Why aren’t copy constructors “chained” like default constructors and destructors?

The copy constructor doesn’t (can’t) really make a copy of the object because Derived::Derived(const Derived&) can’t access pdata to change it.

Sure it can:

Derived(const Derived& d)
    : Base(d)
    cout << "Derived::Derived(const B&)" << endl;

If you don’t specify a base class constructor in the initializer list, its default constructor is called. If you want a constructor other than the default constructor to be called, you must specify which constructor (and with which arguments) you want to call.

As for why this is the case: why should a copy constructor be any different from any other constructor? As an example of a practical problem:

struct Base
    Base() { }
    Base(Base volatile&) { } // (1)
    Base(Base const&)    { } // (2)

struct Derived : Base
    Derived(Derived&) { }

Which of the Base copy constructors would you expect the Derived copy constructor to call?

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