在尝试测试以下函数后,我确定当我尝试运行程序时,注释掉的行给出了seg错误:
uint8_t ll_push_front(struct List *list, int value){
if (list == NULL)
return 1;
struct ListEntry *node = (struct ListEntry *) malloc (sizeof(struct ListEntry));
if (node == NULL) exit (1);
if (list->head_ == NULL || list->tail_ == NULL || list->size_ == 0) {
list->head_ = node;
list->tail_ = node;
node->prev_ = NULL;
node->next_ = NULL;
// =====>> *(node_->val_) = value;
++(list->size_);
return 0;
}
list->head_->prev_ = node;
node->next_ = list->head_;
node->prev_ = NULL;
*(node->val_) = value;
list->head_ = node;
++(list->size_);
return 0;
}
做*(node_->val_) = value
有什么问题,应该如何正确地声明它?
以下是结构:
struct ListEntry {
struct ListEntry * next_; // The next item in the linked list
struct ListEntry * prev_; // The next item in the linked list
int * val_; // The value for this entry
};
/* Lists consist of a chain of list entries linked between head and tail */
struct List {
struct ListEntry * head_; // Pointer to the front/head of the list
struct ListEntry * tail_; // Pointer to the end/tail of the list
unsigned size_; // The size of the list
};
这就是我初始化列表的方式:
void ll_init(struct List **list) {
*list = (struct List *) malloc (sizeof(struct List));
if (list == NULL) exit (1);
(*list)->head_ = 0;
(*list)->tail_ = 0;
(*list)->size_ = 0;
}
当您决定使用指向整数的指针时,您也需要malloc
。
即
struct ListEntry *node = malloc (sizeof(struct ListEntry));
然后
node->val_ = malloc(sizeof(int));
这将使
*(node->val_) = value
工作
或者使用
struct ListEntry {
struct ListEntry * next_; // The next item in the linked list
struct ListEntry * prev_; // The next item in the linked list
int val_; // The value for this entry (no pointer)
};
然后
node->val_ = value
将工作
您可以使用memcpy(node_->val_, &value)
,但您的目的是什么,为什么不将node_->val_
声明为int