C中的双链表，忽略最后一个条目，或者想要一个不存在的条目

我一直在用Kochan的C 4版编程研究C。我试图在将头撞到墙上几个小时和寻求见解之间取得平衡。

我在有关指针的章节中，练习使您构建了许多功能，然后扩展到双向链表。一项是使列表点从前到后和从后到前。我的问题是，从前向后移动时，最后一个条目会在我要跳过该值的位置显示一个值。相同的方向。我一直在使用的方法是在节点的前面和后面都有一个节点条目，以便能够将条目添加到列表的开头和结尾。请注意，我们尚未达到内存分配之类。我也留在我的插入功能，以防万一那里有问题。提前致谢

#include <stdio.h>
#include <stdbool.h>

struct entry
  {
      struct entry *previous;
      int      value;
      struct entry *next;
  };

void insertEntry(struct entry *what, struct entry *where)
{
    what->next = where->next;
    what->previous = where->previous;
    where->next = what;
    where->previous = what;
}

void printEntries_Reverse (struct entry initial)
    {
        while (initial.previous != NULL){
            printf("%i  \n", initial.previous->value);
            initial.previous = initial.previous->previous;
        }
    }

void printEntries(struct entry initial)
{
     while (initial.next != NULL){
        printf("%i   \n", initial.next->value);
        initial.next = initial.next->next;
}
printf("\n");
}

int main (void)
{
  struct entry head, foot, n0, n1, n2, n3, n4, nx;

  struct entry *list_ptr, *list_end;

  list_ptr = &n0;
  list_end = &n4;

    n1.value = 100;
    n2.value = 200;
    n3.value = 300;
    n4.value = 400;


    n0.next = &n1;
    n1.next = &n2;
    n2.next = &n3;
    n3.next = &n4;
    n4.next = NULL;


    n0.previous = NULL;
    n1.previous = &n0;
    n2.previous = &n1;
    n3.previous = &n2;
    n4.previous = &n3;

    nx.value = 250;

    insertEntry(&nx, &n2);
    printEntries(*list_ptr);
    printEntries_Reverse(*list_end);

  return 0;
}```