如何更改结构中的指针指向c中的值？

Question

After trying to test the following function, i have determined the line commented out gives a seg fault when i try to run the program:

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;
}

what is wrong with doing *(node_->val_) = value and how should it be properly declared?

here are the structs:

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
};

This is how i initilize the list:

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;
}

Answer 1

As you have decided to use an pointer to an integer you need to malloc that as well.

ie

struct ListEntry *node =  malloc (sizeof(struct ListEntry));

Then

node->val_  = malloc(sizeof(int));

This will make

*(node->val_) = value 

Work

Alternatively use

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)
};

Then

node->val_ = value

Will work

Answer 2

您可以使用memcpy(node_->val_, &value) ，但是目的是什么，为什么不将node_->val_声明为int