我有以下代码,但它在行上给出错误 - e = list.first();(在代码的末尾)。它说它无法将Node转换为char,有人可以告诉我如何从变量e中的链表返回第一个值。
感谢所有帮助:)
//
// main.cpp
// cprogram1
//
#include <iostream>
using namespace std;
class Node {
char data;
Node* next;
public:
Node() {};
void SetData(int aData) { data = aData; };
void SetNext(Node* aNext) { next = aNext; };
char Data() { return data; };
Node* Next() { return next; };
};
// List class
class List {
Node *head;
public:
List() { head = NULL; };
void Print();
void Append(int data);
void Delete(int data);
Node * First() const;
};
/**
* Print the contents of the list
*/
void List::Print() {
// Temp pointer
Node *tmp = head;
// No nodes
if ( tmp == NULL ) {
cout << "EMPTY" << endl;
return;
}
// One node in the list
if ( tmp->Next() == NULL ) {
cout << tmp->Data();
cout << " --> ";
cout << "NULL" << endl;
}
else {
// Parse and print the list
do {
cout << tmp->Data();
cout << " --> ";
tmp = tmp->Next();
}
while ( tmp != NULL );
cout << "NULL" << endl;
}
}
/**
* Append a node to the linked list
*/
void List::Append(int data) {
// Create a new node
Node* newNode = new Node();
newNode->SetData(data);
newNode->SetNext(NULL);
// Create a temp pointer
Node *tmp = head;
if ( tmp != NULL ) {
// Nodes already present in the list
// Parse to end of list
while ( tmp->Next() != NULL ) {
tmp = tmp->Next();
}
// Point the last node to the new node
tmp->SetNext(newNode);
}
else {
// First node in the list
head = newNode;
}
}
/**
* Delete a node from the list
*/
void List::Delete(int data) {
// Create a temp pointer
Node *tmp = head;
// No nodes
if ( tmp == NULL )
return;
// Last node of the list
if ( tmp->Next() == NULL ) {
delete tmp;
head = NULL;
}
else {
// Parse thru the nodes
Node *prev;
do {
if ( tmp->Data() == data ) break;
prev = tmp;
tmp = tmp->Next();
} while ( tmp != NULL );
// Adjust the pointers
prev->SetNext(tmp->Next());
// Delete the current node
delete tmp;
}
}
Node * List::First() const {
Node *tmp = head;
return head;
}
int main ()
{
char c;
int t = 0;
char e;
List list;
while(t==0)
{
cout << "Please enter your command";
cin >> c;
if(c=='c')
{
cout << "You will need to enter 6 letters, one after the other";
cout << "Please enter the first letter";
cin >> e;
list.Append(e);
cout << "Please enter the second letter";
cin >> e;
list.Append(e);
cout << "Please enter the third letter";
cin >> e;
list.Append(e);
cout << "Please enter the fourth letter";
cin >> e;
list.Append(e);
cout << "Please enter the fifth letter";
cin >> e;
list.Append(e);
cout << "Please enter the sixth letter";
cin >> e;
list.Append(e);
list.Print();
list.Delete('b');
list.Print();
e = list.First();
}
}
}
答案 0 :(得分:1)
当我运行你的代码时,我收到错误:
main.cpp: error: assigning to 'char' from incompatible type 'Node *'
e = list.First();
^ ~~~~~~~~~~~~
我认为解释它。
答案 1 :(得分:1)
试试这个:
e = list.First()->Data();
答案 2 :(得分:1)
让我们来看看您的计划:tested here:
#include <iostream>
using namespace std;
class Node {
char data;
Node* next;
public:
Node() {};
void SetData(int aData) { data = aData; };
void SetNext(Node* aNext) { next = aNext; };
char Data() { return data; };
Node* Next() { return next; };
};
// List class
class List {
Node *head;
public:
List() { head = NULL; };
void Print();
void Append(int data);
void Delete(int data);
Node * First() const;
};
/**
* Print the contents of the list
*/
void List::Print() {
// Temp pointer
Node *tmp = head;
// No nodes
if ( tmp == NULL ) {
cout << "EMPTY" << endl;
return;
}
// One node in the list
if ( tmp->Next() == NULL ) {
cout << tmp->Data();
cout << " --> ";
cout << "NULL" << endl;
}
else {
// Parse and print the list
do {
cout << tmp->Data();
cout << " --> ";
tmp = tmp->Next();
}
while ( tmp != NULL );
cout << "NULL" << endl;
}
}
/**
* Append a node to the linked list
*/
void List::Append(int data) {
// Create a new node
Node* newNode = new Node();
newNode->SetData(data);
newNode->SetNext(NULL);
// Create a temp pointer
Node *tmp = head;
if ( tmp != NULL ) {
// Nodes already present in the list
// Parse to end of list
while ( tmp->Next() != NULL ) {
tmp = tmp->Next();
}
// Point the last node to the new node
tmp->SetNext(newNode);
}
else {
// First node in the list
head = newNode;
}
}
/**
* Delete a node from the list
*/
void List::Delete(int data) {
// Create a temp pointer
Node *tmp = head;
// No nodes
if ( tmp == NULL )
return;
// Last node of the list
if ( tmp->Next() == NULL ) {
delete tmp;
head = NULL;
}
else {
// Parse thru the nodes
Node *prev;
do {
if ( tmp->Data() == data ) break;
prev = tmp;
tmp = tmp->Next();
} while ( tmp != NULL );
// Adjust the pointers
prev->SetNext(tmp->Next());
// Delete the current node
delete tmp;
}
}
Node * List::First() const {
Node *tmp = head;
return head;
}
int main ()
{
char c;
int t = 0;
char e;
List list;
while(t==0)
{
cout << "Please enter your command";
c = 'c';
//cin >> c;
if(c=='c')
{
cout << "You will need to enter 6 letters, one after the other";
cout << "Please enter the first letter";
list.Append('e');
cout << "Please enter the second letter";
//cin >> e;
list.Append('a');
cout << "Please enter the third letter";
//cin >> e;
list.Append('b');
cout << "Please enter the fourth letter";
//cin >> e;
list.Append('f');
cout << "Please enter the fifth letter";
//cin >> e;
list.Append('h');
cout << "Please enter the sixth letter";
//cin >> e;
list.Append(e);
list.Print();
list.Delete('b');
list.Print();
e = list.First()->Data();
t=1;
}
}
}
如您所见,e = list.First()
会返回指向Node
的指针,而不是char
。您已经实现了一个返回char
Data()
的函数。因此,您应该使用e = list.First()->Data();
。
此外,您的循环是无限的,因为t
始终为0.我只是放t=1
来停止循环并查看结果。 (如果您使用此代码,请不要忘记取消注释cin并删除c = 'c'
)
希望对你有所帮助。