排序链表C程序出错 [英] Error in sorted linked list C program
问题描述
大家好，我应该编写一个程序，将整数插入到排序的链表中。然后删除用户选择的任何元素。在代码中我无法将元素插入到我的链表中。有人可以帮帮我吗？
提前致谢
代码
[C]排序链表 - Pastebin.com [ ^ ]
我的尝试：
#include < stdio.h >
#include < string.h >
#include < stdlib.h >
#include < stdbool.h >
struct node {
// int data;
< span class =code-keyword> int key;
struct node * next;
};
struct node * init（）{
struct node * head = 0 ;
return head;
}
int isempty（ struct 节点* head）{
if （head == NULL）
return 1 ;
else
return 2 < /跨度>;
}
// 显示列表
void printList（ struct node * head）{
struct node * ptr = head;
printf（ \ n）;
int n;
n = isempty（head）;
if （n == 1 ）
printf（ list为空）;
// 从头开始
while （ptr！= NULL）{
printf（ ％ d，ptr-> key）;
ptr = ptr-> next;
}
}
void create（ struct 节点* head， int num）{
struct 节点* tmp = head;
struct node * prev = NULL;
struct node * new = malloc（ sizeof （ struct node））;
new - > key = num;
prev = tmp;
int n;
n = isempty（head）;
if （n == 1 ）
head-> next = < span class =code-keyword> new ;
while （tmp！= NULL&& tmp-> key< num）{
prev = tmp;
tmp = tmp-> next;
}
new - > next = tmp;
if （prev！= NULL）
prev-> next = new ;
}
/ * struct node * tmp;
tmp =（struct node *）malloc（sizeof（struct node））;
tmp-> data = data;
tmp-> link = start;
start = tmp; * /
// 删除a链接给定键
struct 节点* 删除（ struct node * head， int del）{
// 从第一个链接开始
struct node * current = head;
struct node * previous = NULL;
struct node * temp = NULL;
// 如果列表为空
int n;
n = isempty（head）;
if （n == 1 ）{
printf（ list为空）;
return NULL;
}
// 在列表中导航
while （current-> key！= del）{
if （current == head） ）{
temp = head;
temp = temp-> next;
免费（头）;
head = temp;
}
// 如果是最后一个节点
if （current-> next == NULL）{
return NULL;
} 其他 {
// 存储对当前链接的引用
previous = current;
// 移至下一个链接
current = current-> next;
}
}
// 找到匹配项，更新link
if （current == head）{
// 首先改为指向下一个链接
head = head-> next;
} 其他 {
// 绕过当前链接
previous-> next = current-> next;
}
返回当前;
}
int main（）{
int op;
int num;
struct node * head;
head = init（）;
执行 {
printf（ \ n Menu \\\
1.Insert \\\
2.delete element \\\
3.display List \\\
4. end program）;
printf（ n \ n \ n请输入选项：）;
scanf（ ％d，& op）;
switch （op）{
case 1 ：
printf（ 输入数据：）;
scanf（ ％d，& num）;
create（head，num）;
break ;
case 2 ：
printf（ 输入数据：）;
scanf（ ％d，& num）;
delete （head，num）;
break ;
case 3 ：
printList（head）;
break ;
case 4 ：
free（head）;
退出（ 0 ）;
默认：
printf（ \ n输入一个选项：）;
}
} while （ 1 ）;
}
我正在向您展示一些正常工作的代码。丢失的部分留作练习。
#include < span class =code-keyword>< stdio.h >
#include < stdlib.h >
＃ include < assert.h >
struct node
{
int 键;
struct node * next;
};
// 不需要init和isempty
// 显示列表中的所有项目
void show（ struct node * head）
{
struct node * ptr = head;
if （！ptr）
{
printf（ < span class =code-string>（empty））;
}
else
{
while （ptr ）
{
printf（ ％d，ptr-> key ）;
ptr = ptr-> next;
}
}
printf（ \ n） ;
}
// 创建一个新节点并将其插入列表中，维护订单
void insert（ struct node ** phead， int key）
{
struct node * 新;
struct node * cur = * phead;
struct node * prev = NULL;
new =（ struct node *）malloc（ sizeof （* new ））;
断言（ new ）;
new - > key = key;
new - > next = NULL;
while （cur&& cur-> key< key）
{
prev = cur ;
cur = cur-> next;
}
if （cur）
new - > next = cur;
if （prev）
prev-> next = new ;
else
* phead = new ;
}
// 删除所有节点，释放分配的内存
void clear（ struct node * head）
{
while （head）
{
struct node * ptr = head;
head = head-> next;
free（ptr）;
}
}
enum
{
INSERT = 1 ，
DELETE，
SHOW，
TERMINATE
};
int main（）
{
int 选项;
struct node * head = NULL;
执行
{
printf（ \ n％d。插入\ n％d。删除\ n％d.Show \ n％d终止\ n，INSERT，DELETE，SHOW，TERMINATE）;
printf（ \ n请输入na选项：）;
if （scanf（ ％d ，& option）！= 1 ）继续;
switch （选项）
{
case INSERT：
{
int 键;
printf（ 请输入数据：）;
if （scanf（ ％d ，& key）== 1 ）
insert（& head，key）;
}
break ;
case DELETE：
printf（ 留下来锻炼\ n）;
break ;
case SHOW：
show（head）;
break ;
case TERMINATE：
clear（head）;
break ;
默认：
printf（ 请输入允许的选项\ n）;
}
} while （选项！= TERMINATE）;
}
编译并不代表你的代码是对的！ ：笑：
将开发过程想象成编写电子邮件：成功编译意味着您使用正确的语言编写电子邮件 - 例如英语而不是德语 - 而不是电子邮件包含您的邮件想发送。
所以现在你进入第二阶段的发展（实际上它是第四或第五阶段，但你将在之后的阶段进入）：测试和调试。
首先查看它的作用，以及它与你想要的有何不同。这很重要，因为它可以为您提供有关其原因的信息。例如，如果程序旨在让用户输入一个数字并将其翻倍并打印答案，那么如果输入/输出是这样的：
输入预期输出实际输出
1 2 1
2 4 4
3 6 9
4 8 16然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。
所以，你可以查看代码和很明显，它在某处：
int Double （ int value ）
{
return value * 值;
}
一旦你知道可能出现的问题，就开始使用调试器找出原因。在方法的第一行放置一个断点，然后运行你的应用程序。当它到达断点时，调试器将停止，并将控制权移交给您。您现在可以逐行运行代码（称为单步执行）并根据需要查看（甚至更改）变量内容（哎呀，您甚至可以更改代码并在需要时再试一次）。 />
在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。
如果没有，为什么不呢？它有什么不同？
希望这可以帮助你找到代码的哪个部分有问题，以及问题是什么。
这是一项技能，它是一个值得开发的，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改进！
学会正确地缩进你的代码，它显示它的结构，它有助于阅读和理解。它还有助于发现结构错误。
#include < span class =code-keyword>< stdio.h >
#include < string.h >
＃ include < stdlib.h >
#include < stdbool.h >
struct node {
// int data;
int key;
struct node * next;
};
struct node * init（）{
struct node * head = 0 ;
return head;
}
int isempty（ struct 节点* head）{
if （head == NULL）
return 1 ;
else
return 2 < /跨度>;
}
// 显示列表
void printList（ struct node * head）{
struct node * ptr = head;
printf（ \ n）;
int n;
n = isempty（head）;
if （n == 1 ）
printf（ list为空）;
// 从头开始
while （ptr！= NULL）{
printf（ ％ d，ptr-> key）;
ptr = ptr-> next;
}
}
void create（ struct 节点* head， int num）{
struct 节点* tmp = head;
struct node * prev = NULL;
struct node * new = malloc（ sizeof （ struct node））;
new - > key = num;
prev = tmp;
int n;
n = isempty（head）;
if （n == 1 ）
head-> next = < span class =code-keyword> new ;
while （tmp！= NULL&& tmp-> key< num）{
prev = tmp;
tmp = tmp-> next;
}
new - > next = tmp;
if （prev！= NULL）
prev-> next = new ;
}
/ * struct node * tmp;
tmp =（struct node *）malloc（sizeof（struct node））;
tmp-> data = data;
tmp-> link = start;
start = tmp; * /
// 删除a链接给定键
struct 节点* 删除（ struct node * head， int del）{
// 从第一个链接开始
struct node * current = head;
struct node * previous = NULL;
struct node * temp = NULL;
// 如果列表为空
int n;
n = isempty（head）;
if （n == 1 ）{
printf（ list为空）;
return NULL;
}
// 在列表中导航
while （current-> key！= del）{
if （current == head） ）{
temp = head;
temp = temp-> next;
免费（头）;
head = temp;
}
// 如果是最后一个节点
if （current-> next == NULL）{
return NULL;
} 其他 {
// 存储对当前链接的引用
previous = current;
// 移至下一个链接
current = current-> next;
}
}
// 找到匹配项，更新link
if （current == head）{
// 首先改为指向下一个链接
head = head-> next;
} 其他 {
// 绕过当前链接
previous-> next = current-> next;
}
返回当前;
}
int main（）{
int op;
int num;
struct node * head;
head = init（）;
执行 {
printf（ \ n Menu \\\
1.Insert \\\
2.delete element \\\
3.display List \\\
4. end program）;
printf（ n \ n \ n请输入选项：）;
scanf（ ％d，& op）;
switch （op）{
case 1 ：
printf（ 输入数据：）;
scanf（ ％d，& num）;
create（head，num）;
break ;
case 2 ：
printf（ 输入数据：）;
scanf（ ％d，& num）;
delete （head，num）;
break ;
case 3 ：
printList（head）;
break ;
case 4 ：
free（head）;
退出（ 0 ）;
默认：
printf（ \ n输入一个选项：）;
}
} while （ 1 ）;
}
专业程序员的编辑器具有此功能，其他功能包括括号匹配和语法高亮。
< a href =https://notepad-plus-plus.org/> Notepad ++ Home [ ^ ]
ultraedit [ ^ ]
-----
引用：在代码中我无法将元素插入到我的链表中。有人能帮帮我吗？
您的代码没有按照您的预期行事，或者您不明白为什么！
有一个几乎通用的解决方案：逐步在调试器上运行代码，检查变量。
调试器在这里向您展示您的代码正在做什么，您的任务是与它应该做什么进行比较做。
调试器中没有魔法，它不知道你的代码应该做什么，它没有找到错误，只是通过向你展示发生了什么来帮助你。当代码没有达到预期的效果时，你就接近了一个错误。
要查看你的代码在做什么：只需设置断点并查看代码是否正常运行，调试器允许你执行第1行第1行并在执行时检查变量。
此解决方案的缺点：
- 这是一个DIY，你是跟踪问题并找到根源的那个，这导致了解决方案。
这个解决方案的优点：
- 它也是一个很好的学习工具，因为它告诉你现实，你可以看到哪种期望与现实相符。
次要效果
- 你会为自己找到虫子感到自豪。
- 你的学习技巧会提高。
你应该很快就会发现什么是错的。
调试器 - 维基百科，免费的百科全书 [ ^ ]
掌握Visual Studio 2010中的调试 - 初学者指南 [ ^ ]
使用Visual Studio 2010进行基本调试 - YouTube [ ^ ]
1.11 - 调试程序（步进和断点）|学习C ++ [ ^ ]
调试器仅显示您的代码正在执行的操作，您的任务是与应该执行的操作进行比较。
Hallo everyone , i am supposed to write a program to insert integers into a sorted linked list. and then delete any element the user chooses to. in the code i am unable to insert elements into my linked list. can someone help me?
Thanks in advance
the code is
[C] sorted linked list - Pastebin.com[^]
What I have tried:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
struct node {
// int data;
int key;
struct node *next;
};
struct node* init(){
struct node *head =0;
return head;
}
int isempty(struct node * head){
if(head == NULL)
return 1;
else
return 2;
}
//display the list
void printList(struct node * head) {
struct node *ptr = head;
printf("\n");
int n;
n=isempty(head);
if(n == 1)
printf("list is empty");
//start from the beginning
while(ptr != NULL) {
printf("%d ",ptr->key);
ptr = ptr->next;
}
}
void create(struct node * head,int num) {
struct node * tmp = head;
struct node * prev = NULL;
struct node* new = malloc(sizeof(struct node));
new->key = num;
prev = tmp;
int n;
n=isempty(head);
if(n == 1)
head->next = new;
while(tmp != NULL && tmp->key < num){
prev = tmp;
tmp = tmp->next;
}
new->next = tmp;
if(prev !=NULL)
prev->next = new;
}
/* struct node *tmp;
tmp=(struct node *)malloc(sizeof(struct node ));
tmp->data=data;
tmp->link=start;
start=tmp;*/
//delete a link with given key
struct node* delete(struct node * head, int del) {
//start from the first link
struct node* current = head;
struct node* previous = NULL;
struct node* temp = NULL;
//if list is empty
int n;
n=isempty(head);
if(n == 1) {
printf("list is empty");
return NULL;
}
//navigate through list
while(current->key != del) {
if (current == head){
temp = head;
temp = temp->next;
free(head);
head = temp;
}
//if it is last node
if(current->next == NULL) {
return NULL;
} else {
//store reference to current link
previous = current;
//move to next link
current = current->next;
}
}
//found a match, update the link
if(current == head) {
//change first to point to next link
head = head->next;
} else {
//bypass the current link
previous->next = current->next;
}
return current;
}
int main() {
int op;
int num;
struct node* head;
head=init();
do{
printf("\n Menu \n 1.Insert \n 2.delete element \n 3.display List \n 4. end program ");
printf("n \n \n please enter an option : ");
scanf("%d",&op);
switch (op) {
case 1:
printf("Enter data:");
scanf("%d",&num);
create(head, num);
break;
case 2:
printf("Enter data:");
scanf("%d",&num);
delete(head,num);
break;
case 3:
printList(head);
break;
case 4:
free(head);
exit(0);
default:
printf("\n enter an option : ");
}
}while(1);
}I'm showing you some working code. Missing parts are left as exercise.
#include <stdio.h> #include <stdlib.h> #include <assert.h> struct node { int key; struct node * next; }; // init and isempty aren't really needed // shows all the items of the list void show( struct node * head ) { struct node * ptr = head; if ( ! ptr ) { printf("(empty)"); } else { while ( ptr ) { printf(" %d ", ptr->key); ptr = ptr->next; } } printf("\n"); } // creates a new node and inserts it into the list, maintaining the order void insert(struct node ** phead, int key ) { struct node * new; struct node * cur = *phead; struct node * prev = NULL; new = (struct node *) malloc( sizeof (*new) ); assert(new); new->key = key; new->next = NULL; while ( cur && cur->key < key) { prev = cur; cur = cur->next; } if ( cur ) new->next = cur; if ( prev ) prev->next = new; else *phead=new; } // removes all the nodes, freeing the allocated memory void clear(struct node * head) { while ( head ) { struct node * ptr = head; head = head->next; free(ptr); } } enum { INSERT = 1, DELETE, SHOW, TERMINATE }; int main() { int option; struct node * head = NULL; do { printf("\n %d.Insert\n %d.Delete\n %d.Show\n %d Terminate\n", INSERT, DELETE, SHOW, TERMINATE); printf("\n Please enter na option: "); if ( scanf("%d", &option) != 1) continue; switch (option) { case INSERT: { int key; printf("Please enter data: "); if ( scanf("%d", &key) == 1) insert( &head, key); } break; case DELETE: printf("Left as exercise\n"); break; case SHOW: show(head); break; case TERMINATE: clear(head); break; default: printf("Please enter an allowed option\n"); } } while ( option != TERMINATE); }
Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.
So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.
Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:
Input Expected output Actual output 1 2 1 2 4 4 3 6 9 4 8 16Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:
int Double(int value) { return value * value; }
Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the method, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!
Learn to indent properly your code, it show its structure and it helps reading and understanding. It also helps spotting structures mistakes.
#include <stdio.h> #include <string.h> #include <stdlib.h> #include <stdbool.h> struct node { // int data; int key; struct node *next; }; struct node* init(){ struct node *head =0; return head; } int isempty(struct node * head){ if(head == NULL) return 1; else return 2; } //display the list void printList(struct node * head) { struct node *ptr = head; printf("\n"); int n; n=isempty(head); if(n == 1) printf("list is empty"); //start from the beginning while(ptr != NULL) { printf("%d ",ptr->key); ptr = ptr->next; } } void create(struct node * head,int num) { struct node * tmp = head; struct node * prev = NULL; struct node* new = malloc(sizeof(struct node)); new->key = num; prev = tmp; int n; n=isempty(head); if(n == 1) head->next = new; while(tmp != NULL && tmp->key < num){ prev = tmp; tmp = tmp->next; } new->next = tmp; if(prev !=NULL) prev->next = new; } /* struct node *tmp; tmp=(struct node *)malloc(sizeof(struct node )); tmp->data=data; tmp->link=start; start=tmp;*/ //delete a link with given key struct node* delete(struct node * head, int del) { //start from the first link struct node* current = head; struct node* previous = NULL; struct node* temp = NULL; //if list is empty int n; n=isempty(head); if(n == 1) { printf("list is empty"); return NULL; } //navigate through list while(current->key != del) { if (current == head){ temp = head; temp = temp->next; free(head); head = temp; } //if it is last node if(current->next == NULL) { return NULL; } else { //store reference to current link previous = current; //move to next link current = current->next; } } //found a match, update the link if(current == head) { //change first to point to next link head = head->next; } else { //bypass the current link previous->next = current->next; } return current; } int main() { int op; int num; struct node* head; head=init(); do{ printf("\n Menu \n 1.Insert \n 2.delete element \n 3.display List \n 4. end program "); printf("n \n \n please enter an option : "); scanf("%d",&op); switch (op) { case 1: printf("Enter data:"); scanf("%d",&num); create(head, num); break; case 2: printf("Enter data:"); scanf("%d",&num); delete(head,num); break; case 3: printList(head); break; case 4: free(head); exit(0); default: printf("\n enter an option : "); } }while(1); }
Professional programmer's editors have this feature and others ones such as parenthesis matching and syntax highlighting.
Notepad++ Home[^]
ultraedit[^]
-----
Quote:in the code i am unable to insert elements into my linked list. can someone help me?
Your code do not behave the way you expect, or you don't understand why !
There is an almost universal solution: Run your code on debugger step by step, inspect variables.
The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't know what your code is supposed to do, it don't find bugs, it just help you to by showing you what is going on. When the code don't do what is expected, you are close to a bug.
To see what your code is doing: Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute.
The downside of this solution:
- It is a DIY, you are the one tracking the problem and finding its roots, which lead to the solution.
The upside of this solution:
- It is also a great learning tool because it show you reality and you can see which expectation match reality.
secondary effects
- Your will be proud of finding bugs yourself.
- Your learning skills will improve.
You should find pretty quickly what is wrong.
Debugger - Wikipedia, the free encyclopedia[^]
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
Basic Debugging with Visual Studio 2010 - YouTube[^]
1.11 — Debugging your program (stepping and breakpoints) | Learn C++[^]
The debugger is here to only show you what your code is doing and your task is to compare with what it should do.
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