在C ++中的链表中间插入节点时出现段错误
Segfault while inserting node at mid of a linked list in c++
我正在创建一个简单的单向链表并尝试在链表的开头、中间和结尾插入一个节点。但是我的 'insertAtMid' 没有正常工作，它使程序崩溃并出错；
“发生异常。分段错误。”
这是整个程序：
/*
Write a program that creates a singly linked list until user wants and displays it. then insert a node at the beginning, middle, and end of the list.
*/
#include <iostream>
using namespace std;
struct node
{
int data;
node *next;
};
class linkedList
{
private:
node *head, *tail;
public:
linkedList()
{
head = NULL;
tail = NULL;
}
void addNode()
{
int item;
cout << "Enter data for node : ";
cin >> item;
node *temp = new node();
temp->data = item;
temp->next = NULL;
if (head == NULL)
{
head = temp;
tail = temp;
}
else
{
tail->next = temp;
tail = tail->next;
}
}
node *getHead()
{
return head;
}
static void display(node *head)
{
int i = 1;
while (head != NULL)
{
cout << "Node " << i << " : " << head->data << endl;
head = head->next;
i++;
}
}
void insertAtBegining()
{
int n;
cout << "Enter data you want to insert : ";
cin >> n;
node *temp = new node();
temp->data = n;
temp-> next = head;
head = temp;
cout << "Inserted Successfully. Updated list is : "<<endl;
}
static void insertAtMid(node * head)
{
int n, length=0, middle;
while (head!=NULL)
{
length ++;
head = head->next;
}
middle = length/2;
while((middle --) > 1)
{
head = head->next;
}
cout << "Enter data you want to insert : ";
cin >> n;
node *temp = new node();
temp->data = n;
temp->next = head->next;
head = temp;
cout << "Inserted Successfully. Updated list is : "<<endl;
}
};
int main()
{
linkedList l1;
char choice;
int menuChoice;
cout << "List is not created yet, please add data to create the list." << endl;
do
{
l1.addNode();
cout << "Node added. Want to add another node (Y/N) : ";
cin >> choice;
} while (choice == 'y' || choice == 'Y');
cout << "List created successfully. Select from options below : "<<endl;
cout << "1. Display list \n2. Insert at Begining \n3. Insert at Mid \n4. Insert at End \n5. exit"<<endl;
cin >> menuChoice;
switch(menuChoice)
{
case 1:
linkedList::display(l1.getHead());
break;
case 2:
l1.insertAtBegining();
linkedList::display(l1.getHead());
break;
case 3:
l1.insertAtMid(l1.getHead());
linkedList::display(l1.getHead());
break;
case 4:
l1.addNode();
cout << "Successfully inserted node at end of list. Updated list is :"<<endl;
linkedList::display(l1.getHead());
break;
case 5:
exit;
break;
default:
cout << "Invalid selection...!!";
}
}
编译器显示错误在第 85 行，那里发生了分段错误。这是错误的屏幕截图：
Exception has occurred
根据您的代码实例，您正在计算节点数并将头指针移动到列表末尾。
在执行以下语句之前，head 当前将指向最后一个节点。
while((middle --) > 1)
{
head = head->next;
}
一旦这条语句 head = head->next; 被执行，那么 head 将指向任何地方，或者你可以说，它指向 NULL。
这就是您遇到分段错误的原因，因为 head->next 试图提供对地址的引用，这是一个垃圾。那里什么都不存在。
代替使用head进行迭代，使用临时节点，将其分配给head然后计算列表的长度。
node *tmp = new node();
tmp = head;
int length = 0;
while(tmp != NULL){
length++;
tmp = tmp->next;
}
middle = ((length % 2) == 0) ? (length /2 ) : (length + 1)/2);
tmp = head;
while((middle --) > 1){
tmp = tmp->next;
}
cout << "Enter data you want to insert : ";
cin >> n;
node *curr = new node();
curr->data = n;
curr->next = temp->next;
temp->next = curr;
我建议您不要使用 head 指针进行迭代，因为它可能会导致悬空引用。
哇，你真的很困惑...首先，你有一个 class，有 private: 个成员 head 和 tail。您不会将 head 或 tail 作为参数传递给您的成员函数。您的 class 成员函数可以访问您的 class 私有成员。
所以你不做:
static void display(node *head)
{
...
相反，只需：
void display ()
{
...
您不要混合应用程序和界面
您的应用程序是您处理从程序用户界面获取的数据的方式。你把两者分开。您的菜单就是您的界面，您的 linked 列表就是您的应用程序（粗略地说）。您在程序界面中执行所有输出和数据输入，并将数据传递给应用程序进行处理。这使您的代码可维护且更易于阅读（并且更容易遵循逻辑）
您不会将数据提示隐藏在 addnode() 函数中。相反，您获取要在 main() 的初始部分或 switch() 语句中添加的数据，然后将整数值传递给 addnode()。示例：
node *addNode (int n) /* choose meaningful return to indicase success/failure */
{
// int item;
// cout << "Enter data for node : "; /* don't mix interface & application */
// cin >> item;
node *temp = new node();
temp->data = n;
temp->next = NULL;
if (head == NULL)
head = tail = temp;
else {
tail->next = temp;
tail = temp;
}
return temp; /* return pointer to allcated node */
}
然后：
/** this is interface, input/output happens here */
std::cout << "List is not created yet, please add data to create the list.\n";
do
{
int n;
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
return 1;
}
if (!l1.addNode (n)) {
std::cerr << "error: adding node.\n";
return 1;
}
std::cout << "Node added. Want to add another node (y/n) : ";
if (!(std::cin >> choice)) {
std::cerr << "error: invalid integer input.\n";
return 1;
}
} while (choice == 'y' || choice == 'Y');
（对于 switch() 语句中的菜单也是如此）
您验证每个用户输入
除非您检查return，否则您无法正确使用任何输入功能。 （从技术上讲，您检查流状态）。您可以通过检查每个输入是否成功或处理错误来做到这一点。永远不要 cin >> data; 并希望 data 是好的。例如（最少）：
int n;
std::cout << "Enter data for node : ";
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
break; /* note: this just lets the menu fail before clearing error */
}
if (l1.insertAtBegining (n)) {
std::cout << "Inserted Successfully. Updated list is : \n";
l1.display();
}
break;
对于使用 iostream 的任何输入，您必须 .clear() 流状态才能进行进一步的输入，并且您必须清空导致错误的输入缓冲区中的内容。一个更完整的例子是：
while (!(std::cin >> menuChoice)) { /* you need to handle error here */
std::cerr << "error: invalid integer input.\n";
std::cin.clear();
std::cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');
std::cout << "\nchoice: ";
}
参见 std::basic_istream::ignore and bookmark the cppreference.com link -- 这是您在 Internet 上最好的参考资料 -- 使用它。
（您应该更新其余代码以正确处理输入）
不要使用 head
进行迭代
您的 head 指针指向列表开头的节点。不要用它来遍历你的列表。 （完成后——它不再指向列表的开头，并且您丢失了指向开头的指针）。相反，你声明一个临时指针来迭代，例如：
void display ()
{
int i = 1;
node *iter = head;
while (iter != NULL)
{
std::cout << "Node " << i << " : " << iter->data << '\n';
iter = iter->next;
i++;
}
}
不要养成 using namespace std; 的习惯，请参阅：Why is “using namespace std;” considered bad practice?. While you are at it, have a look at C++: “std::endl” vs “\n”。
在中间插入
当您在列表中间插入 at 时，您必须天真地跟踪 prev 和 current 节点，以便插入您的节点，以便 prev->next = node_to_insert; 和node_to_insert->next = current;。或者，您可以使用指针正确迭代 address-of-the-node 和 pointer-to-node 以便您可以简单地替换node_to_insert 节点地址的内容，并且 node_to_insert->next 等于 指向节点的指针 ->next。参见 Linus on Understanding Pointers
如果你这样做，事情就会变得容易得多。您的 insertAtMid() 函数变为：
node *insertAtMid (int n)
{
int length=0, middle;
node *iter = head,
**iteradr = &head;
while (iter != NULL) {
length++;
iter = iter->next;
}
middle = length/2;
iter = head;
while (middle--)
{
iteradr = &iter->next;
iter = iter->next;
}
// cout << "Enter data you want to insert : "; /* mixing */
// cin >> n;
node *temp = new node();
temp->data = n;
if (!head)
head = tail = temp;
else {
*iteradr = temp;
temp->next = iter;
}
//cout << "Inserted Successfully. Updated list is : "<<endl; /* mixing */
return temp;
}
总而言之，进行更多（轻微）更改并修复您的“混合应用程序和界面”问题，您可以这样做：
#include <iostream>
#include <limits>
struct node {
int data;
node *next;
};
class linkedList
{
private:
node *head, *tail;
public:
linkedList()
{
head = NULL;
tail = NULL;
}
node *addNode (int n) /* choose meaningful return to indicase success/failure */
{
// int item;
// cout << "Enter data for node : "; /* don't mix interface & application */
// cin >> item;
node *temp = new node();
temp->data = n;
temp->next = NULL;
if (head == NULL)
head = tail = temp;
else {
tail->next = temp;
tail = temp;
}
return temp; /* return pointer to allcated node */
}
node *getHead()
{
return head;
}
void display ()
{
int i = 1;
node *iter = head;
while (iter != NULL)
{
std::cout << "Node " << i << " : " << iter->data << '\n';
iter = iter->next;
i++;
}
}
node *insertAtBegining (int n)
{
// cout << "Enter data you want to insert : "; /* don't mix */
// cin >> n;
node *temp = new node();
temp->data = n;
temp->next = head;
head = temp;
// cout << "Inserted Successfully. Updated list is : "<<endl;
return temp;
}
node *insertAtMid (int n)
{
int length=0, middle;
node *iter = head,
**iteradr = &head;
while (iter != NULL) {
length++;
iter = iter->next;
}
middle = length/2;
iter = head;
while (middle--)
{
iteradr = &iter->next;
iter = iter->next;
}
// cout << "Enter data you want to insert : "; /* mixing */
// cin >> n;
node *temp = new node();
temp->data = n;
if (!head)
head = tail = temp;
else {
*iteradr = temp;
temp->next = iter;
}
//cout << "Inserted Successfully. Updated list is : "<<endl; /* mixing */
return temp;
}
};
int main (void)
{
linkedList l1;
char choice;
int menuChoice;
/** this is interface, input/output happens here */
std::cout << "List is not created yet, please add data to create the list.\n";
do
{
int n;
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
return 1;
}
if (!l1.addNode (n)) {
std::cerr << "error: adding node.\n";
return 1;
}
std::cout << "Node added. Want to add another node (y/n) : ";
if (!(std::cin >> choice)) {
std::cerr << "error: invalid integer input.\n";
return 1;
}
} while (choice == 'y' || choice == 'Y');
std::cout << "List created successfully. Select from options below : \n";
for (;;) { /* loop continually */
std::cout << "\n 1. Display list\n 2. Insert at Begining\n"
" 3. Insert at Mid\n 4. Insert at End\n 5. exit\n\n"
"choice: ";
while (!(std::cin >> menuChoice)) { /* you need to handle error here */
std::cerr << "error: invalid integer input.\n";
std::cin.clear();
std::cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');
std::cout << "\nchoice: ";
}
switch(menuChoice)
{
case 1:
l1.display();
break;
case 2: {
int n;
std::cout << "Enter data for node : ";
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
break;
}
if (l1.insertAtBegining (n)) {
std::cout << "Inserted Successfully. Updated list is : \n";
l1.display();
}
break;
}
case 3: {
int n;
std::cout << "Enter data for node : ";
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
break;
}
if (l1.insertAtMid (n)) {
std::cout << "Inserted Successfully. Updated list is : \n";
l1.display();
}
break;
}
case 4: {
int n;
std::cout << "Enter data for node : ";
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
break;
}
if (l1.addNode (n)) {
std::cout << "Inserted Successfully. Updated list is : \n";
l1.display();
}
break;
}
case 5:
return 0;
break;
default:
std::cout << "Invalid selection...!!\n";
break;
}
}
}
(注意： 你应该添加一个析构函数，以便在你使用 main() 以外的列表时释放列表内存)
并且从不列出一个列表l1（嗯，一个）。你知道这对老眼来说有多难吗？
示例Use/Output
$ ./bin/ll_class_menu
List is not created yet, please add data to create the list.
1
Node added. Want to add another node (y/n) : y
2
Node added. Want to add another node (y/n) : y
3
Node added. Want to add another node (y/n) : y
4
Node added. Want to add another node (y/n) : y
5
Node added. Want to add another node (y/n) : n
List created successfully. Select from options below :
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 1
Node 1 : 1
Node 2 : 2
Node 3 : 3
Node 4 : 4
Node 5 : 5
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 2
Enter data for node : 6
Inserted Successfully. Updated list is :
Node 1 : 6
Node 2 : 1
Node 3 : 2
Node 4 : 3
Node 5 : 4
Node 6 : 5
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 3
Enter data for node : 7
Inserted Successfully. Updated list is :
Node 1 : 6
Node 2 : 1
Node 3 : 2
Node 4 : 7
Node 5 : 3
Node 6 : 4
Node 7 : 5
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 4
Enter data for node : 8
Inserted Successfully. Updated list is :
Node 1 : 6
Node 2 : 1
Node 3 : 2
Node 4 : 7
Node 5 : 3
Node 6 : 4
Node 7 : 5
Node 8 : 8
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 5
这里有很多，慢慢看吧。确保您了解所做的更改以及进行更改的原因。如果您还有其他问题，请告诉我。
我正在创建一个简单的单向链表并尝试在链表的开头、中间和结尾插入一个节点。但是我的 'insertAtMid' 没有正常工作，它使程序崩溃并出错； “发生异常。分段错误。” 这是整个程序：
/*
Write a program that creates a singly linked list until user wants and displays it. then insert a node at the beginning, middle, and end of the list.
*/
#include <iostream>
using namespace std;
struct node
{
int data;
node *next;
};
class linkedList
{
private:
node *head, *tail;
public:
linkedList()
{
head = NULL;
tail = NULL;
}
void addNode()
{
int item;
cout << "Enter data for node : ";
cin >> item;
node *temp = new node();
temp->data = item;
temp->next = NULL;
if (head == NULL)
{
head = temp;
tail = temp;
}
else
{
tail->next = temp;
tail = tail->next;
}
}
node *getHead()
{
return head;
}
static void display(node *head)
{
int i = 1;
while (head != NULL)
{
cout << "Node " << i << " : " << head->data << endl;
head = head->next;
i++;
}
}
void insertAtBegining()
{
int n;
cout << "Enter data you want to insert : ";
cin >> n;
node *temp = new node();
temp->data = n;
temp-> next = head;
head = temp;
cout << "Inserted Successfully. Updated list is : "<<endl;
}
static void insertAtMid(node * head)
{
int n, length=0, middle;
while (head!=NULL)
{
length ++;
head = head->next;
}
middle = length/2;
while((middle --) > 1)
{
head = head->next;
}
cout << "Enter data you want to insert : ";
cin >> n;
node *temp = new node();
temp->data = n;
temp->next = head->next;
head = temp;
cout << "Inserted Successfully. Updated list is : "<<endl;
}
};
int main()
{
linkedList l1;
char choice;
int menuChoice;
cout << "List is not created yet, please add data to create the list." << endl;
do
{
l1.addNode();
cout << "Node added. Want to add another node (Y/N) : ";
cin >> choice;
} while (choice == 'y' || choice == 'Y');
cout << "List created successfully. Select from options below : "<<endl;
cout << "1. Display list \n2. Insert at Begining \n3. Insert at Mid \n4. Insert at End \n5. exit"<<endl;
cin >> menuChoice;
switch(menuChoice)
{
case 1:
linkedList::display(l1.getHead());
break;
case 2:
l1.insertAtBegining();
linkedList::display(l1.getHead());
break;
case 3:
l1.insertAtMid(l1.getHead());
linkedList::display(l1.getHead());
break;
case 4:
l1.addNode();
cout << "Successfully inserted node at end of list. Updated list is :"<<endl;
linkedList::display(l1.getHead());
break;
case 5:
exit;
break;
default:
cout << "Invalid selection...!!";
}
}
编译器显示错误在第 85 行，那里发生了分段错误。这是错误的屏幕截图： Exception has occurred
根据您的代码实例，您正在计算节点数并将头指针移动到列表末尾。
在执行以下语句之前，head 当前将指向最后一个节点。
while((middle --) > 1)
{
head = head->next;
}
一旦这条语句 head = head->next; 被执行，那么 head 将指向任何地方，或者你可以说，它指向 NULL。
这就是您遇到分段错误的原因，因为 head->next 试图提供对地址的引用，这是一个垃圾。那里什么都不存在。
代替使用head进行迭代，使用临时节点，将其分配给head然后计算列表的长度。
node *tmp = new node();
tmp = head;
int length = 0;
while(tmp != NULL){
length++;
tmp = tmp->next;
}
middle = ((length % 2) == 0) ? (length /2 ) : (length + 1)/2);
tmp = head;
while((middle --) > 1){
tmp = tmp->next;
}
cout << "Enter data you want to insert : ";
cin >> n;
node *curr = new node();
curr->data = n;
curr->next = temp->next;
temp->next = curr;
我建议您不要使用 head 指针进行迭代，因为它可能会导致悬空引用。
哇，你真的很困惑...首先，你有一个 class，有 private: 个成员 head 和 tail。您不会将 head 或 tail 作为参数传递给您的成员函数。您的 class 成员函数可以访问您的 class 私有成员。
所以你不做:
static void display(node *head)
{
...
相反，只需：
void display ()
{
...
您不要混合应用程序和界面
您的应用程序是您处理从程序用户界面获取的数据的方式。你把两者分开。您的菜单就是您的界面，您的 linked 列表就是您的应用程序（粗略地说）。您在程序界面中执行所有输出和数据输入，并将数据传递给应用程序进行处理。这使您的代码可维护且更易于阅读（并且更容易遵循逻辑）
您不会将数据提示隐藏在 addnode() 函数中。相反，您获取要在 main() 的初始部分或 switch() 语句中添加的数据，然后将整数值传递给 addnode()。示例：
node *addNode (int n) /* choose meaningful return to indicase success/failure */
{
// int item;
// cout << "Enter data for node : "; /* don't mix interface & application */
// cin >> item;
node *temp = new node();
temp->data = n;
temp->next = NULL;
if (head == NULL)
head = tail = temp;
else {
tail->next = temp;
tail = temp;
}
return temp; /* return pointer to allcated node */
}
然后：
/** this is interface, input/output happens here */
std::cout << "List is not created yet, please add data to create the list.\n";
do
{
int n;
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
return 1;
}
if (!l1.addNode (n)) {
std::cerr << "error: adding node.\n";
return 1;
}
std::cout << "Node added. Want to add another node (y/n) : ";
if (!(std::cin >> choice)) {
std::cerr << "error: invalid integer input.\n";
return 1;
}
} while (choice == 'y' || choice == 'Y');
（对于 switch() 语句中的菜单也是如此）
您验证每个用户输入
除非您检查return，否则您无法正确使用任何输入功能。 （从技术上讲，您检查流状态）。您可以通过检查每个输入是否成功或处理错误来做到这一点。永远不要 cin >> data; 并希望 data 是好的。例如（最少）：
int n;
std::cout << "Enter data for node : ";
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
break; /* note: this just lets the menu fail before clearing error */
}
if (l1.insertAtBegining (n)) {
std::cout << "Inserted Successfully. Updated list is : \n";
l1.display();
}
break;
对于使用 iostream 的任何输入，您必须 .clear() 流状态才能进行进一步的输入，并且您必须清空导致错误的输入缓冲区中的内容。一个更完整的例子是：
while (!(std::cin >> menuChoice)) { /* you need to handle error here */
std::cerr << "error: invalid integer input.\n";
std::cin.clear();
std::cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');
std::cout << "\nchoice: ";
}
参见 std::basic_istream::ignore and bookmark the cppreference.com link -- 这是您在 Internet 上最好的参考资料 -- 使用它。
（您应该更新其余代码以正确处理输入）
不要使用 head
您的 head 指针指向列表开头的节点。不要用它来遍历你的列表。 （完成后——它不再指向列表的开头，并且您丢失了指向开头的指针）。相反，你声明一个临时指针来迭代，例如：
void display ()
{
int i = 1;
node *iter = head;
while (iter != NULL)
{
std::cout << "Node " << i << " : " << iter->data << '\n';
iter = iter->next;
i++;
}
}
不要养成 using namespace std; 的习惯，请参阅：Why is “using namespace std;” considered bad practice?. While you are at it, have a look at C++: “std::endl” vs “\n”。
在中间插入
当您在列表中间插入 at 时，您必须天真地跟踪 prev 和 current 节点，以便插入您的节点，以便 prev->next = node_to_insert; 和node_to_insert->next = current;。或者，您可以使用指针正确迭代 address-of-the-node 和 pointer-to-node 以便您可以简单地替换node_to_insert 节点地址的内容，并且 node_to_insert->next 等于 指向节点的指针 ->next。参见 Linus on Understanding Pointers
如果你这样做，事情就会变得容易得多。您的 insertAtMid() 函数变为：
node *insertAtMid (int n)
{
int length=0, middle;
node *iter = head,
**iteradr = &head;
while (iter != NULL) {
length++;
iter = iter->next;
}
middle = length/2;
iter = head;
while (middle--)
{
iteradr = &iter->next;
iter = iter->next;
}
// cout << "Enter data you want to insert : "; /* mixing */
// cin >> n;
node *temp = new node();
temp->data = n;
if (!head)
head = tail = temp;
else {
*iteradr = temp;
temp->next = iter;
}
//cout << "Inserted Successfully. Updated list is : "<<endl; /* mixing */
return temp;
}
总而言之，进行更多（轻微）更改并修复您的“混合应用程序和界面”问题，您可以这样做：
#include <iostream>
#include <limits>
struct node {
int data;
node *next;
};
class linkedList
{
private:
node *head, *tail;
public:
linkedList()
{
head = NULL;
tail = NULL;
}
node *addNode (int n) /* choose meaningful return to indicase success/failure */
{
// int item;
// cout << "Enter data for node : "; /* don't mix interface & application */
// cin >> item;
node *temp = new node();
temp->data = n;
temp->next = NULL;
if (head == NULL)
head = tail = temp;
else {
tail->next = temp;
tail = temp;
}
return temp; /* return pointer to allcated node */
}
node *getHead()
{
return head;
}
void display ()
{
int i = 1;
node *iter = head;
while (iter != NULL)
{
std::cout << "Node " << i << " : " << iter->data << '\n';
iter = iter->next;
i++;
}
}
node *insertAtBegining (int n)
{
// cout << "Enter data you want to insert : "; /* don't mix */
// cin >> n;
node *temp = new node();
temp->data = n;
temp->next = head;
head = temp;
// cout << "Inserted Successfully. Updated list is : "<<endl;
return temp;
}
node *insertAtMid (int n)
{
int length=0, middle;
node *iter = head,
**iteradr = &head;
while (iter != NULL) {
length++;
iter = iter->next;
}
middle = length/2;
iter = head;
while (middle--)
{
iteradr = &iter->next;
iter = iter->next;
}
// cout << "Enter data you want to insert : "; /* mixing */
// cin >> n;
node *temp = new node();
temp->data = n;
if (!head)
head = tail = temp;
else {
*iteradr = temp;
temp->next = iter;
}
//cout << "Inserted Successfully. Updated list is : "<<endl; /* mixing */
return temp;
}
};
int main (void)
{
linkedList l1;
char choice;
int menuChoice;
/** this is interface, input/output happens here */
std::cout << "List is not created yet, please add data to create the list.\n";
do
{
int n;
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
return 1;
}
if (!l1.addNode (n)) {
std::cerr << "error: adding node.\n";
return 1;
}
std::cout << "Node added. Want to add another node (y/n) : ";
if (!(std::cin >> choice)) {
std::cerr << "error: invalid integer input.\n";
return 1;
}
} while (choice == 'y' || choice == 'Y');
std::cout << "List created successfully. Select from options below : \n";
for (;;) { /* loop continually */
std::cout << "\n 1. Display list\n 2. Insert at Begining\n"
" 3. Insert at Mid\n 4. Insert at End\n 5. exit\n\n"
"choice: ";
while (!(std::cin >> menuChoice)) { /* you need to handle error here */
std::cerr << "error: invalid integer input.\n";
std::cin.clear();
std::cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');
std::cout << "\nchoice: ";
}
switch(menuChoice)
{
case 1:
l1.display();
break;
case 2: {
int n;
std::cout << "Enter data for node : ";
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
break;
}
if (l1.insertAtBegining (n)) {
std::cout << "Inserted Successfully. Updated list is : \n";
l1.display();
}
break;
}
case 3: {
int n;
std::cout << "Enter data for node : ";
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
break;
}
if (l1.insertAtMid (n)) {
std::cout << "Inserted Successfully. Updated list is : \n";
l1.display();
}
break;
}
case 4: {
int n;
std::cout << "Enter data for node : ";
if (!(std::cin >> n)) {
std::cerr << "error: invalid integer input.\n";
break;
}
if (l1.addNode (n)) {
std::cout << "Inserted Successfully. Updated list is : \n";
l1.display();
}
break;
}
case 5:
return 0;
break;
default:
std::cout << "Invalid selection...!!\n";
break;
}
}
}
(注意： 你应该添加一个析构函数，以便在你使用 main() 以外的列表时释放列表内存)
并且从不列出一个列表l1（嗯，一个）。你知道这对老眼来说有多难吗？
示例Use/Output
$ ./bin/ll_class_menu
List is not created yet, please add data to create the list.
1
Node added. Want to add another node (y/n) : y
2
Node added. Want to add another node (y/n) : y
3
Node added. Want to add another node (y/n) : y
4
Node added. Want to add another node (y/n) : y
5
Node added. Want to add another node (y/n) : n
List created successfully. Select from options below :
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 1
Node 1 : 1
Node 2 : 2
Node 3 : 3
Node 4 : 4
Node 5 : 5
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 2
Enter data for node : 6
Inserted Successfully. Updated list is :
Node 1 : 6
Node 2 : 1
Node 3 : 2
Node 4 : 3
Node 5 : 4
Node 6 : 5
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 3
Enter data for node : 7
Inserted Successfully. Updated list is :
Node 1 : 6
Node 2 : 1
Node 3 : 2
Node 4 : 7
Node 5 : 3
Node 6 : 4
Node 7 : 5
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 4
Enter data for node : 8
Inserted Successfully. Updated list is :
Node 1 : 6
Node 2 : 1
Node 3 : 2
Node 4 : 7
Node 5 : 3
Node 6 : 4
Node 7 : 5
Node 8 : 8
1. Display list
2. Insert at Begining
3. Insert at Mid
4. Insert at End
5. exit
choice: 5
这里有很多，慢慢看吧。确保您了解所做的更改以及进行更改的原因。如果您还有其他问题，请告诉我。