FAQ: Why does dynamic_cast only work if a class has at least 1 virtual method?

Question

This does not compile in C++:

class A
{
};

class B : public A
{
};

...

A *a = new B();
B *b = dynamic_cast<B*>(a);

Answer 1

Because dynamic_cast can only downcast polymorphic types, so sayeth the Standard.

You can make your class polymoprphic by adding a virtual destructor to the base class. In fact, you probably should anyway (See Footnote). Else if you try to delete a B object through an A pointer, you'll evoke Undefined Behavior.

class A
{
public:
  virtual ~A() {};
};

et voila!

Footnote

There are exceptions to the "rule" about needing a virtual destructor in polymorphic types.
One such exception is when using boost::shared_ptr as pointed out by Steve Jessop in the comments below. For more information about when you need a virtual destructor, read this Herb Sutter article.

Answer 2

As the other stated: The standard says so.

So why does the standard says so?

Because if the type isn't polymorphic it may (or is? Question to the standard gurus) be a plain type. And for plain types there are many assumptions coming from the C backwards compatibility. One of those is that the type only consists of it's members as the developer declared + necessary alignment bytes. So there cannot be any extra (hidden) fields. So there is no way to store in the memory space conserved by A the information that it really is a B.

This is only possible when it is polymorphic as then it is allowed to add such hidden stuff. (In most implementations this is done via the vtable).

Answer 3

From 5.2.7 (Dynamic cast) :

The result of the expression dynamic_cast<T>(v) is the result of converting the expression v to type T.

[ ... multiple lines which refer to other cases ... ]

Otherwise v shall be a pointer to or an lvalue of a polymorphic type (10.3).

From 10.3 (Virtual functions) :

A class that declares or inherits a virtual function is called a polymorphic class.