作为我在内核空间中编写的程序的一部分,我创建了一个链表,该链表的节点中有另一个链表。 节点可以是两种类型,一种是仅具有 int 值和 char* 值的 channel ,要么是具有 int 值和 channel 链表的设备文件。但是我在我的 freeList 函数中得到了 NULL 指针引用。
我得到的错误是:无法处理内核 NULL 指针取消引用
知道如何解决这个问题吗?
struct node {
int val;
char* msg;
struct node* headOfIdList;
struct node* next;
};
static void addChannel(struct node* head, int id, char* msg) {
printk(KERN_INFO "creating channel\n");
struct node *curr ;
curr=head;
while (curr->next != NULL) {
curr = curr->next;
}
curr->next = kmalloc(sizeof(struct node), GFP_KERNEL);
curr->next->val = id;
curr->next->msg = msg;
curr->next->headOfIdList = NULL;
curr->next->next = NULL;
printk(KERN_INFO "channel created\n");
}
static void addFile(struct node* head, int minor) {
printk(KERN_INFO "creating file\n");
struct node *curr ;
curr=head;
while (curr->next != NULL) {
curr = curr->next;
}
curr->next = kmalloc(sizeof(struct node), GFP_KERNEL);
curr->next->val = minor;
curr->next->msg = NULL;
curr->next->headOfIdList = NULL;
curr->next->next = NULL;
printk(KERN_INFO "file created\n");
}
static struct node* find(struct node* head, int val) {
printk(KERN_INFO "looking for node\n");
struct node *curr ;
curr=head;
while (curr != NULL) {
if (curr->val == val) {
return curr;
}
curr = curr->next;
}
return NULL;
}
static void freeList(struct node* head) {
printk(KERN_INFO "freeing list\n");
struct node *curr ;
curr=head;
while (curr != NULL) {
struct node *tmp = curr->next;
if (curr->headOfIdList != NULL) {
freeList(curr->headOfIdList);
}
kfree(curr);
curr = tmp;
//curr=curr->next;
}
}
最佳答案
如果在 loop-while 之外声明 tmp 变量? struct node *curr, *tmp;
例如。
关于C - 如何释放在其节点中具有链表的链表?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47910169/