Question

I have’t coded in c++ for some time and I got stuck when I tried to compile this simple snippet:

class A
{
  public:
    void f() {}
};

int main()
{
  {
    A a;
    a.f(); // works fine
  }

  {
    A *a = new A();
    a.f(); // this doesn't
  }
}

Answer 1

It’s a pointer, so instead try:

a->f();

Basically the operator . (used to access an object’s fields and methods) is used on objects and references, so:

A a;
a.f();
A& ref = a;
ref.f();

If you have a pointer type, you have to dereference it first to obtain a reference:

A* ptr = new A();
(*ptr).f();
ptr->f();

The a->b notation is usually just a shorthand for (*a).b.

A note on smart pointers

The operator-> can be overloaded, which is notably used by smart pointers. When you’re using smart pointers, then you also use -> to refer to the pointed object:

auto ptr = make_unique<A>();
ptr->f();