I was going through the linked list and I have confusion. maybe I am mixing the concept or something else. Here is the code of creating the header node.
struct node
{
int data;
struct node *link;
};
node *head;
void main()
{
head = new node;
}
1)The first thing I want to know is how can we write struct node *link; in inside the same node structure? because using first the node structure is created then we can declare pointer of that.
2) node *head; will already declare a memory of size node, then we need to do again head = new node;?
The self referential struct can hold a pointer to itself. Please do not confuse with size of pointer to size of structure. Size of pointer is constant irrespective of data type.
struct node
{
int data;
struct node *link;
};
Had it the struct node *link be something else like struct node link, It will not compile just like you think.
regarding why allocation by using the new is required, when we do node *head, it says that head points to memory location of actual node with area for data and link.
It might be useful to read pointer concept again
A link list is a chain of objects. What you are doing here is creating a struct with two variables. One is the date to store in this node. The other is a recursive struct. This where link lists get their name. One struct links to the next. When this node is created, link has no value but you can add nodes by creating a new node and storing it in link.
As for the rest of your code, I don't think you are doing it right. Usually nodes are wrapped in a link list class so that you can control the link list by adding, deleting and searching the nodes. When you are controlling a link list you need at least two pointers. One to point to the first node in the list (also called the "head"). The second pointer is the search pointer that will start at the head and go node by node until you find what you are looking for.
Now to answer your second question when you write node* head you are only declaring the pointer. You are not declaring "a memory of size node" so in the initialize function of the link list you need to create the first node and have the head point to it head = new node;
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 am new in Data Structure.
I came up with a question when I was trying to write code using linked list using C++.
In linked list, is pointer next, previous defined by the compiler? Or we have to name our own and it does not matter how we call it?
next and previous are not predefined. We have to define it ourself. Usually for a single linked list, we come up with a node structure which consists of a data field and anext pointer which is a pointer to a node of the same type. A simple example is as follows:
struct node{
int data; //data part
struct node *next; // next pointer - points to a node of the same type
}
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.