问题
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.
回答1:
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.
回答2:
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>::addwas returning an iterator to the newly added objects, likeinsertmethods do in the STL (and you could rename it insert too) - in the implementation of
List<T>::addyou assign memory totempthen perform a bunch of operations, if any throws, you have leaked memory - the
setNextNullcall should not be necessary: the constructor ofNodeshould initialize all the data member to meaningfull values, includedm_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 ?
回答3:
That line should read
Node<T>* temp = new Node<T>;
Same for the next pointer in the Node class.
回答4:
As said, the solution is
Node<T>* temp = new Node<T>;
... because Node itself is not a type, Node<T> is.
回答5:
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.
回答6:
And you will need to specify the template parameter for the Node *temp in printlist also.
回答7:
// 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;
}
}
回答8:
You Should add new node in this way
Node<T>* temp=new node<T>;
Hope you Solved :)
回答9:
#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);
}
来源：https://stackoverflow.com/questions/2079296/c-templates-linkedlist