Question

我正在尝试编写一段将元素添加到列表中的代码。

cqlsh:test> insert into address_table (name, addresses) values ('Caroline', null);
cqlsh:test> update address_table set addresses = null where name = 'Dexter' ;
cqlsh:test> select * from address_table;

 name     | addresses
----------+-----------
 Caroline |      null
    Alice |      null

(2 rows)

但是，当我尝试打印列表时，没有任何输出。因此，即使我输入了1 2 3 4..etc，打印功能也不会输出任何东西

typedef struct things {
    int value;
    struct things *next;
} something;


int main()
{
    int input = 0;
    something *head = NULL;
    something *current = NULL;
    current = head; //current points to head address
    while(input != -1)
    {
        scanf("%d", &input);
        while(current != NULL) //loop until current node returns NULL
        {
            current = current->next; //go to next node
        }

        current = malloc(sizeof(something)); //allocate memory for new node assuming current is NULL
        current->value = input;
        current->next = NULL; //next points to NULL
    }
    current=head; //current points back to head
    while(current != NULL)
    {
        printf("%d -> ", current->value);
        current = current->next;
    }
    puts("NULL");

    return 0;
}

我期望输出类似while(current != NULL) { printf("%d -> ", current->value); current = current->next; } puts("NULL");。我刚刚开始学习链接列表，因此请多多指教。

Answer 1

您在任何时候都不会更新head的值。或将列表中的最后一个节点指向新创建的节点。

检查是否首先设置了head，如果没有，请填充它。否则，找到列表的最后一个节点，并将新节点添加为该节点的“下一个”节点，如下所示。

if(head == NULL)
{
    head = malloc(sizeof(something));
    head->value = input;
    head->next = NULL; //next points to NULL
}
else
{
    current = head;
    while(current->next != NULL) //loop until current node returns NULL
    {
        current = current->next; //go to next node
    }

    current->next = malloc(sizeof(something)); //allocate memory for new node assuming current is NULL
    current->next->value = input;
    current->next->next = NULL; //next points to NULL
}

Answer 2

您当前的方法不适用于单个指针。在current分配内存的地方不会将节点插入列表。

只需将current用作指针，如下所示，您的方法就可以使用。

int input = 0;
something *head = NULL;
something **current = NULL;
current = &head; //current points to head address
while(input != -1)
{
    scanf("%d", &input);
    while(*current != NULL) //loop until current node returns NULL
    {
        current = &(*current)->next; //go to next node
    }

    *current = malloc(sizeof(something)); //allocate memory for new node assuming current is NULL
    (*current)->value = input;
    (*current)->next = NULL; //next points to NULL
}
current=&head; //current points back to head
while(*current != NULL)
{
    printf("%d -> ", (*current)->value);
    current = &(*current)->next;
}
puts("NULL");

我不知道为什么我的链表指针不动

2 个答案: