如何释放链表中的信息？

Question

我有一个包含 *head 和 **ptail 的双向链表。我编写了添加到列表和从列表中删除的代码，但我的问题是释放信息。

这是我的节点和链表的声明：

struct tcb_t { //Node
    int         thread_id;
    int         thread_priority;
    ucontext_t *thread_context;
    struct tcb_t *next;
}; typedef struct tcb_t tcb_t;

struct queue_t { //Linked List
    tcb_t *head, **ptail; 
}; typedef struct queue_t queue_t;

这是我初始化双向链表的代码：

struct queue_t* queue_create() { //Good
  struct queue_t *q = (queue_t *) calloc(1,sizeof(queue_t));
  q->head = NULL;
  q->ptail = &q->head; // problem
  return q;
}

我的问题源于以下函数。此函数旨在释放列表中的所有节点，但 while 循环是无限的。我认为这是由于链表创建时尾部指向头部，但我不确定是否有办法在不重写 queue_create().

的情况下修复它

void t_shutdown() { //Fix
  if(ready != NULL){
    tcb_t *helper = ready->head;
    while(helper->next != NULL){ 
      tcb_t *temp = helper;
      helper = helper->next;
      if(temp->thread_id > 0){
        free(temp->thread_context->uc_stack.ss_sp);
      }
      free(temp->thread_context);
      free(temp);
    }
    free(ready);
  }
  
  ready = NULL;
}

我想遍历列表并释放所有数据，但 helper->next 永远是 NULL。

如有任何帮助，我们将不胜感激。

编辑 1

这些函数展示了如何在列表中添加和删除数据：

void queue_add(struct queue_t *q, tcb_t *ptr) { //Good
    *q->ptail = ptr;
    q->ptail = &ptr->next;
}

tcb_t *queue_remove(struct queue_t *q) { //Good
    struct tcb_t *ptr = q->head;
    if (ptr) {
        q->head = ptr->next;
        if (q->ptail == &ptr->next) {
            q->head == NULL;
            q->ptail = &q->head; // problem
        }
    }
    return ptr;
}

Answer 1

快速修复：

#include <stdlib.h>

typedef int ucontext_t;

typedef struct tcb_tag {
    int         thread_id;
    int         thread_priority;
    ucontext_t *thread_context;
    struct tcb_tag *prev;
    struct tcb_tag *next;
} tcb_t;

typedef struct queue_tag {
    tcb_t *head;
    tcb_t *tail;
} queue_t;

queue_t queue_create()
{
    queue_t queue = { NULL, NULL };
    return queue;
}

void queue_add(queue_t *queue, tcb_t *node)  // aka push_back()
{
    if (queue->tail) {
        queue->tail->next = node;
        node->prev = queue->tail;
        node->next = NULL;  // just to make sure. Idk where those nodes come from.
    }

    queue->tail = node;

    if (!queue->head)
        queue->head = queue->tail;
}

tcb_t* queue_remove(queue_t *queue)  // aka pop_front()
{
    if (!queue->head)
        return NULL;

    tcb_t *old_head = queue->head;
    queue->head = old_head->next;

    if (queue->head)
        queue->head->prev = NULL;
    else queue->tail = NULL;  // No head - no tail.

    return old_head;
}

void t_shutdown(queue_t *queue)
{
    for (tcb_t *curr_node = queue->head, *next_node; curr_node; curr_node = next_node) {
        next_node = curr_node->next;

        // do what you have to with curr_node->thread_context here

        free(curr_node);
    }
}

int main(void)
{
    queue_t q = queue_create();
    queue_add(&q, calloc(1, sizeof(tcb_t)));
    queue_add(&q, calloc(1, sizeof(tcb_t)));
    queue_add(&q, calloc(1, sizeof(tcb_t)));
    queue_add(&q, calloc(1, sizeof(tcb_t)));
    queue_add(&q, calloc(1, sizeof(tcb_t)));
    queue_remove(&q);
    queue_remove(&q);
    queue_remove(&q);
    t_shutdown(&q);
}

如何释放链表中的信息？

How can I free information in a linked list?

c

free

linked-list

编辑 1