator find(const T& item) const;
void sort();
//Operators
List operator = (const List& list);
private:
class Node;
Node* head;
Node* tail;
int size;
};
template <class T>
class List<T>::Node
{
public:
//Constructors and destructors
Node(const T& _data, const Node* _next) : data(_data), next(_next) {}
~Node(); //Destructor
//Methods
//Operators
Node operator = (const Node& node);
private:
T data;
Node* next;
};
I'm writing a function to insert data into a list like this:
template<class T>
void List<T>::insert(const T& data)
{
Node newNode = new Node(data, NULL);
if (head == NULL)
{
head = &newNode;
tail = &newNode;
}
else
{
(*tail)->next = &newNode;
tail = &newNode;
}
size++;
}
However what I find strange is that if I swap (*tail)->next = &newNode; to (*tail).next = &newNode; it still compiles. Why, and what is the correct way of doing it?

Answer 1

The definitions of your classes can be (for the purposes of this question) simplified into:
class List {
...
private:
Node* head;
Node* tail;
};
class Node {
...
private:
Node* next;
};
Now in your List::insert method:
Node newNode = new Node(data, NULL);
(*tail)->next = &newNode;
...when you use new expression, the result will be pointer to the newly allocated memory.
What you should do is:
Node* newNode = new Node(data, NULL);
tail->next = newNode; // <-- equivalent to (*tail).next = newNode;

Answer 2

Using Node->tail is short form of writing (*Node).tail. Both forms are valid. Strangeus is the fact that you say that (*Node)->tail compiles. To this happens, Node must be defined as a double pointer, i.e.:
Node **tail;
But your code has some others bugs in. In this line:
Node newNode = new Node(data, NULL);
you are define a local object and assing a dynamic memory to it. The correct way is:
Node *newNode = new Node(data, NULL); // defining it as a pointer
and instead of assing as:
head = &newNode;
do:
head = newNode;
As a final note, consider using smart pointer instead of raw pointer. The former is safer than the last

Answer 3

The -> operator will automatically derefference a pointer for you then call the method to the right. So:
tail->next
would also work but
tail.next
wouldn't because tail is a pointer. To use the . operator you have to defrence the pointer first as in
(*tail).next
(*tail)
turns your pointer into an object. At that point you can use either -> or .
A . will not work on a pointer but -> will.
Generally, just for easy of typing I use -> because it is shorter then using (*) to turn a pointer into an object just so I can use a dot but they are equivalent operations.

Answer 4

You have noticed that (*tail)->next = &newNode and (*tail).next = &newNode both compile, which strikes you as odd.
But somehow you might also have noticed that this line also compiles!
Node newNode = new Node(data, NULL);
That is the thing that you should give you pause.
You are inside of a template here. Lots of things "compile".
Did you try instantiating the template?
ADDENDUM:
Here just to show you how crazy things can be, check out this program:
#include <iostream>
using namespace std;
template <class T>
class C {
void f();
};
template <class T>
void C<T>::f() {
int x = new int;
}
int main() {
std::cout << "Hello, world\n";
}
Now check this out:
$ g++ template-example.cpp && ./a.out
Hello, world
But now notice
#include <iostream>
using namespace std;
int main() {
int x = new int;
std::cout << "Hello, world\n";
}
which yields:
$ g++ hello.cpp
hello.cpp: In function ‘int main()’:
hello.cpp:4: error: invalid conversion from ‘int*’ to ‘int’
TL;DR: WHEN YOU ARE IN A TEMPLATE, THINGS THAT SHOULD NOT COMPILE SOMETIMES "DO"! (They're not really compiling -- YET!)