所有，在GNU c中，我有一个循环的双向链表我正在尝试实现一个delete_node函数。它适用于除节点0之外的所有节点。它确实删除（free（））节点0，但第一次在删除节点0后遍历列表时，它仍然存在于第一次传递，导致条件停止迭代到失败。实施的基础是：
struct record
{
char *line;
int lineno;
int linetype;
struct record *prev;
struct record *next;
};
typedef struct record rec;
void
iterfwd (rec *list) {
rec *iter = list; // second copy to iterate list
if (iter == NULL) {
fprintf (stdout,"%s(), The list is empty\n",__func__);
} else {
do {
printf ("%2d - prev: %p cur: %p next: %p\n", iter->lineno, iter->prev, iter, iter->next);
iter = iter->next;
} while (iter != list);
}
}
void
delete_node (rec *list, int num) {
rec *iter = list; // second copy to iterate list
int cnt = 0;
int found = 0;
if (iter == NULL) {
fprintf (stdout,"%s(), The list is empty\n",__func__);
} else {
// traverse list forward (check cnt == num, else if end -> Out Of Range)
do {
if (cnt == num) {
found=1;
(iter->prev)->next = iter->next;
(iter->next)->prev = iter->prev;
free (iter);
break;
}
iter = iter-> next;
cnt++;
} while (iter != list);
if (found != 1) {
fprintf (stderr, "%s(), Error: record to delete is out of range (%d)\n", __func__, num);
}
}
}
int main (int argc, char *argv[]) {
struct record *textfile = NULL; // instance of record, pointer to list
int node = 0;
node = (argc >= 2) ? atoi (argv[1]) : 0;
textfile = fillrecord (); // fill textfile circular linked-list
iterfwd (textfile);
delete_node (textfile, node);
iterfwd (textfile);
return 0;
}
完整列表位于：http://www.3111skyline.com/dl/dev/prg/src/ll-double-cir.c.txt
列表中有50条用于测试的数据记录，我已插入printf语句来确认指针操作。删除除节点0之外的任何节点按预期工作（以下是针对iter的指针地址 - ＆gt; prev，iter，iter-＆gt; next用于删除节点10的受影响行[删除前和删除后]）：
9 - prev: 0x603490 cur: 0x603520 next: 0x6035b0
10 - prev: 0x603520 cur: 0x6035b0 next: 0x603640 <-- delete_node
11 - prev: 0x6035b0 cur: 0x603640 next: 0x6036d0
9 - prev: 0x603490 cur: 0x603520 next: 0x603640
10 - prev: 0x603520 cur: 0x6035b0 next: 0x603640 <-- (node deleted)
11 - prev: 0x603520 cur: 0x603640 next: 0x6036d0
在列表的下一个遍历中，所有都按预期工作：
7 - prev: 0x603370 cur: 0x603400 next: 0x603490
8 - prev: 0x603400 cur: 0x603490 next: 0x603520
9 - prev: 0x603490 cur: 0x603520 next: 0x603640
11 - prev: 0x603520 cur: 0x603640 next: 0x6036d0
12 - prev: 0x603640 cur: 0x6036d0 next: 0x603760
但是，如果删除节点0，则delete_node会正确处理指针：
49 - prev: 0x604b10 cur: 0x604ba0 next: 0x603010
0 - prev: 0x604ba0 cur: 0x603010 next: 0x6030a0 <-- delete_node
1 - prev: 0x603010 cur: 0x6030a0 next: 0x603130
49 - prev: 0x604b10 cur: 0x604ba0 next: 0x6030a0
0 - prev: 0x604ba0 cur: 0x603010 next: 0x6030a0 <-- (node deleted)
1 - prev: 0x604ba0 cur: 0x6030a0 next: 0x603130
但是在删除后第一次尝试遍历列表时，节点0出现在第一遍中，导致迭代器条件＆＃39; while（iter！= list）＆＃39;失败并陷入困境：
0 - prev: 0x604ba0 cur: 0x603010 next: 0x6030a0
1 - prev: 0x604ba0 cur: 0x6030a0 next: 0x603130
2 - prev: 0x6030a0 cur: 0x603130 next: 0x6031c0
3 - prev: 0x603130 cur: 0x6031c0 next: 0x603250
4 - prev: 0x6031c0 cur: 0x603250 next: 0x6032e0
<snip>
47 - prev: 0x6049f0 cur: 0x604a80 next: 0x604b10
48 - prev: 0x604a80 cur: 0x604b10 next: 0x604ba0
49 - prev: 0x604b10 cur: 0x604ba0 next: 0x6030a0
1 - prev: 0x604ba0 cur: 0x6030a0 next: 0x603130
2 - prev: 0x6030a0 cur: 0x603130 next: 0x6031c0
3 - prev: 0x603130 cur: 0x6031c0 next: 0x603250
如上所示，在迭代器遍历0-49之后，删除的节点0消失并再次开始正确遍历1-49，但此时它处于循环中，因为条件（iter！= list）总是为true（节点0消失，防止它等同于列表）。这是一个纯循环列表，没有HEAD或TAIL节点设置为null，end-＆gt; next指向列表的开头，first-＆gt; prev指向结尾。让delete_node（）函数为节点0工作的诀窍是什么，以便删除后的第一次迭代以1开始而不是旧的0然后消失？
答案 0 :(得分：1)
当请求 it 指向的节点作为删除请求时，您不会修改调用者的指针。以下是一些代码的简化版本，演示了一种方法：
#include <stdio.h>
#include <stdlib.h>
typedef struct record rec;
struct record
{
int data;
rec *prev, *next;
};
void delete_node (rec ** pp, int num)
{
if (!*pp)
return;
// find the num'th node
while (num-- && *pp)
pp = &(*pp)->next;
// setup victim
rec *victim = *pp;
// non-self-reference node means just rewire
if (victim && (victim != victim->next))
{
victim->prev->next = victim->next;
victim->next->prev = victim->prev;
*pp = victim->next;
}
else
{ // deleted node was self-referenced. last node
*pp = NULL;
}
free(victim);
}
void iterfwd(const rec* list)
{
const rec *p = list;
printf("list: %p\n", list);
if (p)
{
for (; p; p = (p->next != list ? p->next : NULL))
printf("prev: %p, self:%p, next:%p, data = %d\n", p->prev, p, p->next, p->data);
}
puts("");
}
void insert(rec **pp, int data)
{
// setup new node
rec *newp = malloc(sizeof(*newp));
newp->data = data;
if (!*pp)
{
newp->next = newp->prev = newp;
*pp = newp;
}
else
{ // insert between prev and head.
newp->next = *pp;
(*pp)->prev->next = newp;
newp->prev = (*pp)->prev;
(*pp)->prev = newp;
}
}
int main()
{
rec *list = NULL;
int i;
for (i=1; i<=5; ++i)
insert(&list, i);
iterfwd(list);
// delete fourth node (0-based)
delete_node(&list, 3);
iterfwd(list);
// delete first node (0-based)
delete_node(&list, 0);
iterfwd(list);
// delete first node (0-based)
delete_node(&list, 0);
iterfwd(list);
// delete first node (0-based)
delete_node(&list, 0);
iterfwd(list);
// delete first node (0-based)
delete_node(&list, 0);
iterfwd(list);
return 0;
}
输出（显然取决于系统）
注意在删除请求0元素时如何修改传入的指针（按地址传递）。
list: 0x100103af0
prev: 0x100103b70, self:0x100103af0, next:0x100103b10, data = 1
prev: 0x100103af0, self:0x100103b10, next:0x100103b30, data = 2
prev: 0x100103b10, self:0x100103b30, next:0x100103b50, data = 3
prev: 0x100103b30, self:0x100103b50, next:0x100103b70, data = 4
prev: 0x100103b50, self:0x100103b70, next:0x100103af0, data = 5
list: 0x100103af0
prev: 0x100103b70, self:0x100103af0, next:0x100103b10, data = 1
prev: 0x100103af0, self:0x100103b10, next:0x100103b30, data = 2
prev: 0x100103b10, self:0x100103b30, next:0x100103b70, data = 3
prev: 0x100103b30, self:0x100103b70, next:0x100103af0, data = 5
list: 0x100103b10
prev: 0x100103b70, self:0x100103b10, next:0x100103b30, data = 2
prev: 0x100103b10, self:0x100103b30, next:0x100103b70, data = 3
prev: 0x100103b30, self:0x100103b70, next:0x100103b10, data = 5
list: 0x100103b30
prev: 0x100103b70, self:0x100103b30, next:0x100103b70, data = 3
prev: 0x100103b30, self:0x100103b70, next:0x100103b30, data = 5
list: 0x100103b70
prev: 0x100103b70, self:0x100103b70, next:0x100103b70, data = 5
list: 0x0