初学者请教链表中&n1->val和&n1的区别

#include <iostream>
using namespace std;

struct point {
    int val;
    point* next;
    point(int x) :val(x), next(NULL) {}
};
int main() {
    point *n1=new point(2);
    point *n2=new point(3);
    point *n3=new point(4);
    point *n4=new point(5);
    n1->next = n2; n2->next = n3; n3->next = n4; n4->next = n1;
    cout << n1->val << endl;
    cout << n2->val << endl;
    cout << n3->val << endl;
    cout << n4->val << endl;
    cout << "the address of n4 point to ; " << n4-> next << endl;
    cout << "the address of n2 is ; " << &n1->val << endl;
    cout << "&n1 is " << &n1 << endl;
    return 0;
}

请问这道题中为什么&n1和&n1->val的值不相等

&n1->val是获得n1的成员变量val的地址
&n1是获得n1指针的地址

&n1 是结构体 n1 的内存地址； &n1->val 是结构体 n1 中 val 的内存地址。