c++ - 无法在成员函数本身中声明派生数据类型指针变量

Question

我正在尝试创建一个函数来打印两个链表的公共(public)元素。我将指针作为参数(head1 和 head2)传递给两个列表的头部。

我还尝试在成员函数本身中声明两个指针等于 head1 和 head2(分别为 curr1 和 curr2)，这样我就可以在不更改两个列表的头指针的情况下执行所需的操作。但我无法做到这一点。

类(class):

class SimilarEle {

    private: struct Node {
                int data;
                Node* next;
              };

    public: Node* head = NULL;
            void AddNode (int addData);
            bool Common_ele (Node*head1, Node*head2);

           /*
           // this declaration is valid and no error.
            Node* curr1 = NULL;
            Node* curr2 = NULL;
           */
 };

和函数

bool SimilarEle :: Common_ele(Node*head1, Node*head2) {

     bool flag2 = false;
     Node* curr1 = head1, curr2 = head2;    //error occured by this declrn'
        while (curr1  != NULL) {
            while (curr2  != NULL){
                if (curr1 -> data == curr2 -> data) {
                    cout << curr1 -> data << " ";
                    flag2 = true;
                }
                curr2 = curr2 -> next;
            }
            curr1 = curr1 -> next;
            curr2 = head2;
            }
        return flag2;
    }

此外，如果我在类本身中声明相同的指针(curr1 和 curr2)，我就能够编译该程序。以下是发生的错误。

    conversion from 'SimilarEle::Node' to non-scalar type 'SimilarEle::Node' 
    requested Node curr1 = head1, curr2 = head2; 

感谢任何帮助。

Answer 1

当你将变量声明为

Node* curr1 = head1, curr2 = head2;    //error occured by this declrn'

只有 curr1 被声明为指针类型。 curr2 实际上被声明为 Node 结果是

Node* curr1 = head1; 
Node curr2 = head2; 

我建议单独拼写每个变量以提高可读性。如果你真的想在一行中完成，你需要做

Node *curr1 = head1, *curr2 = head2;