【Redis】对通用双向链表实现的理解(二)
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return NULL;
node->value = value;
if (list->len == 0) {
list->head = list->tail = node;
node->prev = node->next = NULL;
} else {
node->prev = NULL;
node->next = list->head;
list->head->prev = node;
list->head = node;
}
list->len++;
return list;
}
/* 把包含指针指向的值的节点插入链表尾部，
* 如发生错误,将返回NULL并且不对链表进行任何操作,
*
* 如成功则返回该链表指针.*/
list *listAddNodeTail(list *list, void *value)
{
listNode *node;
if ((node = zmalloc(sizeof(*node))) == NULL)
return NULL;
node->value = value;
if (list->len == 0) {
list->head = list->tail = node;
node->prev = node->next = NULL;
} else {
node->prev = list->tail;
node->next = NULL;
list->tail->next = node;
list->tail = node;
}
list->len++;
return list;
}
/* 把新节点插入链表某个节点的前或后，
* 如发生错误,将返回NULL并且不对链表进行任何操作,
*
* 如成功则返回该链表指针.*/
list *listInsertNode(list *list, listNode *old_node, void *value, int after) {
listNode *node;
if ((node = zmalloc(sizeof(*node))) == NULL)
return NULL;
node->value = value;
if (after) { // after!=0则表示节点插入的位置在旧节点之后，否则在其之前
node->prev = old_node;
node->next = old_node->next;
if (list->tail == old_node) {
list->tail = node;
}
} else {
node->next = old_node;
node->prev = old_node->prev;
if (list->head == old_node) {
list->head = node;
}
}
if (node->prev != NULL) {
node->prev->next = node;
}
if (node->next != NULL) {
node->next->prev = node;
}
list->len++;
return list;
}
/* 从链表中删除某个节点.
* 会调用底层函数把节点的空间释放.
*
* 该方法不能失败. */
void listDelNode(list *list, listNode *node)
{
if (node->prev)
node->prev->next = node->next;
else
list->head = node->next;
if (node->
next)
node->next->prev = node->prev;
else
list->tail = node->prev;
if (list->free) list->free(node->value);
zfree(node);
list->len--;
}
/* 获取迭代器对象'iter'. 在初始化之后
*每次调用listNext()都会返回链表的下一个元素.
*
* 该方法不能失败. */
listIter *listGetIterator(list *list, int direction)
{
listIter *iter;
if ((iter = zmalloc(sizeof(*iter))) == NULL) return NULL;
if (direction == AL_START_HEAD)
iter->next = list->head;
else
iter->next = list->tail;
iter->direction = direction;
return iter;
}
/* 释放迭代器对象的空间 */
void listReleaseIterator(listIter *iter) {
zfree(iter);
}
/* 将迭代器指针重新指向表头 */
void listRewind(list *list, listIter *li) {
li->next = list->head;
li->direction = AL_START_HEAD;
}
/* 将迭代器指针重新指向表尾 */
void listRewindTail(list *list, listIter *li){
li->next = list->tail;
li->direction = AL_START_TAIL;
}
/* 获取迭代器的下一个元素.
* 可以通过listDelNode()方法来删除当前返回的节点,但不能删除其他的节点。
*
* 如果成功则返回迭代器的下一个元素,否则返回NULL;
* 因此推荐以下这样使用:
*
* iter = listGetIterator(list,);
* while ((node = listNext(iter)) != NULL) {
* doSomethingWith(listNodeva lue(node));
* }
*
* */
listNode *listNext(listIter *iter)
{
listNode *current = iter->next;
if (current != NULL) {
if (iter->direction == AL_START_HEAD)
iter->next = current->next;
else
iter->next = current->prev;
}
return current;
}
/* 复制整个链表.
* 成功则返回复制的链表指针，否则返回NULL.
*
* 复制的方法通过listSetDupMethod()来指定，
* 如果没有指定dup方法则会完整拷贝原始节点的值.
*
* 原始链表不会给更改. */
list *listDup(list *orig) // 有个疑问: 既然需要保持原始链表的不被修改，为什么不加const修饰
{
list *copy;
listIter *iter;
listNode *node;
if ((copy = listCreate()) == NULL)
return NULL;
copy->dup = orig->dup;
copy->free = orig->free;
copy->match = ori