CODEWITHSUNDEEP
c++pointersstruct
livetolearn
livetolearn

Reputation: 221

Accessing address of members of a struct variable using pointer to that structure

#include<iostream>
    using namespace std;
    int main()
    {
    struct node
    {
        int data;
        struct node *next;
    };

    struct node *node1;

    struct node node2;

    node2.data = 200;
    node2.next = NULL;

    cout<<"address of node2: "<<&node2<<endl;

    cout<<"address of node2.data: "<<&node2.data<<endl;

    cout<<"address of node2.next: "<<&node2.next<<endl;

    cout<<"value of node2 data: "<<node2.data<<endl;

    cout<<"value of node2 next is: "<<node2.next<<endl;

    node1 = (struct node*)malloc(sizeof(node));

    node1->data = 100;

    node1->next = NULL;

    cout<<"value of node1 data: "<<node1->data<<endl;

    cout<<"value of node1 next: "<<node1->next<<endl;

    cout<<"address of node1 variable is: "<<&node1<<endl;

    cout<<"address of node1 data variable is: "<<&node1->data<<endl;

    cout<<"address of node1 next variable is: "<<&node1->next<<endl;

    cout<<"value stored at node1 variable is: "<<node1<<endl;

    }

I wanted to print the address of the members of a struct variable using the pointers to that structure. As it can be seen in my above code example, I have used &node1->next and &node1->data to print the addresses. It seems to be printing the correct addresses because I am able to access the values by dereferencing the address returned by &node1->next and &node1->data. *(&node1->next) and *(&node1->data) returns the values correctly.

But I do not understand how the notations "&node1-> data" and "&node1->next" are returning the address of the members of the struct variable. I accidentally discovered that &node1->data and &node1->next printed the addresses. Whereas with other notations like &node2.data and &node2.next, I was able to logically come up with the notations to print the addresses, but while using pointer to struct to print the address, I accidentally discovered them instead of being able to logically come up with the correct notation.

I want to know whether whatever I have come up with is the correct usage to print the addresses of the member variables and if yes how is it the correct representation ?

Upvotes: 1

Views: 5230

Answers (1)

eerorika
eerorika

Reputation: 238291

I want to know whether whatever I have come up with is the correct usage to print the addresses of the member variables and if yes how is it the correct representation ?

Yes, it is correct.

The indirection operator or "arrow" -> returns a reference to a member of the pointed object. So, the argument of the address-of-operator & is the member. The adderess-of returns the address of that object i.e. the address of the member. Therefore, logically the correct way to get the address of the member is to first apply arrow operator to get the object, and then the address-of-operator like so:

auto* pointer_to_object = get_the_object();
auto* address_of_member = &pointer_to_object->name_of_member;

It seems to be printing the correct addresses because I am able to access the values by dereferencing the address returned by &node1->next and &node1->data. *(&node1->next) and *(&node1->data) returns the values correctly.

This should be quite obvious. The dereference operation is the inverse of address-of operation, so applying those two in sequence will cancel each other out. *&node1->data is equivalent to node1->data, which indeed returns you the value of the member.

Upvotes: 1

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