I would like to check the type of a superclass A
against the type of a subclass B
(with a method inside the superclass A
, so that B
will inherit it).
Here's what I thought did the trick (that is, the use of forward declaration):
#include <iostream>
#include <typeinfo>
using namespace std;
class B;
class A {
public:
int i_;
void Check () {
if (typeid (*this) == typeid (B))
cout << "True: Same type as B." << endl;
else
cout << "False: Not the same type as B." << endl;
}
};
class B : public A {
public:
double d_;
};
int main () {
A a;
B b;
a.Check (); // should be false
b.Check (); // should be true
return 0;
}
However this code does not compile. The error I get is:
main.cc: In member function ‘void A::Check()’:
main.cc:12: error: invalid use of incomplete type ‘struct B’
main.cc:6: error: forward declaration of ‘struct B’
How could I solve this problem?
I think that the problem you are trying to solve is much better handled by a virtual method:
class A
{
public:
virtual bool Check() { return false; };
}
class B : public A
{
public:
// override A::Check()
virtual bool Check() { return true; };
}
Methods in the base class A should not need to know whether the object is "really" an A or a B. That's a violation of basic object-oriented design principles. If the behavior needs to change when the object is a B, then that behavior should be defined in B and handled by virtual method calls.