c-如何将一个结构指向另一个结构

Question

我需要帮助将*user_playlist指向播放列表节点。我该怎么做？

struct playlist_ {
  int album;
  int track_num;
  struct playlist_ *next;
};
typedef struct playlist_  playlists;
struct users_ {
  int user_ID;
  playlists *user_playlist;
  struct users_ *next;
};
typedef struct users_ users;
p_users = (users *)malloc(sizeof(users *));
p_users -> user_ID = account_number;
head_users = p_users;
head_users -> next = NULL;
users_pointer = head_users;
p_playlists = (playlists *)malloc(sizeof(playlists));
curr_playlists = p_playlists;
curr_playlists -> album = NULL;
curr_playlists -> track_num = NULL;
curr_playlists -> next = NULL;
curr_users -> user_playlist = curr_playlists;
 users *head_users,*curr_users,*p_users,*users_pointer;
 playlists *head_playlists,*curr_playlists,*p_playlists,*playlist_pointer;

Answer 1

这是一个问题：

p_users = (users *)malloc(sizeof(users *));

它只分配指针的大小而不分配users结构的大小。更改为：

p_users = malloc(sizeof(users));

或：

p_users = malloc(sizeof(*p_users));

铸造malloc()的返回值是不必要的，并且具有潜在的危险性。

编辑：

取消引用统一指针或NULL指针将导致分段错误。这声明了，但不初始化curr_users（与列出的其他变量相同）：

users *head_users,*curr_users,*p_users,*users_pointer;

然后尝试访问curr_users:

curr_users->user_playlist = curr_playlists; /* Segmentation fault. */

使用前将curr_users初始化为有效的users结构或NULL，并在延迟前检查是否为非空：

users *head_users = NULL,*curr_users = NULL,
    *p_users = NULL,*users_pointer = NULL;
if (curr_users)
{
    curr_users->user_playlist = curr_playlists;
}