我有一个带有双链列表的文件,其中包含一组进程标识符和一些状态信息。
struct pr7_process
{
pid_t pid; /* process ID, supplied from fork() */
/* if 0, this entry is currently not in use */
int state; /* process state, your own definition */
int exit_status; /* supplied from wait() if process has finished */
struct pr7_process *next; // a pointer to the next process
struct pr7_process *prev;
};
/* the process list */
struct process_list
{
struct pr7_process *head;
struct pr7_process *tail;
};
我有一种方法可以删除列表中的元素:
{
struct pr7_process *cur;
for(cur = list->head; cur != NULL; cur = cur->next)
{
if (cur->pid == pid)
{
printf("cur pid: %dn", cur->pid);
cur->state = STATE_NONE;
if(list->head == list->tail)
{
free(cur);
}
else
{
cur->prev->next = cur->next;
cur->next->prev = cur->prev;
free(cur);
}
break;
}
}
}
我的删除功能有什么问题?当我尝试打印我的列表时,我似乎得到了一个无限循环。以前我认为这是我使用 free() 的方式,但显然不是从回复中:)
谢谢!
当您将节点集next
添加到NULL
时。
然后当你释放所有时,自由直到下一个 == NULL。
删除节点时。更新链接和空闲节点。
另外;在NULL 上免费是一个 noop。
瓦尔格林德在做这些事情时是一个宝贵的工具。
相信你必须做更多的检查;即:
struct pr7_process {
int pid;
...
} const new_proc = {
0, 44, 0, NULL, NULL
};
void del(struct process_list *list, int pid)
{
struct pr7_process *cur;
for (cur = list->head; cur != NULL; cur = cur->next) {
if (cur->pid == pid) {
printf("cur pid: %dn", cur->pid);
if(list->head == list->tail) {
free(cur);
list->head = NULL;
list->tail = NULL;
} else if (cur == list->head) {
list->head = list->head->next;
free(cur);
list->head->prev = NULL;
} else if (cur == list->tail) {
list->tail = cur->prev;
free(cur);
list->tail->next = NULL;
} else {
cur->prev->next = cur->next;
cur->next->prev = cur->prev;
free(cur);
}
break;
}
}
}
假设您构建的列表类似于:
int push(struct process_list *list, int pid, int state)
{
if (list->head == NULL) { /* or move this to where ever you see fit */
if ((list->head = malloc(sizeof(struct pr7_process))) == NULL)
return -1;
list->tail = list->head;
*list->tail = new_proc;
} else {
if ((list->tail->next = malloc(sizeof(struct pr7_process))) == NULL)
return -1;
*list->tail->next = new_proc;
list->tail->next->prev = list->tail;
list->tail = list->tail->next;
}
list->tail->pid = pid;
list->tail->state = state;
return 0;
}
void wipe(struct process_list *list)
{
struct pr7_process *node = list->tail;
while (node != list->head) {
node = list->tail->prev;
free(list->tail);
list->tail = node;
}
free(list->head);
list->head = NULL;
list->tail = NULL;
}
void prnt(struct process_list list, int dir)
{
if (dir == 1) {
while (list.head != NULL) {
printf("%4d: %dn", list.head->pid, list.head->state);
list.head = list.head->next;
}
} else {
while (list.tail != NULL) {
printf("%4d: %dn", list.tail->pid, list.tail->state);
list.tail = list.tail->prev;
}
}
}
int main(void)
{
struct process_list list = {NULL, NULL};
push(&list, 331, 2); /* if(push() != -1) ... */
push(&list, 332, 66);
push(&list, 333, 47);
prnt(list, 1);
del(&list, 332);
prnt(list, 1);
wipe(&list);
prnt(list, 1);
return 0;
}
我知道你不能释放()不是由malloc分配的东西,我该如何克服这个问题?
有什么要克服的? 要么是动态分配的,您需要free()
它,要么是按自动存储持续时间分配的,而您没有。 这里没有问题。
通常,使用这样的灯,您将malloc
所有内容,以便您可以可靠地释放事物。 否则,您不知道它们是如何分配的,并且可能会遇到未定义的行为。