What is the actual meaning of keyword "this"?

Question

I have two questions related to C++:

In many textbooks, the keyword this is a pointer to the calling object. Correct?

As i like to play with coding, i wrote the following simple code:

struct Base
{
    void g();
    virtual void f();
};

void Base::f() {
    cout << "Base::f()" << endl;
}

void Base::g() {
    cout << "Base::g()" << endl;
    cout << "sizeof(*this) : " << sizeof(*this) << endl;
    this->f();
}

struct Derived : public Base
{
    int d;
    void f();
};

void Derived::f()
{
    cout << "Derived::f()" << endl;
}

int main()
{
    Base a;
    Derived b;

    cout << "sizeof(a) : " << sizeof(a) << endl;
    cout << "sizeof(b) : " << sizeof(b) << endl;

    a.g();
    b.g();
}

The above code produces the following output:

sizeof(a) : 4
sizeof(b) : 8
Base::g()
sizeof(*this) : 4
Base::f()
Base::g()
sizeof(*this) : 4   // why 4 bytes not 8 bytes?????????
Derived::f()

If this is pointing to the calling object, should the second line of sizeof(*this) print 8 instead of 4 since the calling object is b? What actually is happening here? Is this has been demoted?!!!!

If this has been demoted to type Base, how this->f() invokes the correct function? I am really confused.

Answer 1

void Base::g() {
    cout << "Base::g()" << endl;
    cout << "sizeof(*this) : " << sizeof(*this) << endl;
    this->f();
}

The important distinction that needs to be made is that sizeof is a compile-time operator, not a runtime operator. The compiler interprets the expression sizeof(*this) as "the size of the object pointed to by this", which, in the scope of Base::g would be an object of type Base. The compiler will essentially rewrite that statement as this, because it knows that the size of Base is four bytes:

cout << "sizeof(*this) : " << 4 << endl;