I need help returning the next node of a doubly linked list in C-CodePudding

I created a doubly linked list and in the function that adds values to the list, it calls the function list_next which is supposed to return the next node in the list but I'm unable to figure out just how to do that. I feel like ListPos Pos is a double pointer hence why I wrote my code like that, but it's apparently wrong. How should I go about writing it and how would I avoid that mistake in the future?

struct node
{
    struct node *next;
    struct node *prev;
    char *value;
};
typedef struct list_pos
{
    struct node *node;
} ListPos;

ListPos list_next(ListPos pos)
{
  return &(*pos.node)->next;
}

CodePudding user response：

&(*pos.node)->next doesn't make any sense.

You probably want this:

ListPos list_next(ListPos pos)
{
  ListPos next;
  next.node = pos.node->next;
  return next;
}

CodePudding user response：

Here's your code with some changes, explained in comments, which hopefully will help you:

struct node
{
    struct node *next;
    struct node *prev;
    char *value;
};

// typedefs are more confusing than they are useful
struct list_pos
{
    struct node *node;
};

struct list_pos list_next(struct list_pos pos)
{
  // don't be afraid to use temp variables to make your code clearer
  struct node *current = pos.node;
  assert(current != NULL); // one way to indicate assumption
  // you could also check if current is NULL and return NULL in that case
  struct node *next = current->next;
  return (struct list_pos){next}; // compound literal to avoid temp variable
}

CodePudding user response：

What Jabberwocky suggested is the right answer. I just wanted to add that you could replace

typedef struct list_pos
{
    struct node *node;
} ListPos;

with

typedef struct node* ListPos;

because you looked like you were thinking that ListPos is a pointer rather than separate struct.