CODEWITHSUNDEEP
c++polymorphism
Qiang Xu
Qiang Xu

Reputation: 4803

More than 1 address for derived class object?

In Item 27 of "Effective C++" (3rd Edition, Page 118), Scott Meyers said:

class Base { ... };
class Derived: public Base { ... };
Derived d;
Base *pb = &d;

Here we're just creating a base class pointer to a derived class object, but sometimes, the two pointers will not be the same. When that's the case, an offset is applied at runtime to the Derived* pointer to get the correct Base* pointer value.

This last example demonstrates that a single object (e.g., an object of type Derived) might have more than one address (e.g., its address when pointed to by a Base* pointer and its address when pointed to by a Derived* pointer).

Here is a bit hard to understand. I know that a pointer to the base class can point to an object of the derived class at runtime, this is called polymorphism or dynamic binding. But does the derived class object really have more than 1 address in the memory?

Guess I have some misunderstanding here. Could someone give some clarification? Maybe this has something to do with how polymorphism is implemented in the C++ compiler?

Upvotes: 14

Views: 1526

Answers (2)

James Kanze
James Kanze

Reputation: 154017

Just try it:

class B1
{
    int i;
};

class B2
{
    int i;
};

class D : public B1, public B2
{
    int i;
};

int
main()
{
    D aD;
    std::cout << &aD << std::endl;
    std::cout << static_cast<B1*>( &aD ) << std::endl;
    std::cout << static_cast<B2*>( &aD ) << std::endl;
    return 0;
}

There's no possible way for the B1 sub-object to have the same address as the B2 sub-object.

Upvotes: 17

Pete Becker
Pete Becker

Reputation: 76458

An object has exactly one address; that's where it's located in memory. When you create a pointer to a base subobject you get the address of that subobject, and that doesn't have to be the same as the address of the object that contains it. A simpler example:

struct S {
    int i;
    int j;
};

S s;

The address of s will be different from the address of s.j.

Similarly, the address of a base subobject does not have to be the same as the address of the derived object. With single inheritance is usually is, but when multiple inheritance comes into play, and ignoring empty base classes, at most one of the base subobjects can have the same address as the derived object. So when you convert a pointer to the derived object into a pointer to one of its bases you don't necessarily get the same value as the address of the derived object.

Upvotes: 5

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