c - 链表删除root之后的节点

Question

我正在做链表练习，遇到了一个问题。链表是尾部，所以我想删除根节点上方第一个元素。无论如何，我不想删除根节点。代码是:

struct goods {
    char* section;
    char* pname;
    int value;
    int customer_num;
    int cash;
    struct goods* next;
};

void delete_goods(int customer_num, struct goods* root) {
    struct goods *current = root;
    struct goods *previous = NULL;
    while (current != NULL) {
        if (current->customer_num == customer_num) {
            if (previous == NULL) {
                current = current->next;
                free(root);
            } else {
                previous->next = current->next;
                free(current);
                current = previous->next;
            }
        } else {
            previous = current;
            current = current->next;
        }
    }
}

int main() {
    root = malloc(sizeof(struct goods));
    root->next = NULL;
    printf("give the number of starting customers\n");
    scanf("%d", &customers);
    inform_list(customers);     // adds element to list



    else if (r == 'r' && customers != 0) {
        printf("removing...\n");
        delete_goods(customers, root);
        customers -= 1;
        printf("customers:\t%d\n", customers);
        print();
    }
}

我没有发布完整的代码(它包括一些向链接列表添加元素的功能，以方便您，如果您愿意，我可以这样做。我需要修复我的删除功能，以便它满足我上面提到的要求。以下是列表的示例输出:
customers: 2
客户:2
元素值(value):32
产品名称:阿斯达
客户:1
元素值(value):43
产品名称: sdsad
客户:0
元素值(value):0
产品名称:(无)
我需要的是我的删除功能，如果需要的话删除客户 1，然后删除客户 2 等。

Answer 1

正如其他人提到的，您希望保留根节点，因此您希望从 root->next 开始(即 root 将始终为非空) .

这应该有效[请原谅无偿的风格清理]:

void
delete_goods(int customer_num, struct goods *root)
{
    struct goods *current = root->next;
    struct goods *previous = NULL;
    struct goods *next;

    for (;  current != NULL;  current = next) {
        next = current->next;

        if (current->customer_num == customer_num) {
            if (previous != NULL)
                previous->next = next;
            else
                root->next = next;

            free(current);
        }
        else
            previous = current;
    }
}

这是一个稍微更紧凑的版本:

void
delete_goods(int customer_num, struct goods *root)
{
    struct goods *current = root->next;
    struct goods *previous = root;
    struct goods *next;

    for (;  current != NULL;  current = next) {
        next = current->next;

        if (current->customer_num == customer_num) {
            previous->next = next;
            free(current);
        }
        else
            previous = current;
    }
}