Re: why dynamic_cast fail here?

From:

Kai-Uwe Bux <jkherciueh@gmx.net>

Newsgroups:

comp.lang.c++

Date:

Thu, 15 Mar 2007 08:37:17 -0400

Message-ID:

<etbem1$q18$1@murdoch.acc.Virginia.EDU>

pietromas@gmail.com wrote:

In the example below, why does the dynamic_cast fail (return NULL)? It
should be able to cast between sibling classes ...

No, it should not and is not. See [5.2.7] for the conversions supported by
dynamic_cast<>. Specifically [5.2.7/8]

  The run-time check logically executes as follows:
  ? If, in the most derived object pointed (referred) to by v, v points
    (refers) to a public base class subobject of a T object, and if only one
    object of type T is derived from the sub-object pointed (referred) to by
    v, the result is a pointer (an lvalue referring) to that T object.
  ? Otherwise, if v points (refers) to a public base class sub-object of the
    most derived object, and the type of the most derived object has a base
    class, of type T, that is unambiguous and public, the result is a
    pointer (an lvalue referring) to the T sub-object of the most derived
    object.
  ? Otherwise, the run-time check fails.

#include <iostream>

class A
{
    public:
        virtual const int get() const = 0;
};

class B : public A
{
    public:
        virtual const int get() const { return 0; }
};

class C : public A
{
    public:
        virtual const int get() const { return 1; }
};

int main()
{
    A *a;

    a = new B();
    std::cout << a->get() << std::endl;

    a = dynamic_cast<C*>(a);

The dynamic type of *a is B. The dynamic_cast<> is specifically designed to
catch the error in trying to regard this object as anything that it not
truly is. Therefore, dynamic_cast<> will fail if you cast the object to any
type that is not a base class of its dynamic type.

    if(a)
        std::cout << a->get() << std::endl;

    return 0;
}

Best

Kai-Uwe Bux