Member Selection for structs and references to structs

We showed that you can use the member selection operator (.) to select a member from a struct object:

#include <iostream>

struct Employee
{
    int id {};
    int age {};
    double wage {};
};

int main()
{
    Employee joe { 1, 34, 65000.0 };

    // Use member selection operator (.) to select a member from struct object
    ++joe.age; // Joe had a birthday
    joe.wage = 68000.0; // Joe got a promotion

    return 0;
}

Since references to an object act just like the object itself, we can also use the member selection operator (.) to select a member from a reference to a struct:

#include <iostream>

struct Employee
{
    int id{};
    int age{};
    double wage{};
};

void printEmployee(const Employee& e)
{
    // Use member selection operator (.) to select member from reference to struct
    std::cout << "Id: " << e.id << '\n';
    std::cout << "Age: " << e.age << '\n';
    std::cout << "Wage: " << e.wage << '\n';
}

int main()
{
    Employee joe{ 1, 34, 65000.0 };

    ++joe.age;
    joe.wage = 68000.0;

    printEmployee(joe);

    return 0;
}

Member selection for pointers to structs

However, use of the member selection operator (.) doesn't work if you have a pointer to a struct:

#include <iostream>

struct Employee
{
    int id{};
    int age{};
    double wage{};
};

int main()
{
    Employee joe{ 1, 34, 65000.0 };

    ++joe.age;
    joe.wage = 68000.0;

    Employee* ptr{ &joe };
    std::cout << ptr.id << '\n'; // Compile error: can't use operator. with pointers

    return 0;
}

With normal variables or references, we can access object directly. However, because pointers hold addresses, we first need to dereference the pointer to get the object before we can do anything. So one way to access a member from a pointer to as struct is as follows: