Copy constructor and dynamic allocation - c++

I would like to ask you how to write a copy constructor (and operator = ) for the following classes.
Class Node stores coordinates x,y of each node and pointer to another node.
class Node
{
private:
double x, y;
Node *n;
public:
Node (double xx, double yy, Node *nn) : x(xx), y(yy), n(nn) {}
void setNode (Node *nn) : n(nn) {}
...
};
Class NodesList (inherited from std:: vector) stores all dynamically allocated Nodes
class NodesList : public std::vector<Node *>
{}
The main program:
int main()
{
Node *n1 = new Node(5,10,NULL);
Node *n2 = new Node(10,10,NULL);
Node *n3 = new Node(20,10,NULL);
n1->setNode(n2);
n2->setNode(n3);
n3->setNode(n2);
NodesList nl1;
nl1.push_back(n1);
nl1.push_back(n2);
nl1.push_back(n3);
//Copy contructor is used, how to write
NodesList nl2(nl1);
//OPerator = is used, how to write?
NodesList nl3 = nl1;
}
I do not want to create a shallow copy of each node but a deep copy of each node. Could I ask you for a sample code with copy constructor?
Each node can be pointed more than once. Let us have such situation, when 3 nodes n[1], n[2], n[3] are stored in the NodesList nl1:
n[1] points to n[2]
n[2] points to n[3]
n[3] points to n[2]
A] Our copy constructor process the node n[1]. It creates a new object n[1]_new represented by the copy of the old object n[1]_old. The node n[2] pointed from n[1]_old still does not exist, so n[2]_new must be also created... The pointer from n1_new to n2_new is set.
B] Then second point n[2] is processed. It can not be created twice, n[2]_new was created in A]. But pointed node n[3] does not exist, so the new object n[3]_new as a copy of an old object n[3]_old is created. The pointer from n2_new to n3_new is set.
C] Node n[3]_new has already been created and n[2]_new. The pointer from n3_new to n2_new is set and no other object will be created...
So the copy constructor should check whether the object has been created in the past or has not...
Some reference counting could be helpful...

There is my solution of the problem. A new data member n_ref storing a new verion of the node n was added:
class Node
{
private:
double x, y;
Node *n, *n_ref;
public:
Node (double xx, double yy, Node *nn) : x(xx), y(yy), n(nn) {n_ref = NULL;}
Node * getNode() {return n;}
Node * getRefNode () {return n_ref;}
void setNode (Node *nn) {this->n = nn;}
void setRefNode (Node *nn) {this->n_ref = nn;}
The copy constructor creates a shallow copy of the node:
Node (const Node *node)
{
x = node->x;
y = node->y;
n = node->n;
n_ref = node->n_ref;
}
The copy constructor for NodesList
NodesList::NodesList(const NodesList& source)
{
const_iterator e = source.end();
for (const_iterator i = source.begin(); i != e; ++i) {
//Node* n = new Node(**i);
//Node n still has not been added to the list
if ((*i)->getRefNode() == NULL)
{
//Create node
Node *node = new Node(*i);
//Add node to the list
push_back(node);
//Set this note as processed
(*i)->setRefNode(node);
//Pointed node still has not been added to the list
if ((*i)->getNode()->getRefNode() == NULL)
{
//Create new pointed node
Node *node_pointed = new Node ((*i)->getNode());
//Add node to the list
push_back(node_pointed);
//Set pointer to n
node->setNode(node_pointed);
//Set node as processed
((*i)->getNode())->setRefNode(node_pointed);
}
//Pointed node has already been added to the list
else
{
//Set pointer to node n
node->setNode((*i)->getRefNode());
}
}
//Node n has already been added to the list
else
{
//Get node
Node * node = (*i)->getRefNode();
//Pointed node still has not been added
if ((*i)->getNode()->getRefNode() == NULL)
{
//Create new node
Node *node_pointed = new Node ((*i)->getNode());
//Add node to the list
push_back(node_pointed);
//Set pointer to n
node->setNode(node_pointed);
//Set node as processed
((*i)->getNode())->setRefNode(node_pointed);
}
//Pointed node has already been added to the list
else
{
//Set pointer to n
node->setNode((*i)->getNode()->getRefNode());
}
}
}
}

Perform a shallow copy in NodeList::NodeList(const NodeList&) and you don't have to worry about cycles breaking the copy operation. Disclaimer: the following is untested, incomplete and may have bugs.
class NodeList {
private:
typedef std::vector<Node*> Delegate;
Delegate nodes;
public:
NodeList(int capacity=16) : nodes() { nodes.reserve(capacity); }
NodeList(const NodeList& from);
virtual ~NodeList();
NodeList& operator=(const NodeList& from);
/* delegated stuff */
typedef Delegate::size_type size_type;
typedef Delegate::reference reference;
typedef Delegate::const_reference const_reference;
typedef Delegate::iterator iterator;
typedef Delegate::const_iterator const_iterator;
size_type size() const { return nodes.size(); }
iterator begin() { return nodes.begin(); }
const_iterator begin() const { return nodes.begin(); }
iterator end() { return nodes.end(); }
const_iterator end() const { return nodes.end(); }
// ...
};
NodeList::NodeList(const NodeList& from)
: nodes(from.size()), flags(NodeList::owner)
{
std::map<Node*, Node*> replacement;
Delegate::const_iterator pfrom;
Delegate::iterator pto;
// shallow copy nodes
for (pfrom=from.begin(), pto=nodes.begin();
pfrom != from.end();
++pfrom, ++pto)
{
replacement[*pfrom] = *pto = new Node(**pfrom);
}
// then fix nodes' nodes
for (pto = nodes.begin(); pto != nodes.end(); ++pto) {
(*pto)->setNode(replacement[(*pto)->getNode()]);
}
}
NodeList::operator=(const NodeList&) can use the copy-swap idiom, the same as Tronic's Node::operator=(const Node&).
This design has a potential memory leak in that a copied NodeList is (initally) the only place that references its nodes. If a temporary NodeList goes out of scope, a poor implementation will leak the Nodes the list contained.
One solution is to proclaim that NodeLists own Nodes. As long as you don't add a Node to more than one NodeList (via NodeList::push_back, NodeList::operator[] &c), NodeList's methods can delete nodes when necessary (e.g. in NodeList::~NodeList, NodeList::pop_back).
NodeList::~NodeList() {
Delegate::iterator pnode;
for (pnode = nodes.begin(); pnode != nodes.end(); ++pnode) {
delete *pnode;
}
}
void NodeList::pop_back() {
delete nodes.back();
nodes.pop_back();
}
Another solution is to use smart pointers, rather than Node*. NodeList should store shared pointers. Node::n should be a weak pointer to prevent ownership cycles.

I would just use std::list<Node> instead of NodesList. Well, let's code...
NodesList::NodesList(const NodesList& source)
{
const_iterator e = source.end();
for (const_iterator i = source.begin(); i != e; ++i) {
Node* n = new Node(**i);
push_back(n);
}
}

Apparently each Node is only allowed to point to another Node in the same list? Otherwise the "deep copy" of a list needs more definition. Should it not be connected to the original NodeList? Should it not be connected to any original Node? Are copies of Nodes not in the list being copied added to some other list or free-floating?
If all the Node-to-Node pointers are constrained within the NodeList, then perhaps you should store indexes instead of pointers, then no special handling is required.

You should not inherit from standard library containers (because they lack virtual destructors). Instead, include them as member variables in your classes.
Since you want a deep copy, you need these: (rule of three)
Node(Node const& orig): x(orig.x), y(orig.y), n() {
if (orig.n) n = new Node(*orig.n);
}
Node& operator=(Node const& orig) {
// The copy-swap idiom
Node tmp = orig;
swap(tmp); // Implementing this member function left as an exercise
return *this;
}
~Node() { delete n; }
A better idea might be to avoid using pointers entirely and just put your nodes in a suitable container.

Related

deep copy in struct pointers

in this code, I think it must do deep copy because I'm passing pointers, but it doesn't.
I think it must print 3, but print 0. what should I do to solve this? i want to have a deep copy instead of a shallow copy.
struct node{
int number = 0;
struct node* right_child = NULL;
struct node* left_child = NULL;
};
void test(struct node* node1 , struct node* node2){
node1 = node2;
}
int main(){
struct node* a1 = new struct node;
struct node* a2 = new struct node;
a2->number = 3;
test(a1 , a2);
cout << a1->number;
}
The simple C-ish way: We ad a function that recursively clones the nodes
#include <iostream>
struct node{
int number = 0;
node* right_child = nullptr; // nullptr safer than NULL
node* left_child = nullptr;
};
node * clone(const node * src){
if (src) { // there is a node. Copy it.
return new node{src->number,
clone(src->right_child), // clone right
clone(src->left_child)}; // clone left
}
else { // no node. Nothing to do and end the branch
return nullptr;
}
}
void test(node*& node1, // reference to pointer
node* node2){
delete node1; // Free storage currently used by node1
// But what of the nodes that could be in its tree?
// Think on this. Ask another question if necessary.
// this is where the simple way starts to fall down
node1 = clone(node2);
}
int main(){
struct node* a1 = new node; // wasted memory. We're about to replace it.
// need a more complicated tree to prove deep copy works
struct node* a2 = new node{3,
new node{4, nullptr, nullptr}, // a right node
nullptr}; // no left node
// a2->number = 3; done above now
test(a1 , a2);
std::cout << a1->number << ' '
<< a1->right_child->number;
delete a1; // clean up
delete a2;
}
More complicated canonical C++ way using The Rule Of Three and The Copy and Swap Idiom
#include <iostream>
#include <algorithm> // for std::swap
struct node{
int number = 0;
node* right_child = nullptr; // nullptr safer than NULL
node* left_child = nullptr;
static node * clone(node * src)
{
if (src)
{
return new node(*src);
}
return nullptr;
}
// generic constructor
node(int number = 0,
node* right_child = nullptr,
node* left_child = nullptr):
number(number),
right_child(right_child),
left_child(left_child)
{
}
//copy constructor
node (const node & src):
number(src.number),
right_child(clone(src.right_child)),
left_child(clone(src.left_child))
{
}
// assignment operator via copy and swap.
node & operator=(node src)
{
std::swap(number, src.number);
std::swap(right_child, src.right_child);
std::swap(left_child, src.left_child);
return *this;
}
// destructor
~node()
{
delete right_child;
delete left_child;
}
};
void test(node* node1,
node* node2){
*node1 = *node2; // now, thanks to all of the infrastructure above, we can
// assign nodes with the dumb old = operator. All we have to do
// is dereference the pointers.
}
int main(){
struct node* a1 = new node; // wasted memory. We're about to replace it.
// need a more complicated tree to prove deep copy works
struct node* a2 = new node{3,
new node{4, nullptr, nullptr}, // a right node
nullptr}; // no left node
// a2->number = 3; done above now
test(a1 , a2);
std::cout << a1->number << ' '
<< a1->right_child->number;
delete a1; // clean up
delete a2;
}
All of the nodes are now self-managing.
But, my opinion, the nodes should be kept as dumb as they are in the simple example. They shouldn't need to know about the tree at all. Knowledge of the tree should be in a Tree class that uses the nodes as simple containers. The Tree class should do all of the management of the nodes--creating, copying, cloning, deleting--because only it knows everything necessary to represent the tree. The users of the tree shouldn't even know nodes exist. They put data into the tree, take data out of the tree, and observe the tree without caring how the tree works and being able to so something stupid like
delete a_node;
and blow the crap out of data the tree isn't done with yet.
The Tree class preferably works iteratively rather than recursively so that the trees can be arbitrarily sized. Unless you're really careful recursing through a large tree to clone it, delete it, display it, or pretty much anything else you do with a tree runs the risk of exhausting automatic storage and causing what's typically called a Stack Overflow.
just use
void test(struct node *node1, struct node *node2) { *node1 = *node2; }
instead of
void test(struct node *node1, struct node *node2) { node1 = node2; }
and it will print 3.
This is because...
when you do node1 = node2; int the test1, you assign the pointer itself, not the structure pointed to by the pointer.When the function ends, the parameters node1 and node2 will be destroyed, so you have done nothing...

Iterate over elements where not all the elements should have a value

I'm not sure how to describe this. I have to iterate through objects which are connected by pointers with each other, However the first element shouldn't have values stored in its class just the pointer to the next element.
I came up with a small class hierarchy. If I use it like that I have to cast the base class to the derived element. It only seems to work with a dirty cast.
Is there a better solution?
Here the example code:
#include <iostream>
struct Basenode {
Basenode* next;
};
struct Skipnode : Basenode {
Skipnode(int in_key, int in_value);
int key;
int value;
};
Skipnode::Skipnode(int in_key, int in_value)
: key{ in_key }, value{ in_value }
{
}
int main()
try {
Basenode head; // no key and value
Skipnode first(4, 2); // key and value
Skipnode second(8, 2);
Basenode* p = &head;
head.next = &first; // fine
first.next = &second; // fine
// p = p->next->key; // not accesible because is Basenode not derrived Skipnode
std::cout << static_cast<Skipnode*>(p->next)->key; // fine but diryt cast slicing ?
std::cin.get();
}
catch (std::runtime_error& e) {
std::cerr << e.what() << "\n";
std::cin.get();
}
catch (...) {
std::cerr << "unknown error " << "\n";
std::cin.get();
}
Edit: it was asked int the comments why i need this anyway. I think i gave a to limited example.
I need it to implement a skiplist. Many algorithms of it require to start on a element before the first element. The head element. I could make it a normal node and put in dummy values for the values but it didnt seem right. So now i came up with this ugly solution from youre suggestions to start on the head elements.
Heres a snippet with the copy constructor as example.
class Skiplist {
public:
//...
Skiplist(const Skiplist& other); // copy constructor
//...
private:
struct Skipnode; // forward declaration so Basenode can have Skiplist*
struct Basenode { // Empty node, mainly created to represent head element.
// Is there a way to get a empty head with no key / values without using this ?
Basenode(int in_level);
Basenode(const std::vector<Skipnode*>& in_next);
std::vector <Skipnode*> next;
};
struct Skipnode : Basenode { // derived so with Basenode* we can start the iteration of the node on head
Skipnode(value_type val, int in_level);
Skipnode(value_type val, const std::vector<Skipnode*>& in_next);
value_type value; // first key / second mapped type = value
//key_type key;
//mapped_type value;
};
Basenode head{ 0 }; // element before first element containg pointers to all the first elements of each level
//...
};
Skiplist::Skiplist(const Skiplist& other) // copy constructor
:head{ other.head }, top_level{ other.top_level }, random_engine{ other.random_engine }
// on the first level let the other Skiplist present its elements and make a deep copy of them
// now still the higher levels point to the other node so this is fixed in the second part
// then the next level pointers are installed linked to the elements of the new node
{
if (top_level == 0) return; // no elements are present so dont bother to allocate nodes
{
// installment of lowest level, each element is located here
Skipnode* other_node = other.head.next[0];
Basenode* current_position = &head;
while (other_node != nullptr) {
Skipnode* new_node = new Skipnode{ other_node->value,other_node->next };
current_position->next[0] = new_node;
current_position = current_position->next[0];
other_node = other_node->next[0];
}
current_position->next[0] = nullptr;
}
// installment of the other levels
for (size_type curr = 1; curr < top_level; ++curr) {
Basenode* current_position = &head; // the current position of the level[curr]
Skipnode* next_position = current_position->next[curr]; // next position after curr containing still pointers to the other skiplist
Basenode* lowest_position = &head; // lowest level position used to find the new pointers and attach them "behind" current
while (lowest_position != nullptr && next_position != nullptr) {
if (lowest_position->next[0]->value.first == next_position->value.first) { // check by unique key, address of next pos is still of the other skiplist
current_position->next[curr] = lowest_position->next[0]; // lowest is the valid address of new node
current_position = current_position->next[curr];
next_position = next_position->next[curr]; // go to next element of other node
if (next_position == nullptr) { // case end is reached
current_position->next[curr] = nullptr;
current_position = current_position->next[curr];
}
}
else { // forward position of lowest level until other key == next position key
lowest_position = lowest_position->next[0];
}
}
}
}
See here for a basic explanation how a skiplist is organized:
https://en.wikipedia.org/wiki/Skip_list
The whole code is on codereview:
https://codereview.stackexchange.com/questions/197752/non-generic-skip-list-implementation-in-c-version-2
All the things #SomeProgrammerDude is saying, or:
I don't see a need for class BaseNode at all. Why can't we just have (all other things being equal):
SkipNode *head = &first;
...
Or better yet a class called (for example) SkipNodeList that handles all aspects of managing and iterating through a list of SkipNodes.
Of course, this is all a bit silly anyway, just use std::list (or std::forward_list) for this and benefit from all that STL goodness.
Or you can derive from one of these to add your own functionality (such as a mutex to make the list threadsafe or keeping track of the number of elements currently in the list, as suggested by #iMajuscule).
Yes, there is a better way:
Forward-declare Skipnode, and in BaseNode, use a pointer to SkipNode, this way you don't have to cast:
struct Skipnode;
struct Basenode {
Skipnode* next;
};
Also, to illustrate how this design where Skipnode inherits from Basenode could make sense (related to the discussion in the comments), we can imagine having a member in Basenode counting how many elements are next (counting the one in the next member and its successors)

Circular double linked list with smart pointers in c++

Is it possible to create a circular doubly-linked list using smart pointers in C++
struct Node {
int val;
shared_ptr<Node> next;
weak_ptr prev;
};
shared_ptr<Node> head;
But this will have a circular reference of shared pointers and thus not deallocate correctly.
Make the circular linked list a class itself (with whatever operations you need to build it, like append). Have its destructor break the link by setting tail->next = nullptr. It should not matter which link you break, so if you're not using a head and tail, just set any one of them nullptr, and you're good.
In my testing, I made a circular linked list, and the nodes did not destruct. Then at the end, I added tail->next = nullptr before it exited, and all the destructors fired correctly.
My original posted answer was rather light on details. This one gives a proper explanation of how you can achieve a circular linked list without a memory leak and still adhere to the Rule of Zero. The answer is basically the same, using a sentinel, but the mechanism is a little more involved than I had originally let on.
The trick is to use a sentinel type that behaves just like a list node, but in fact does not really have a shared pointer to the head of the list. To achieve this, the node class should be separated into a behavior object and a state object.
class NodeState {
std::shared_ptr<Node> next_;
std::weak_ptr<Node> prev_;
int value_;
NodeState (int v) : value_(v) {}
NodeState (std::shared_ptr<Node> p) : next_(p), prev_(p) {}
//...
};
class Node {
virtual ~Node () = default;
virtual NodeState & state () = 0;
std::shared_ptr<Node> & next () { return state().next_; }
std::weak_ptr<Node> & prev () { return state().prev_; }
int & value () { return state().value_; }
void insert (const std::shared_ptr<Node> &p) {
//...
}
};
Now, you can define a node implementation and a sentinel implementation.
class NodeImplementation : public Node {
NodeState state_;
NodeState & state () { return state_; }
NodeImplementation (int v) : state_(v) {}
//...
};
class NodeSentinel : public Node {
List &list_;
NodeSentinel (List &l) : list_(l) {}
NodeState & state () { return list_.sentinel_state_; }
};
The list itself contains a NodeState used by the sentinel object. Upon initialization, the list creates a sentinel object and initializes its state.
class List {
//...
NodeState sentinel_state_;
std::shared_ptr<Node> head () { return sentinel_state_.next_; }
std::shared_ptr<Node> sentinel () {
return std::shared_ptr<Node>(head()->prev());
}
//...
public:
List () : sentinel_state_(std::make_shared<NodeSentinel>(*this)) {}
//...
void push_front (int value) {
head()->insert(std::make_shared<NodeImplementation>(value));
}
void push_back (int value) {
sentinel()->insert(std::make_shared<NodeImplementation>(value));
}
//...
};
So, what does this organization do? It avoids the issue of a circular reference by using a sentinel node to act as the break. While the tail of the list points to the sentinel object, the sentinel object itself does not point to anything. Instead, it uses the state of the list itself to determine its next and previous neighbors.
Thus, the circular shared pointers only persists as long as the list exists. Once the list is destroyed, Item A loses its reference, and via the domino effect, Sentinel itself will be destroyed.
A fundamental point is that the sentinel object itself must never be exposed to the user of the list interface directly. It should remain internal to the list object at all times. It essentially represents end() in an STL like container, and logically, it can never be removed from the list (until the list itself is destroyed). In practice, this means removal operations on the list need to exit early if the passed in iterator represents the sentinel.
Demo
Try It Online
It is also possible to define a member function next() which can select between a shared or weak pointer.
#include <iostream>
#include <memory>
using namespace std;
struct T {
int n_;
shared_ptr<T> next_;
weak_ptr<T> weaknext_;
T(shared_ptr<T> next, int n) : next_(next), n_(n) {};
auto next() {
if (next_ == nullptr)
return shared_ptr<T>(weaknext_);
return next_;
}
~T() { cout << n_ << "ok\n"; }
};
int main() {
auto p0 = make_shared<T>(nullptr, 1);
auto p1 = make_shared<T>(p0, 2);
auto p2 = make_shared<T>(p1, 3);
p0->weaknext_ = p2; //makes the list circular
auto p = p2;
for (int i = 0; i < 5; ++i) {
cout << p->n_ << "\n";
p = p->next();
}
}

C++ pointers not working?

I have a problem with working with c++ pointers. I'm trying to code a splay tree by using a Node struct and Tree struct. However, upon testing, I have encountered a problem. The portion of my code that's not working is below:
struct Node {
Node* l, *r, *p;
int v;
Node() {}
Node(int _v, Node* _p) : v(_v), p(_p) {}
};
struct Tree {
Node* root;
Tree() : root(0) {}
//...
void insert(int k) {
if (!root) {
root = new Node(k, 0);
return;
}
Node* cur = new Node();
cur->v = root->v;
while (1) {
int x = cur->v;
cout << x << endl;
return;
if (k <= x) {
//cout << x << endl;
//return;
if (!cur->l) {
cur->l = new Node(k, cur);
//splay(cur->l);
return;
} else cur = cur->l;
} else {
if (!cur->r) {
cur->r = new Node(k, cur);
//splay(cur->r);
return;
} else cur = cur->r;
}
}
}
//...
};
int main() {
Tree t = Tree();
t.insert(1);
t.insert(5);
return 0;
}
First, I inserted a node with value 1 in the tree; since there was no root, the tree assigned its root as a new node with value 1. Then, when I inserted 5 into the tree, something weird happened. If you leave the code like it is (keeping the first cout), then it will print out 1 for x. However, if you comment out the first cout and return and uncomment the second cout and return, you'll find that it prints out a random garbage number for x, even though no modifications were made. Can somebody tell me what's wrong?
C++ does not initialize class members automatically.
struct Node {
Node* l, *r, *p;
int v;
Node() {}
Node(int _v, Node* _p) : v(_v), p(_p) {}
};
When you create a new node in your code C++ allocates a piece of memory for the Node but it will not clear it. So the values of l, r & p will be whatever was there.
In your algorithm the tests: if (!cur->r) & (!cur->l) currently fail because there is uninitialized garbage in the nodes and not NULL.
As a result when you try to insert the second node the algorithm thinks that there is a valid node to the right of root. And tries to read the memory there and the value there which is the junk x you see. Depending on the value of the junk it may also crash for some people running the code :)
Also I'm 99.9% certain that Node* cur should be a pointer to a Node in the tree and not a new node so:
Node* cur = new Node(); cur->v = root->v; is wrong and should be Node* cur = root;
Proper Initialization -
In c++11 you can do:
struct Node {
Node* l = nullptr;
Node *r = nullptr;
Node *p = nullptr;
int v = 0;
Node() {}
Node(int _v, Node* _p) : v(_v), p(_p) {}
};
Otherwise
struct Node {
Node* l;
Node *r;
Node *p;
int v;
Node() : l(NULL), r(NULL), p(NULL), v(0){}
Node(int _v, Node* _p) : l(NULL), r(NULL), p(_p), v(_v) {}
};
You should initialize members of a class in the same order they were defined.
Now there are a lot of other things that are problematic in the code:
Tree seems to allocate lots of nodes but does not release any memory. (easiest to just use unique_ptr for l and r and root Node)
Is tree the owner of subnodes? Or should it be Node owning and allocating left and right? (goes away if you use std::unique_ptr for left and right)
You are not initializing the members in the order they are defined. This can cause all kind of errors. (since the compiler reorders initialization without telling you)
Node and Tree handle raw pointers but do not define a proper operator=, copy ctor (or delete them) (goes away if you use unique_ptr)
Tree is missing a dtor to clean allocated memory (goes away if you use unique_ptr)

Looking for help when creating a copy constructor for a LinkedList class in C++

First, I realize that there are multiple topics about this on StackOverflow. I'm having a slight bit of trouble understanding the responses in some of them. I'm creating this in hopes that someone can help me understand the process.
I apologize if this question seems relatively novice, but I am doing my best to understand it.
I'm learning about data structures and have been asked to create a LinkedList implementation file based on a header that was provided.
This is homework, so please no 'here is the exact code' type answers. Pseudocode, steps, and hints are welcome.
Here is the part of the header that I'm working on at the moment:
typedef std::string ElementType;
class LinkedList
{
private:
class Node
{
public:
ElementType data;
Node * next;
Node( )
: data( ElementType( ) ), next( NULL )
{ }
Node( ElementType initData )
: data( initData ), next( NULL )
{ }
}; // end of Node class
typedef Node * NodePointer;
public:
LinkedList( );
/* Construct a List object
Precondition: none.
Postcondition: An empty List object has been constructed.
*/
LinkedList( const LinkedList &source );
/* Construct a copy of a List object.
Precondition: None.
Postcondition: A copy of source has been constructed.
*/
~LinkedList( );
/* Destroys a List object.
Precondition: None.
Postcondition: Any memory allocated to the List object has been freed.
*/
const LinkedList & operator=( const LinkedList &rightSide );
/* Assign a copy of a List object to the current object.
private:
NodePointer first;
int mySize;
};
So far, I've created the destructor, can you check and make sure it is correct?
//Destructor
LinkedList::~LinkedList()
{
NodePointer ptr = first;
while(ptr != 0 ) {
NodePointer next = ptr->next;
delete ptr;
ptr = next;
}
first = 0;
}
Now here is the part where I'm lost...What are the basic steps of creating the copy constructor? I've finished the default constructor which was simple, but I'm a bit confused with what I should be doing on the copy constructor.
I'm also slightly confused about overloading the = operator, I assume it will be very similar to the copy constructor.
Edit
My first attempt at the copy constructor:
LinkedList::LinkedList(const LinkedList & source)
{
//create a ptr to our copy
Node * copy_node = source.first;
//where we will be inserting our copy
Node * insert_node = first;
while(copy_node != nullptr)
{
//insert our new copy node
insert_node = new Node(copy_node->data);
//move to our next node
copy_node = copy_node->next;
if(copy_node != nullptr) {
insert_node = insert_node->next;
} else {
insert_node = first;
}
//update size
mySize++;
}
}
I feel like something is missing there.
What are the basic steps of creating the copy constructor? I've finished the default constructor which was simple, but I'm a bit confused with what I should be doing on the copy constructor.
Well, you need to copy the source, obviously. If the source is a list of N nodes then you need to construct another list of N nodes, with each node being a copy of the corresponding one in the source.
So loop over the source nodes and create copies of them.
I'm also slightly confused about overloading the = operator, I assume it will be very similar to the copy constructor.
Yes, except you need to dispose of the current nodes first. However, a simple and safe way to implement assignment is copy-and-swap, so define a correct swap member:
void swap(LinkedList& other)
{
std::swap(first, other.first);
std::swap(size, other.size);
}
Then use it to implement assignment:
LinkedList& operator=(const LinkedList& source)
{
LinkedList(source).swap(*this);
return *this;
}
This creates a temporary that is a copy of source, then swaps it with *this, so the old contents of *this get destroyed by the temporary, and *this ends up with the copied data.
N.B. the return type should be non-const, it is not idiomatic to return a const-reference from an assignment operator.
The most easiest way is to implement a function which adds new nodes into the list and call it in the loop inside the constructor:
LinkedList(const LinkedList& rhs)
{
Node* node = rhs.first;
while (node) {
add(node.data);
node = node->next;
}
}
void add(ElementType data)
{
Node* node = new Node(data);
// add node somehow
// I'd recommend to keep pointer to the tail
}
Please note this implementation is not exception safe!
EDIT: added copy function for the copy constructor, the first step to the exception safety:
LinkedList(const LinkedList& rhs)
{
copy(rhs.first, first);
}
void copy(Node* src, Node*& dest)
{
// handle the head element separately
if (!src) {
return; // empty list
} else {
dest = new Node(src->data); // equivalent to first = new Node...
src = src->next;
}
// copy the rest of the list
Node* p = dest;
while (src) {
p->next = new Node(src->data);
src = src->next;
p = p->next;
}
}