I have my node defined something like:
class LLNode
{
public:
std::shared_ptr<LLNode> prev;
std::shared_ptr<LLNode> next;
std::shared_ptr<int> data;
LLNode(void)
: prev(std::shared_ptr<LLNode>(nullptr)),
next(std::shared_ptr<LLNode>(nullptr)),
data(std::shared_ptr<int>(nullptr))
{
}
LLNode(const LLNode &node)
: prev(std::shared_ptr<LLNode>(node.prev == nullptr?nullptr:new LLNode(node.prev))),
next(std::shared_ptr<LLNode>(node.next == nullptr?nullptr:new LLNode(node.next))),
data(std::shared_ptr<int>(new int(node.data)))
{
}
};
However, if I have a node which is linked to another node (which obviously will often be the case), copying node A will instantiate a copy of the next node B, which in turn will try to instantiate a copy of node A, which will try to copy node B, etc. etc. until there's a stackoverflow or memory error. This could be fixed by only instantiating a new copy of next (or prev), but then nothing linked previously (or next) to this node will be copied.
Is there a good way to copy a doubly linked list node?
You are doing the mistake that you are trying to copy the whole chain/list from a single node. That does not make that much sense to do in copy ctor of a list node. Make the copy ctor just copy the members' values, do not recurse. Copying the whole chain/list is the job for a LinkedList class.
Just set next and prev to null, regardless of the next and prev values of the node being copied. Write a separate function that copies the node and all it's children, which would be used to copy the whole list.
Related
I am new to c++ and am working on an assignment involving vectors and a doubly linked list. I am given this struct as such.
struct Node {
std::vector<T> data;
Node<T>* next;
Node<T>* prev;
Node(): next(nullptr), prev(nullptr){} };
I am now required in another class to create a constructor for this Node to be used for various methods. I understand that the struct already has an initialization list for next and prev, I think I am just overthinking what the constructor should be.
Class LinkedVector {
Node<T>* head;
LinkedVector<T>::LinkedVector(){
head = NULL;
}
}
Is this the correct way of constructing the linked list? Again I am new to c++ and any help pointing me in the right direction is most helpful. Thank you and have a great day.
Yes, that is almost correct, you want to set head to nullptr instead of NULL of course.
Then when you create methods to add to the list you will need to check if it is empty and if so initialize head rather than adding an element elsewhere.
Now I know that why pointers are used in defining linked lists. Simply because structure cannot have a recursive definition and if there would have been no pointers, the compiler won't be able to calculate the size of the node structure.
struct list{
int data;
struct list* next; // this is fine
};
But confusion creeps up when I declare the first node of the linked list as:
struct list* head;
Why this has to be a pointer? Can't it be simply declared as
struct list head;
and the address of this used for further uses? Please clarify my doubt.
There's no definitive answer to this question. You can do it either way. The answer to this question depends on how you want to organize your linked list and how you want to represent an empty list.
You have two choices:
A list without a "dummy" head element. In this case the empty list is represented by null in head pointer
struct list* head = NULL;
So this is the answer to your question: we declare it as a pointer to be able to represent an empty list by setting head pointer to null.
A list with a "dummy" head element. In this case the first element of the list is not used to store actual user data: it simply serves as a starting "dummy" element of the list. It is declared as
struct list head = { 0 };
The above represents an empty list, since head.next is null and head object itself "does not count".
I.e. you can declare it that way, if you so desire. Just keep in mind that head is not really a list element. The actual elements begin after head.
And, as always, keep in mind that when you use non-dynamically-allocated objects, the lifetime of those objects is governed by scoping rules. If you want to override these rules and control the objects' lifetimes manually, then you have no other choice but to allocate them dynamically and, therefore, use pointers.
You can declare a list such a way
struct list head = {};
But there will be some difficulties in the realization of functions that access the list. They have to take into account that the first node is not used as other nodes of the list and data member of the first node data also is not used.
Usually the list is declared the following way
struct List
{
// some other stuff as for example constructors and member functions
struct node
{
int data;
struct node* next; // this is fine
} head;
};
and
List list = {};
Or in C++ you could write simply
struct List
{
// some other stuff as for example constructors and member functions
struct node
{
int data;
struct node* next; // this is fine
} head = nullptr;
};
List list;
Of course you could define the default constructor of the List yourself.
In this case for example to check whether the list is empty it is enough to define the following member function
struct List
{
bool empty() const { return head == nullptr; }
// some other stuff as for example constructors and member functions
struct node
{
int data;
struct node* next; // this is fine
} head;
};
In simple terms, if your head is the start node of the linked list, then it will only contain the address of the first node from where linked list will begin. This is done to avoid confusion for a general programmer. Since the head will contain only address, hence, it is declared as a pointer. But the way you want to declare is also fine, just code accordingly. Tip: If you later on want to make some changes in your linked list, like deletion or insertion operations at the beginning of the linked list, you will face problems as you will require another pointer variable. So its better to declare the first node as pointer.
I have a uni assignment in which I have to implement a singly linked list that contains different objects that are derived from a common abstract base class called Shape.
I'll link to GitHub for the class implementation: shapes.h , shapes.cpp. So far it consists of Shape and its derived class Circle. There'll also be Rectangle, Point and Polygon later.
I should now implement a singly linked list of these different kinds of shapes. So far I've come up with the following class prototype for the List-class and the Node-class:
class Node
{
public:
Node() {}
friend class ShapeList;
private:
Shape* data;
Node* nextNode;
};
class ShapeList
{
public:
ShapeList(){head = NULL;}
void Append(Shape& inData);
private:
Node* head;
};
Adding elements void Append(Shape& inData) to a ShapeList-object should be able to be called from main in the following style:
ShapeList list1;
list1.Append( Circle(5,5,5) );
list1.Append( Rectangle( 4, 10, 2, 4) );
Given this information, how should I go about implementing void Append(Shape& inData)? I've tried several different approaches, but haven't come up with the correct solution so far.
It's also completely possible that the parameter to Append should be something else than (Shape& inData).
edit:
I've implemented Append(Shape& inData) but it works only sometimes:
Circle circle1;
ShapeList list1;
list1.Append( circle1 );
but not with
ShapeList list1;
list1.Append ( Circle(5,5,5) )
So far my Append()-implementation looks as follows:
void ShapeList::Append(Shape& inData)
{
//Create a new node
Node* newNode = new Node();
newNode->data=&inData;
newNode->nextNode=NULL;
//Create a temp pointer
Node *tmp = head;
if (tmp != NULL)
{
//Nodes already present in the list
//Traverse to the end of the list
while(tmp->nextNode != NULL)
tmp = tmp->nextNode;
tmp->nextNode=newNode;
}
else
head=newNode;
}
Does that look ok to you guys?
Since this is tagged under 'homework', I will only point you to the good direction. This may be too basic or maybe it is enough for your needs...
In a typical situation, you would simply use a container that is already written such as std::list.
But for implementing your own linked list
When you start from the head member of the ShapeList, you should be able to traverse the entire list and find a node for which 'nextNode' has never been assigned.
This is where you want to add a new node.
Now thee a a few tricks to be make things work:
1- In C++, variables are not automatically initialized. You must therefore initialize the many values when you create a new node, especially the next node pointer.
2- Instead of having pointers to references, I suggest that either you create copies of Shapes, of use some kind of smart pointers to avoid copying.
3- Don't forget about memory management, when you destroy your linked list, you will have to destroy all nodes individually since.
One very nice implementation of the singly linked list is as a circular list with the "head" pointer pointing at the tail. This makes it easy to insert at either the front or append to the end: in either case you create a fresh node, make the current tail point to it, and make it point to the current head, and then in the insert case make the head pointer point to the new node.
What you appear to be missing (other than what's already been pointed out: allocating, deallocating, and copying the nodes) is a way to know that you've actually created the list. So you'll want to add in some sort of output - either an operator << or a print() routine, which will walk the list, and call your graphical objects' printing mechanisms in order.
You say that it is possible that the argument to Append might not be Shape &data. Given the requirement of the calling convention specified, it should be:
Append( const Shape &data ) // provided shapes have copy constructors
{
Node *newNode = new Node( data ); // requires a constructor of Node that copies data to a freshly allocated location and sticks a pointer to that location in its data field - then Node's destructor needs to release that pointer.
... ( and the code to manipulate the existing list and newNode's next pointer )
}
Among other things this makes responsibility for management clear and simple.
If you have a Node constructor that takes both a pointer to a Node and a Shape, you should be able to do Append in two lines - one allocating the new Node and calling the constructor appropriately, and one modifying a pointer to point to the new node.
I would add - based on your edit - that you absolutely need to do the allocation and copy inside Append.
You probably want Node to be nested inside of ShapeList so its full name will be ShapeList::Node, not just ::Node.
Since Node will own some data remotely, you probably need to define the big three for it.
In line with that, when you push something onto the list, the list will hold a dynamically allocated copy, not the original object.
Edit: Append should take a Shape const & rather than a Shape &. A reference to const can bind to a temporary object, but a reference to non-const cannot, so the calls using parameters that create temporary objects (e.g., list.Append(Circle(5,5,5))) won't compile if the parameter is a reference to non-const object.
I'd also change Node::Node to require that you pass it a parameter or two. As-is, your linked-list code is dealing with the internals of a Node more than I'd like. I'd change it to something like:
Node::Node(Shape const *d, Node *n=NULL) : data(d), nextNode(n) {}
Then in append, instead of:
Node* newNode = new Node();
newNode->data=&inData;
newNode->nextNode=NULL;
You'd use something like:
Node *newNode = new Node(&inData); // or, probably, `... = new Node(inData.clone());`
...and Node's ctor would handle things from there.
Also note that it's easier to add to the beginning of a linked list than to the end (it saves you from walking the whole list). If you really want to add to the end, it's probably worthwhile to save a pointer to the last node you added, so you can go directly to the end, rather than walking the whole list every time.
Here is one way to handle the polymorphic requirement (std::shared_ptr), demonstrated with the STL singly linked list...
typedef forward_list<shared_ptr<Shape>> ShapeList;
ShapeList list1;
list1.push_back(make_shared<Circle>(5,5,5));
list1.push_back(make_shared<Rectangle>(4, 10, 2, 4));
Here is how it would effect Node:
class Node
{
public:
Node() {}
friend class ShapeList;
private:
shared_ptr<Shape> data;
Node* nextNode;
};
and ShapeList...
class ShapeList
{
public:
ShapeList(){head = NULL;}
void Append(const shared_ptr<Shape>& inData);
private:
Node* head;
};
The way I know how to represent a linked list is basically creating a Node class (more preferably a struct), and the creating the actual linkedList class. However, yesterday I was searching for the logic of reversing a singly linked list operation and almost 90% of the solutions I've encountered was including that the function, returning data type Node* . Thus I got confused since if you want to reverse a list no matter what operation you done, wouldn't it be in the type of linkedList again? Am I doing it the wrong way?
The linked list implementation I do all the time;
#include <iostream>
using namespace std;
struct Node
{
int data;
Node *next;
};
class linkedList
{
public:
Node* firstPtr;
Node* lastPtr;
linkedList()
{
firstPtr=lastPtr=NULL;
}
void insert(int value)
{
Node* newNode=new Node;
newNode->data=value;
if(firstPtr==NULL)
firstPtr=lastPtr=newNode;
else {
newNode->next=firstPtr;
firstPtr=newNode;
}
}
void print()
{
Node *temp=firstPtr;
while(temp!=NULL)
{
cout<<temp->data<<" ";
temp=temp->next;
}
}
};
You approach isn't wrong, but you might be giving too much emphasis on your linkedList class.
What does that class actually contain? A pointer to the first node, and a pointer to the last node (which is redundant information, since you can find the last node by only knowing the first one). So basically linkedList is just a helper class with no extra information.
The member functions from linkedList could easily be moved inside Node or made free functions that take a Node as parameter.
Well, what is a linked list but a pointer to the first node? A list is fully accessible provided you can get to the first node, and all you need for that is a pointer to the first node.
Unless you want to store extra control information about the list (such as its length for example), there's no need for a separate data type for the list itself.
Now some implementations (such as yours) may also store a pointer to the last node in the list for efficiency, allowing you to append an item in O(1) instead of O(n). But that's an extra feature for the list, not a requirement of lists in general.
Those functions might be returning of type Node* because after reversing the linked-list they will return the pointer to the First node of the list.
Alright, so I'm trying out C++ for the first time, as it looks like I'll have to use it for an upcoming course in college. I have a couple years of programming under my belt, but not much in the non-garbage-collected world.
I have a class, a Node for use in a doubly linked list. So basically it has a value and two pointers to other Nodes. The main constructor looks like Node(const std::string & val, Node * prev, Node * next). The exercise includes a copy constructor that does a shallow copy of another Node, with a comment above it that says to change it to make a deep copy.
Here is what I thought that meant:
Node(const Node & other)
: value(other.value)
{
prev = new Node(other.prev->value, other.prev->prev, other.prev->next);
next = new Node(other.next->value, other.next->prev, other.next->next);
}
This seems to accomplish the goal of making it so that changing the copied Node doesn't affect the new Node. However, when I do it this way, I am allocating new stuff on the heap. This worries me, because I think it means that I should also be deleting it in the Node's destructor. But this is now inconsistent with the other constructor, where pointers to the Nodes are just passed in, already pointing to something. I can't rightly go deleteing next and prev in the destructor with that going on, right?
I'm really confused, guidance appreciated!
EDIT: Here is the code (before my above change to it), as requested:
#include <string>
//! Node implements a doubly-linked list node
class Node {
friend class LinkedList; //!< LinkedList can access private members of Node
public:
//! Constructor
Node(const std::string & v, Node * p, Node * n) :
value(v), prev(p), next(n)
{
}
//! Change to deep copy
Node(const Node & other) :
value(other.value), prev(other.prev), next(other.next)
{
}
//! Read-only public methods for use by clients of the LinkedList class
const std::string & GetValue() const
{
return value;
}
Node * GetPrevious()const
{
return prev;
}
Node * GetNext()const
{
return next;
}
//! Change to deep copy
Node & operator=(const Node & other)
{
if(this!=&other)
{
value=other.value;
prev=other.prev;
next=other.next;
}
return *this;
}
private:
std::string value; //!< value stored in the node
Node * prev; //!< pointer to previous node in the list
Node * next; //!< pointer to next node in the list
};
First of all, I'm not really sure how the objective of the exercise should be understood. How deep should the copy be? In a solution like yours, this->next->next and other.next->next would be still the same thing. Should this object also be duplicated? And the rest of the list? Where does it end? One could of course deep-copy the whole list, but this would be a quite unexpected behavior of a copy constructor of a single node, I think.
Is maybe the value member variable a pointer, that is supposed to be deep copied? That would make much more sense for me.
But back to your interpretation:
Node a(...);
// ... more code that adds a whole list to a
Node b(a);
There are two problems with your implementation. For one b->next->prev points to a, while I suspect it should point back to b. Secondly you need to think about the corner cases, where a might be the first or last node in the list.
And to your main question: you are of course right, somewhere the newly created objects need to be deleted again. No matter if you just copy the prev and next nodes or the whole list, I would say the user of that copy is responsible to delete all the copied nodes again. I assume with a normal, not-copied list, the user of that list would walk through all the nodes and delete them manually one after another, once he's done with the list. He wouldn't not assume the destructor of one node to delete the whole list. And the same goes for copies, they should behave the same. The user of the copied stuff should delete all the copies. (In practice you would probably have a list class, that does all that node management for you).
But again, if the copy constructor of the node copies the whole list, or even just several of it's nodes, this would be very unexpected and all the time people would forget to clean up all these copies. But that's not your node class' fault, but the exercise requirements'.
Usually "deep copy" involves traversing the data structure and copying the entire thing. In your case, given a Node, make a complete copy of the list.
A deep copy makes an entire copy of a structure. What I mean by structure is a collection of objects that work together to perform a task. If you had a car class that had an object for each wheel and the body - a deep copy would make a copy of the entire car (and make copies of both the wheels and the body).
In your case, the "Entire Structure" is the list. A deep copy operation would only make sense if performed at the "list level." A deep copy of a node would copy the data that the node points to - but would not assign itself to be part of a list (as a Node should be unaware of the "master" list object).
List* List::CopyList()
{
List* nlist = new List();
ListNode* node = NULL, prev = NULL;
ListNode* newNodes = new ListNode[m_nodeCount];
int i = 0;
while ((node == NULL && node = m_first) || (node = node->Next()) != NULL)
{
newNodes[i] = node->CopyNode(); // also makes a new copy of the node's data
newNodes[i]->SetNext(NULL);
newNodes[i]->SetPrev(prev);
if (prev) prev->SetNext(newNodes[i]);
prev = newNodes[i];
++i;
}
if (m_len > 0)
nlist->SetFirst(newNodes[i]);
if (m_len > 1)
nlist->SetLast(newNodes[m_len - 1]);
return nlist;
}
Note: I just pulled that code out of my ass so it's not tested. Hope it helps though :)
You are correct in worrying.
By passing pointers into the Node constructor there is no information about ownership passed with the pointer. This is a poor design of the constructor. Either you should pass in a reference indicating you don't own the next node or pass a std::auto_ptr<> which indicates that you must take ownership. One could argue that the next or prev could be NULL (beginning or end of the list) and thus you can not use references, but this can be overcome by having alternative constructors.
Of course there are exceptions:
Is the Node class a private member of another class. If this is the case the use of the Node class is completely controlled by the owner and thus its correct usage would be controlled by the owning class.
What you have not provided is the definition of the destructor. With this we will be able to tell if the node is actually taking ownership of the pointer that are passed in to the constructors (or if the next and prev are already smart pointers)?
If every node makes copies of the nodes it points to then you can safely delete the objects in the destructors. If you are passing pointers (as the constructor Node(const std::string & v, Node * p, Node * n)) does then you do not "own" the pointers and should not delete them. If this is part of a linked list class then that class should own the pointers and delete the objects as necessary. You could also make Node a private subclass of the linked list class to avoid users (or yourself) messing with your pointers.
You've made a mistake in the recursion in your implementation as well, the copy constructor contains one level of a deep copy and calls the "normal" constructor, which takes pointers, making it shallow. What this means is that your deep copying is only one level deep. It should repeatedly call the copy constructor, something like this:
Node(const Node & other) : value(other.value)
{
prev = new Node(*(other.prev));
next = new Node(*(other.next));
}
AFAIK there is no benefit to using a deep copy here though, the only practical application I can think of is when copying the entire list, which could be handled more effectively in the class representing said list.