带有继承的 C++ 动态转换

Question

#include <iostream>

using namespace std;

class A
{
public:
    void foo() { cout << "foo in A" << endl; }
};

class B : public A
{
public:
    void foo() { cout << "foo in B" << endl; }
};


int main() {
    A* a = new B;
    a->foo(); // will print "foo in A" because foo is not virtual
    B* b = new B;
    b->foo(); // will print "foo in B" because static type of b is B

    // the problem
    A* ab;
    ab = dynamic_cast<B*>(new B);
    ab->foo(); // will print "foo in A" !!!!!
}

Does the 'dynamic_cast' not change the static type of ab? I mean, logically, it seams equivalent to B* ab = new B; because of the casting.. But it does not.
I thought that dynamic cast changes the static type of the object, am I wrong? And if so, what is the difference between:

A* ab = dynamic_cast<B*>(new B);

and

A* ab = new B;  

Thanks

Answer 1

You are dynamic_cast ing to B, but at time of assignment to ab, you are implicitely casting back to A, so the dynamic_cast gets lost again.

The actual type of the object ab is pointing to still remains B, but the pointer the object is accessed with is of type A, so A::foo is selected. It would have been different if foo was virtual, though.

Answer 2

If you call the foo() function from A pointer, foo() of class A will be called. I believe you're looking for a virtual behavior. If that is the case, declare foo() of class A as:

virtual void foo() { cout << "foo in A" << endl; }