如何使函数在c ++中返回结构指针

Question

我想让函数返回特定节点的地址.但是编译器没有检测到我创建的节点数据类型结构.

struct node
{
int data;

node *link;
};

node *header,*current;
node traverse(int pos);


node *Linkedlist::traverse(int pos)
{
    int location = 0;  
    current->link = header->link;
    node *address = new node;
    address->data = NULL;
    address->link = NULL;


    while(current->link != NULL)
    {

        if(location == pos)
        {
            cout <<current->link->data <<" "<< endl; 
            address->link=current->link;
        }
        location ++;
        current->link = current->link->link;

    }


    return  address->link;
}

Answer 1

更改

return  *address;

至

return  address;