Question

The size of a class with no data members is returned as 1 byte, even though there is an implicit ‘this’ pointer declared. Shouldn’t the size returned be 4 bytes(on a 32 bit machine)? I came across articles which indicated that ‘this’ pointer is not counted for calculating the size of the object. But I am unable to understand the reason for this.
Also, if any member function is declared virtual, the size of the class is now returned as 4 bytes. This means that the vptr is counted for calculating the size of the object. Why is the vptr considered and ‘this’ pointer ignored for calculating the size of object?

Answer 1

The this pointer is not a member of the class. It’s just a construct that is used in methods belonging to the class to refer to the current instance.

If you have a class like this:

class IntPair
{
public:
  IntPair(int a, int b) : _a(a), _b(b) { }

  int sum() const { return _a + _b; }

public:
  int _a;
  int _b;
};

This class only needs space for two instances of int for each instance. Once you’ve created an instance and are running the sum() method, that method is called with a pointer to the instance, but that pointer always comes from somewhere else, it isn’t stored in the object instance.

For example:

IntPair *fib12 = new IntPair(89, 144);

cout << fib12->sum();

Notice how the variable that becomes the this pointer is stored outside the object, in the scope that created it.

You could, in fact, always transform a method like the one above into:

static int sum2(const IntPair* instance)
{
  return instance->_a + instance->_b;
}

If the above is defined inside the class (so it can access the private members), there’s no difference. In fact, this is how methods are implemented behind the scene; the this pointer is just a hidden argument to all member methods.

The call would become:

IntPair* fib12 = new IntPair(89, 144);

cout << IntPair::sum2(fib12);