Reversing Linked List with Recursion, using STL - c++

Code for Reversing Linked List with Recursion, using STL
#include<iostream>
#include<conio.h>
#include<list>
using namespace std;
template<typename T>
class node
{
public:
T data;
node<T> *next;
node(){ next = NULL; }
node(const T& item, node<T> *nextnode = NULL)
{
data = item;
next = nextnode;
}
};
template<typename T>
class Reverse_list
{
private:
node<T> *head;
void reverse(node<T> *front);
public:
Reverse_list(){ head = NULL; }
//template<typename T>
void Reverse();
template<typename T>
void Display( list<T>& alist );
};
int main()
{
Reverse_list <int> rl;
list<int> intlist;
int size, no;
cout << "Size of List ?? ";
cin >> size;
for (int i = 1; i <= size; i++)
{
cout << "Enter the " << i <<" "<< "element";
cin >> no;
intlist.push_front(no);
}
rl.Display(intlist);
rl.Reverse();
rl.Display(intlist);
_getch();
return 0;
}
template<typename T>
void Reverse_list<T>::Display(list<T>& alist)
{
list<int>::iterator iter = alist.begin();
while (iter != alist.end())
{
cout << *iter << " ";
iter++;
}
}
template<typename T>
void Reverse_list<T>::reverse(node<T> *front)
{
if (front->next == NULL)
{
head = front;
return;
}
reverse(front->next);
node<int> *back = front->next;
back->next = front;
front->next = NULL;
}
template<typename T>
void Reverse_list<T>::Reverse()
{
reverse(head);
}
The above code generates 2 errors.
Error 1) No instance of function template matches the argument list. ( No error number.)
If I remove line 1 ( mentioned in a code ) then above error is no more. ( Why? )
Error 2) C2783: 'void Reverse_list::Reverse1(void)' : could not deduce template argument for 'T'
How to solve above errors.
In above program , I wanted to pass " head" node ( which is private ) as
argument to Reverse function. But we can not access private member outside of the class. So I passed indirectly. Is this a correct way of passing ?? Or there is some other way of accessing private data ??

I'm not sure to understand your intentions but...
You're trying to declare a method (reverse()) inside another method (Reverse()) ? Uhmmm....
We return to this later.
Imagine that the following instruction is correct instruction of Reverse_list<T>::Reverse()
node<T> *back = front->next;
Why you declare back as a pointer to a generic Node<T> when you assign front->next (so a specific Node<int>) to it?
If you define back as a node<int> pointer, the method Reverse() has no longer reason to be a template (dependant from T) method. And you can avoid both errors.
With your actual code, when you call
rl.Reverse();
you call a template method but the compiler doesn't know how to determine the type T. You could explicit it in this way
rl.Reverse<int>();
but, as written before, I thik it's better if you remove the whole template part.
Or, alternatively, you can transform the whole class in a template class; where head is a pointer to a generic Node<T>, not a specifica Node<int>.
Something like (if I understand correctly your intentions)
template <typename T>
class Reverse_list
{
private:
node<T> *head;
void reverse (node<T> * front);
public:
Reverse_list() : head(NULL)
{ }
void Reverse();
void Display(list<T>& alist);
};
template<typename T>
void Reverse_list<T>::reverse (node<T> * front)
{
if (front->next == NULL)
{
head = front;
return;
}
reverse(front->next);
node<T> *back = front->next;
back->next = front;
front->next = NULL;
}
template<typename T>
void Reverse_list<T>::Reverse()
{ reverse(head); }
In this case, in main(), rl should be declared as
Reverse_list<int> rl;
fixing T as int, and the call to Reverse() should be
rl.Reverse();
--- EDIT 2016.05.10 ---
With the "template Reverse_list" solution, you should correct three points (at last).
1) in Reverse_list class declaration, you have commented the template<typename T> row before void Reverse(); good; you should delete (comment) the same line (for the same reason) before void Display( list<T>& alist );; so the class become
template<typename T>
class Reverse_list
{
private:
node<T> *head;
void reverse(node<T> *front);
public:
Reverse_list(){ head = NULL; }
//template<typename T>
void Reverse();
//template<typename T>
void Display( list<T>& alist );
};
2) Display() now is a method of a templated class; so the line
list<int>::iterator iter = alist.begin();
become
list<T>::iterator iter = alist.begin();
3) reverse() now is a method of a templated class; so the line
node<int> *back = front->next;
become
node<T> *back = front->next;

Related

Adding function templates to classes

So far, I have implemented a basic LinkedList. This works, but only for integers, and I would like it to work for any type.
I'm trying to get it to work for first any same type (i.e a LinkedList of just strings, or then just ints). After, I would like it to find a way of making it a LinkedList of anything (containing strings, then ints, then longs, all in one list).
#include <iostream>
struct Node{
Node(int value);
Node *next;
int data;
};
Node::Node(int value){
this->data = value;
this->next = nullptr;
}
struct LinkedList{
Node *head;
LinkedList();
void push_back(int value);
void print();
};
LinkedList::LinkedList(){
this->head = nullptr;
}
void LinkedList::push_back(int value){
Node *n = new Node(value);
if(this->head == nullptr){
this->head = n;
} else {
Node *cursor = this->head;
while (cursor->next != nullptr){
cursor = cursor->next;
}
cursor->next = n;
}
}
void LinkedList::print(){
Node *cursor = this->head;
while(cursor != nullptr){
std::cout << cursor->data << '\n';
cursor = cursor->next;
}
}
int main(){
LinkedList l = LinkedList();
l.push_back(1);
l.push_back(2);
l.print();
}
The above works, however, only for ints.
I'm knew, but I think the way is to use templates, however, doing so, I seem to be doing overkill? and it doesn't compile? Is there a cleaner was to do this?
#include <iostream>
template <typename T>
struct Node {
Node(T value);
int data;
Node<T> *next;
};
template <typename T>
Node<T>::Node(T value){
this->next = nullptr;
this->data = value;
}
template <typename T>
class LinkedList{
public:
LinkedList();
Node<T> *head;
void push_back(T data);
void print();
};
template <typename T>
LinkedList<T>::LinkedList(){
this->head = nullptr;
}
template <typename T>
void LinkedList<T>::push_back(T data){
Node *n = new Node(data);
if(this->head == nullptr){
this->head = n;
} else {
Node *cursor = this->head;
while(cursor->next != nullptr){
cursor = cursor->next;
}
cursor->next = n;
}
}
template <typename T>
void LinkedList<T>::print(){
Node *cursor = this->head;
while(cursor != nullptr){
std::cout << cursor->data << '\n';
cursor = cursor->next;
}
}
int main(){
LinkedList<T> *list = new LinkedList<T>();
list->push_back(1);
list->push_back(2);
list->push_back(3);
}
When declaring template classes, you use the "T" as a 'type placeholder' in the declaration and implementation (as you have done). However, when you want to actually use an object of the templated class, you replace the "T" with the actual type you want.
So, in your main (assuming you want an int type), you would have code like this:
int main(){
LinkedList<int> *list = new LinkedList<int>(); // THIS object uses "int" wherever "T" occurs in the declaration/implementation
list->push_back(1);
list->push_back(2);
list->push_back(3);
}
I also noticed a 'possible/probable error' in your struct declaration, where you specified that the data member is of (fixed) type int; maybe (almost certainly, actually, as you later assign a "T"-type value to it) you want this to vary according to the actual type requested? If so, make the following change:
template <typename T>
struct Node {
Node(T value);
// int data;
T data; // Data will be whatever "T" is when an object is created.
Node<T> *next;
};
Feel free to ask for further clarification and/or explanation.

C++ pass by reference then set a pointer to the object

I am creating a class LinkedList. I am having difficulty adding another node to my list.
Here is what I have so far:
template<typename T>
class LinkedList
{
private:
T element;
T *next;
public:
LinkedList();
LinkedList(T element);
void add(LinkedList<T> &otherList);
void print();
};
template<typename T>
LinkedList<T>::LinkedList()
{
next = NULL;
}
template<typename T>
LinkedList<T>::LinkedList(T element)
{
this->element = element;
next = NULL;
}
template<typename T>
void LinkedList<T>::add(LinkedList<T> &otherList)
{
next = &otherList;
}
template<typename T>
void LinkedList<T>::print()
{
LinkedList<T> *current = this;
while (current != NULL)
{
std::cout << current->element;
current = current->next;
}
}
int main()
{
LinkedList<std::string> myFirst("First");
LinkedList<std::string> mySecond("Second");
myFirst.add(mySecond);
myFirst.print();
return 0;
}
This works however if I make the change:
void add(const LinkedList<T> &otherList);
template<typename T>
void LinkedList<T>::add(const LinkedList<T> &otherList)
{
next = &otherList; //now an error right here
}
Then I get an error stating:
Assigning to 'LinkedList<std::__1::basic_string<char> > *' from incompatible type 'const LinkedList<std::__1::basic_string<char> > *'
Why is it I get this error?
next is a T*, and you're trying to assign a const LinkedList<T>* to it.
I suppose you meant something like next = &(otherList.element) (though even then I think your list semantics are somewhat broken — elements shouldn't typically be shared by multiple containers unless you're very, very clear about the ownership semantics).
Contrary to your claims, your first program doesn't work either for the same reason.

template using another template

Here is a cut-down version of a template List code (adapted from http://www.daniweb.com/software-development/cpp/threads/237391/c-template-linked-list-help)
List complains (compile error) that "Node is not a type". Why is this, and what is the fix?
I tried replacing the "class Node" with a "struct Node" (and related changes), and the struct version worked fine. So the main question seems to be: how does a template-class access another template-class?
#include <iostream>
using namespace std;
template <typename T>
class Node
{
public:
Node(){}
Node(T theData, Node<T>* theLink) : data(theData), link(theLink){}
Node<T>* getLink( ) const { return link; }
const T getData( ) const { return data; }
void setData(const T& theData) { data = theData; }
void setLink(Node<T>* pointer) { link = pointer; }
private:
T data;
Node<T> *link;
};
template <typename T>
class List {
public:
List() {
first = NULL;
last = NULL;
count = 0;
}
void insertFirst(const T& newData) {
first = new Node(newData, first);
++count;
}
void printList() {
Node<T> *tempt;
tempt = first;
while(tempt != NULL){
cout << tempt->getData() << " ";
tempt = tempt->getLink();
}
}
~List() { }
private:
Node<T> *first;
Node<T> *last;
int count;
};
int main() {
List<int> myIntList;
cout << "Inserting 1 in the list...\n";
myIntList.insertFirst(1);
myIntList.printList();
cout << endl;
List<double> myDoubleList;
cout << "Inserting 1.5 in the list...\n";
myDoubleList.insertFirst(1.5);
myDoubleList.printList();
cout << endl;
}
You are using
new Node(newData, first);
within the List template. At that point, Node does not refer to a type, but to a template. But of course to create an instance of a type with new, you need a type there.
The most probable thing you want to do is to make it a type by instantiating the template, i.e.
new Node<T>(newData, first);

Implicit instantiation of undefined template

Here's a template (queue) I'm trying to write:
#include <iostream>
using namespace std;
template <typename T>
class Queue
{
friend ostream& operator<< (ostream &, const Queue<T> & );
private:
template<class> class Node;
Node<T> *front;
Node<T> *back;
public:
Queue() : front(0), back(0) {}
~Queue();
bool Empty()
{
return front == 0;
}
void Push(const T& NewEl)
{
Node<T&> *El = new Node<T> (NewEl);
if (Empty())
front=back=El;
else
{
back-> next = El;
back = El;
}
}
void Pop()
{
if (Empty())
cout << "Очередь пуста." << endl;
else
{
Node<T> *El = front;
front = front -> next;
delete El;
}
}
void Clear()
{
while (! Empty())
Pop();
}
};
template <typename T>
class Node
{
friend class Queue<T>;
public:
Node() {next = 0;}
Node(T nd) {nd=node; next=0;}
T& getsetnode(){return node;}
Node<T>*& getsetnext(){return next;}
private:
T front;
T back;
T node;
Node<T> *next;
};
template <class T> ostream& operator<< (ostream &, const Queue<T> & );
int main()
{
Queue<int> *queueInt = new Queue<int>;
for (int i = 0; i<10; i++)
{
queueInt->Push(i);
cout << "Pushed " << i << endl;
}
if (!queueInt->Empty())
{
queueInt->Pop();
cout << "Pop" << endl;
}
queueInt->Front();
queueInt->Back();
queueInt->Clear();
cout << "Clear" << endl;
return 0;
}
At these lines:
Node<T&> *El = new Node<T> (NewEl);
front = front -> next;
delete El;
I get Implicit instantiation of undefined template 'Queue<int>::Node<int>'. What am I doing wrong? After reading this post I tried changing int to const int to see if that was the problem, but apparently it isn't, because I get the same error.
I'm using XCode with LLVM compiler 4.2. When I switch to GCC I get more errors:
template<class> class Node; gets Declaration of 'struct Queue<int>::Node<int>',
Node<T&> *El = new Node<T> (NewEl); gets Invalid use of incomplete type,
and anything dealing with assignment of anything to El can't convert <int&>* to <int>* (but deleting reference doesn't change anything for LLVM).
template <typename T>
class Queue
{
private:
template<class> class Node;
/* ... */
This is a forward declaration of Queue::Node. The latter defined class Node is in the global namespace, so they aren't the same and any usage of Queue::Node will result in an incomplete-type error. Since you don't provide an interface to the interior nodes anyway just scrap the global definition of Node and stick it into Queue:
template <typename T>
class Queue
{
private:
class Node
{
public:
Node() {next = 0;}
/* ... */
};
/* ... */
};

C++ Templates - LinkedList

EDIT -- Answered below, missed the angled braces. Thanks all.
I have been attempting to write a rudimentary singly linked list, which I can use in other programs. I wish it to be able to work with built-in and user defined types, meaning it must be templated.
Due to this my node must also be templated, as I do not know the information it is going to store. I have written a node class as follows -
template <class T> class Node
{
T data; //the object information
Node* next; //pointer to the next node element
public:
//Methods omitted for brevity
};
My linked list class is implemented in a seperate class, and needs to instantiate a node when adding new nodes to the end of the list. I have implemented this as follows -
#include <iostream>
#include "Node.h"
using namespace std;
template <class T> class CustomLinkedList
{
Node<T> *head, *tail;
public:
CustomLinkedList()
{
head = NULL;
tail = NULL;
}
~CustomLinkedList()
{
}
//Method adds info to the end of the list
void add(T info)
{
if(head == NULL) //if our list is currently empty
{
head = new Node<T>; //Create new node of type T
head->setData(info);
tail = head;
}
else //if not empty add to the end and move the tail
{
Node* temp = new Node<T>;
temp->setData(info);
temp->setNextNull();
tail->setNext(temp);
tail = tail->getNext();
}
}
//print method omitted
};
I have set up a driver/test class as follows -
#include "CustomLinkedList.h"
using namespace std;
int main()
{
CustomLinkedList<int> firstList;
firstList.add(32);
firstList.printlist();
//Pause the program until input is received
int i;
cin >> i;
return 0;
}
I get an error upon compilation however - error C2955: 'Node' : use of class template requires template argument list - which points me to the following line of code in my add method -
Node* temp = new Node<T>;
I do not understand why this has no information about the type, since it was passed to linked list when created in my driver class. What should I be doing to pass the type information to Node?
Should I create a private node struct instead of a seperate class, and combine the methods of both classes in one file? I'm not certain this would overcome the problem, but I think it might. I would rather have seperate classes if possible though.
Thanks, Andrew.
While the answers have already been provided, I think I'll add my grain of salt.
When designing templates class, it is a good idea not to repeat the template arguments just about everywhere, just in case you wish to (one day) change a particular detail. In general, this is done by using typedefs.
template <class T>
class Node
{
public:
// bunch of types
typedef T value_type;
typedef T& reference_type;
typedef T const& const_reference_type;
typedef T* pointer_type;
typedef T const* const_pointer_type;
// From now on, T should never appear
private:
value_type m_value;
Node* m_next;
};
template <class T>
class List
{
// private, no need to expose implementation
typedef Node<T> node_type;
// From now on, T should never appear
typedef node_type* node_pointer;
public:
typedef typename node_type::value_type value_type;
typedef typename node_type::reference_type reference_type;
typedef typename node_type::const_reference_type const_reference_type;
// ...
void add(value_type info);
private:
node_pointer m_head, m_tail;
};
It is also better to define the methods outside of the class declaration, makes it is easier to read the interface.
template <class T>
void List<T>::add(value_type info)
{
if(head == NULL) //if our list is currently empty
{
head = new node_type;
head->setData(info);
tail = head;
}
else //if not empty add to the end and move the tail
{
Node* temp = new node_type;
temp->setData(info);
temp->setNextNull();
tail->setNext(temp);
tail = tail->getNext();
}
}
Now, a couple of remarks:
it would be more user friendly if List<T>::add was returning an iterator to the newly added objects, like insert methods do in the STL (and you could rename it insert too)
in the implementation of List<T>::add you assign memory to temp then perform a bunch of operations, if any throws, you have leaked memory
the setNextNull call should not be necessary: the constructor of Node should initialize all the data member to meaningfull values, included m_next
So here is a revised version:
template <class T>
Node<T>::Node(value_type info): m_value(info), m_next(NULL) {}
template <class T>
typename List<T>::iterator insert(value_type info)
{
if (m_head == NULL)
{
m_head = new node_type(info);
m_tail = m_head;
return iterator(m_tail);
}
else
{
m_tail.setNext(new node_type(info));
node_pointer temp = m_tail;
m_tail = temp.getNext();
return iterator(temp);
}
}
Note how the simple fact of using a proper constructor improves our exception safety: if ever anything throw during the constructor, new is required not to allocate any memory, thus nothing is leaked and we have not performed any operation yet. Our List<T>::insert method is now resilient.
Final question:
Usual insert methods of single linked lists insert at the beginning, because it's easier:
template <class T>
typename List<T>::iterator insert(value_type info)
{
m_head = new node_type(info, m_head); // if this throws, m_head is left unmodified
return iterator(m_head);
}
Are you sure you want to go with an insert at the end ? or did you do it this way because of the push_back method on traditional vectors and lists ?
Might wanna try
Node<T>* temp = new Node<T>;
Also, to get hints on how to design the list, you can of course look at std::list, although it can be a bit daunting at times.
You need:
Node<T> *temp = new Node<T>;
Might be worth a typedef NodeType = Node<T> in the CustomLinkedList class to prevent this problem from cropping up again.
That line should read
Node<T>* temp = new Node<T>;
Same for the next pointer in the Node class.
As said, the solution is
Node<T>* temp = new Node<T>;
... because Node itself is not a type, Node<T> is.
And you will need to specify the template parameter for the Node *temp in printlist also.
// file: main.cc
#include "linkedlist.h"
int main(int argc, char *argv[]) {
LinkedList<int> list;
for(int i = 1; i < 10; i++) list.add(i);
list.print();
}
// file: node.h
#ifndef _NODE_H
#define _NODE_H
template<typename T> class LinkedList;
template<typename T>class Node {
friend class LinkedList<T>;
public:
Node(T data = 0, Node<T> *next = 0)
: data(data), next(next)
{ /* vacio */ }
private:
T data;
Node<T> *next;
};
#endif//_NODE_H
// file: linkedlist.h
#ifndef _LINKEDLIST_H
#define _LINKEDLIST_H
#include <iostream>
using namespace std;
#include "node.h"
template<typename T> class LinkedList {
public:
LinkedList();
~LinkedList();
void add(T);
void print();
private:
Node<T> *head;
Node<T> *tail;
};
#endif//_LINKEDLIST_H
template<typename T>LinkedList<T>::LinkedList()
: head(0), tail(0)
{ /* empty */ }
template<typename T>LinkedList<T>::~LinkedList() {
if(head) {
Node<T> *p = head;
Node<T> *q = 0;
while(p) {
q = p;
p = p->next;
delete q;
}
cout << endl;
}
}
template<typename T>LinkedList<T>::void add(T info) {
if(head) {
tail->next = new Node<T>(info);
tail = tail->next;
} else {
head = tail = new Node<T>(info);
}
}
template<typename T>LinkedList<T>::void print() {
if(head) {
Node<T> *p = head;
while(p) {
cout << p->data << "-> ";
p = p->next;
}
cout << endl;
}
}
You Should add new node in this way
Node<T>* temp=new node<T>;
Hope you Solved :)
#include<iostream>
using namespace std;
template < class data > class node {
private :
data t;
node<data > *ptr;
public:
node() {
ptr = NULL;
}
data get_data() {
return t;
}
void set_data(data d) {
t = d;
}
void set_ptr(node<data > *p) {
ptr = p;
}
node * get_ptr() {
return ptr;
}
};
template <class data > node < data > * add_at_last(data d , node<data > *start) {
node< data > *temp , *p = start;
temp = new node<data>();
temp->set_data(d);
temp->set_ptr(NULL);
if(!start) {
start = temp;
return temp;
}
else {
while(p->get_ptr()) {
p = p->get_ptr();
}
p->set_ptr(temp);
}
}
template < class data > void display(node< data > *start) {
node< data > *temp;
temp = start;
while(temp != NULL) {
cout<<temp->get_data()<<" ";
temp = temp->get_ptr();
}
cout<<endl;
}
template <class data > node < data > * reverse_list(node<data > * start) {
node< data > *p = start , *q = NULL , *r = NULL;
while(p->get_ptr()) {
q = p;
p = p->get_ptr();
q->set_ptr(r);
r = q;
}
p->set_ptr(r);
return p;
}
int main() {
node < int > *start;
for(int i =0 ; i < 10 ; i ++) {
if(!i) {
start = add_at_last(i , start);
}
else {
add_at_last(i , start);
}
}
display(start);
start = reverse_list(start);
cout<<endl<<"reverse list is"<<endl<<endl;
display(start);
}