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C 释放一个双链接列表_C_Linked List_Free_Doubly Linked List - Fatal编程技术网

C 释放一个双链接列表

C 释放一个双链接列表,c,linked-list,free,doubly-linked-list,C,Linked List,Free,Doubly Linked List,我正在尝试释放一个双链接列表,我的问题是是否还需要释放每个节点中的所有数据和指针。多谢各位 功能: static void free_list(Room *head, Room *head2) { Room *tmp = head; Room *tmp2 = head2; Room *store; Room *store2; tmp = head2; tmp2 = head; printf("\nFreeing trap lis

我正在尝试释放一个双链接列表,我的问题是是否还需要释放每个节点中的所有数据和指针。多谢各位

功能:

static void free_list(Room *head, Room *head2) {
    Room *tmp = head;
    Room *tmp2 = head2;
    Room *store;
    Room *store2;
    tmp = head2;
    tmp2 = head;

    printf("\nFreeing trap list...\n");
    sleep(2);
    while (tmp != NULL) {
        store = tmp->pNext;
        free(tmp);
        tmp = store;
    }

    printf("\nFreeing rooms list...\n");
    sleep(2);
    while (tmp2 != NULL) {
        store2 = tmp2->pNext;
        free(tmp2);
        tmp2 = store2;
    }
}
结构:

typedef struct Room {
    struct Room *forward;
    struct Room *left;
    struct Room *right;
    struct Room *previous;
    struct Room *pPrev;
    struct Room *pNext;
    Room_Type Room_Type;
    bool emergency_call;
} Room;

那么,在本例中,我是否还需要释放前向指针以及其他类型的指针
head
head2
是两个不同的指针,每个指针指向两个不同列表的开头。

这种定义容器的方法非常混乱:

typedef struct Room{
  struct Room* forward;
  struct Room* left;
  struct Room* right;
  struct Room* previous;
  struct Room* pPrev;
  struct Room* pNext;
  Room_Type Room_Type;
  bool emergency_call;
} Room;
分而治之:

typedef struct Node {
  struct Node* pPrev;
  struct Node* pNext;
  Room_Type Room_Type;
  bool emergency_call;
} Node;

typedef struct List {
  struct Node* pHead;
  struct Node* pTail;
} List;
使用这种方法,一个循环就足够了:

void free_list(List *list)
{
    Node *node = list->pHead;

    while (node != NULL)
    {
        Node *next = node->pNext;

        free(node);
        node = next;
    }
    free(list);
}

这种定义容器的方法非常混乱:

typedef struct Room{
  struct Room* forward;
  struct Room* left;
  struct Room* right;
  struct Room* previous;
  struct Room* pPrev;
  struct Room* pNext;
  Room_Type Room_Type;
  bool emergency_call;
} Room;
分而治之:

typedef struct Node {
  struct Node* pPrev;
  struct Node* pNext;
  Room_Type Room_Type;
  bool emergency_call;
} Node;

typedef struct List {
  struct Node* pHead;
  struct Node* pTail;
} List;
使用这种方法,一个循环就足够了:

void free_list(List *list)
{
    Node *node = list->pHead;

    while (node != NULL)
    {
        Node *next = node->pNext;

        free(node);
        node = next;
    }
    free(list);
}

“如果我还需要释放所有数据”-->释放或不释放通常与列表的创建方式以及数据添加到列表的方式有关。不幸的是,没有显示这些内容。“如果我还需要释放所有数据”-->释放或不释放通常与列表的创建方式以及数据添加到列表的方式有关。很遗憾,这些没有显示。现在我有了更好的视图,谢谢。现在我有了更好的视图,谢谢