无法在C中创建正确的链接列表

Question

我想构建仅具有add_to_end和show_list函数的链接列表，但当我想要显示head时，尽管item工作（查看代码），但列表崩溃。

#include <stdio.h> 
#include <stdlib.h> 

typedef struct Data 
{ 
    int x; 
    int y; 
    struct Data * next; 
}List; 

void AddEnd(List * item, List * head); 
void Show(List * head); 

int main(void) 
{ 
    int choice; 
    List item; 
    List * head; 
    List * temp; 
    head = NULL; 
    while(printf("q - quit - Enter 1 add end, 2 show: "), scanf("%d", &choice)) 
    { 
     switch(choice) 
     { 
      case 1: 
       AddEnd(&item, head); 
       break; 
      case 2: 
       printf("X = %d y= %d\n", item.x, item.y);    /*prints 1 2*/ 
       printf("x = %d y = %d\n", head->x, head->y);   /*crash of program...should print 1 2*/ 
       Show(head); 
       break; 
      default: 
       printf("TRY AGAIN\n"); 
       break;   
     } 
    } 
    temp = head; 
    while(temp) 
    { 
     free(temp); 
     temp = head->next; 
     head = temp->next; 
    } 
    return 0; 
} 

void AddEnd(List * item, List * head) 
{ 
    List * node; 
    node = (List *)malloc(sizeof(List)); 
    printf("Enter x and y: "); 
    scanf("%d %d", &node->x, &node->y); 
    if(head == NULL) 
    { 
     node->next = NULL; 
     head = node; 
     * item = * head; 

    } 
    else 
    { 
     item->next = node; 
     node->next = NULL; 
    } 
} 

void Show(List * head) 
{ 
    List * node; 
    node = head; 
    while(node) 
    { 
     printf("x = %d y = %d\n", node->x, node->y); 
     node = node->next; 
    } 
} 

Answer 1

你写的代码是一个完整的混乱。我不确定为什么你需要在那里使用变量item。找到下面修改的代码并尝试执行它。在主函数中，声明了一个指向列表头的指针，将其设置为NULL，然后将NULL值作为参数发送给AddEnd函数。这是行不通的。您需要将&head作为参数发送给函数，以便将更改反映回调用函数。

#include <stdio.h> 
#include <stdlib.h> 

typedef struct Data 
{ 
    int x; 
    int y; 
    struct Data * next; 
}List; 

void AddEnd(List * item, List ** head); 
void Show(List * head); 

int main(void) 
{ 
    int choice; 
    List item; 
    List * head; 
    List * temp; 
    head = NULL; 
    while(printf("q - quit - Enter 1 add end, 2 show: "), scanf("%d", &choice)) 
    { 
     switch(choice) 
     { 
      case 1: 
       AddEnd(&item, &head); 
       break; 
      case 2: 
       // printf("X = %d y= %d\n", item.x, item.y);    /*prints 1 2*/ 
       // printf("x = %d y = %d\n", head->x, head->y);   /*crash of program...should print 1 2*/ 
       Show(head); 
       break; 
      default: 
       printf("TRY AGAIN\n"); 
       break;   
     } 
    } 
    temp = head; 
    while(temp) 
    { 
     free(temp); 
     temp = head->next; 
     head = temp->next; 
    } 
    return 0; 
} 

void AddEnd(List * item, List ** head) 
{ 
    List * node,*first,*second; 
    node = (List *)malloc(sizeof(List)); 
    printf("Enter x and y: "); 
    scanf("%d %d", &node->x, &node->y); 
    if(*head == NULL) 
    { 
     node->next = NULL; 
     *head = node; 
     // * item = **head; 

    } 
    else 
    { 
     for(first=*head;first!=NULL;first=first->next)//traverse to the end of the list 
      second=first; 
     node->next = NULL; 
     second->next=node; 
    } 
} 

void Show(List * head) 
{ 
    List * node; 
    node = head; 
    while(node) 
    { 
     printf("x = %d y = %d\n", node->x, node->y); 
     node = node->next; 
    } 
} 

Answer 2

void AddEnd(List * item, List * head) 
{ 
    List * node; 
    node = (List *)malloc(sizeof(List)); 
    printf("Enter x and y: "); 
    scanf("%d %d", &node->x, &node->y); 
    if(head == NULL) 
    { 
     head=node; 
     head->next=NULL; 
    // head->x=item->x; 
    // head->y=item->y; you don'tneed this as you have it in node. 
    } 
    else 
    { 
    node->next = head->next; 
    head->next = node; 
    // node->x=item->x; 
    // node->y=item->y; 
    } 
} 

Answer 3

的问题是这样的

AddEnd(List * item,List* head)

在这里，当你调用加数像这样

head = null; addEnd(&item,head);

的malloc分配地址将被复制到本地变量head在AddEnd中，而不是局部变量head的main()。在main（）中的头不变。

溶液：

变化加数作为这样 AddEnd(List * item,List ** head)和在加数改变代码相若方式也。

呼叫AddEnd(&item,&head);