c++ - 从方法获取链表中的位置

Question

在这个函数中(我正在处理的一个更大的通讯簿程序的一部分)它显示了一个链接列表中的项目，或者在我的例子中是一个联系人。我想得到这个项目在链接列表中的位置，以便删除联系人并将其添加回链接列表。

void display_contact(std::string first, std::string last)
{
        bool found;

        node *curr = head;
        found = false;

        while (curr != NULL & !found)
        {
            if (curr->firstName == first)
            {
                found = true;
            }
            else if (curr->lastName == last)
            {
                found = true;
            }
            else
            {
                curr = curr->next;
            }

        }
        if (found)
        {
            std::cout << "First Name: " << curr->firstName << "\n";
            std::cout << "Last Name: " << curr->lastName << "\n";
            std::cout << "Phone Number: " << curr->phoneNumber << "\n";
            std::cout << "Editing contacts is coming soon.\n";
        }
        else
        {
            std::cout << "\n" << first << " " << last << " was not found.\n";
        }
}

我现在得到的实际结果只是联系信息(链表中的变量)

Answer 1

node *curr = head;
found = false;
int idx = -1;
while (curr != NULL & !found)
{
    idx++;
    ....
 }

如果找到idx有索引