C 中的基本链表

Question

我正在用 C 编写一个基本的链表程序，在删除时遇到了一些麻烦。这是我拥有的：

#include <stdio.h>

struct node * delete(struct node * head, struct node * toDelete);
void print(struct node * head);

struct node {
    int value;
    struct node *next;
};

int main(int argc, const char * argv[]) {

    struct node node1, node2, node3;
    struct node *head = &node1;

    node1.value = 1;
    node1.next = &node2;

    node2.value = 2;
    node2.next = &node3;

    node3.value = 3;
    node3.next = (struct node *) 0;

    print(head);

    delete(head, &node3);

    print(head);

    return 0;
}

struct node * delete(struct node * head, struct node * toDelete) {
    //if to delete is head
    if (head == toDelete) {
        head = head->next;

    } else {
        //find node preceding node to delete
        struct node *current = head;
        while (current->next != toDelete) {
            current = current->next;
        }
        current = current->next->next;
    }
    return head;
}

void print(struct node * head) {
    struct node *current = head;

    while (current != (struct node *) 0) {
        printf("%i\n", current->value);
        current = current->next;
    }
}

问题 #1： 所以我试着写：

delete(head, node3);

但是xCode要我在"node3"前面加上“&”。是不是一般我定义函数取指针的时候，需要传入内存地址？

问题 #2：

我的打印函数可以打印出 3 个节点的值。在调用 delete 并尝试删除 node3 后，它仍然打印出 3 个节点。我不确定我哪里出错了。我找到我要删除的节点之前的节点，并将其下一个指针设置为节点之后的节点（非正式地：node.next = node.next.next）。

有什么想法吗？

感谢您的帮助，克莱曼

Answer 1

只需尝试将 current = current->next->next; 更改为 current->next=current->next->next。如果它不起作用，请告诉我。

Answer 2

你应该通过它 &node3。要删除，请更改您的代码 current = current->next->next; 至 current->next = current->next->next;

Answer 3

but xCode wanted me to add "&" in front of "node3". Is it generally true that
when I define a function to take a pointer, I need to pass in the memory 
address?

是的，如果你声明函数接受一个指针，你必须给它传递一个指针。

此外，当从链表中删除一个值时，您将要更改

current->next = current->next->next

Answer 4

Is it generally true that when I define a function to take a pointer, I need to pass in the memory address?

是的，xCode是对的。 node3 是一个 struct node，但是你的函数 delete 将 struct node * 作为第二个参数，所以你必须将指针传递给 node3，而不是变量本身.

After calling delete and trying to delete node3, it still prints out the 3 nodes.

这是因为您没有更改 next 的值。另外，为了内存安全，不要忘记检查指针是否为 NULL:

while ((current->next != toDelete) && (current->next != NULL)) {
    current = current->next;
}
if (current->next != NULL)
    current->next = current->next->next;

C 中的基本链表

Basic Linked List in C

c

linked-list