Re: Compiler Warning C4373 about virtual methods

Thanks Igor,

I have verified you are correct. Here is my verification code and related
compile error message.

The point you mentioned in your below reply makes me confused about my
previous understanding about virtual function. In my previous understanding,
it does not matter the type of pointer, what matters is the type of class.

But in your code, you showed the type of pointer matters... Confused. It is
appreciated if you could share your point of how to decide which function is
called in the world of virtual. :-)

[Code]
#include <iostream>

using namespace std;

class Base {
public:
    virtual void foo() {cout << "Base " << endl; }
};

class Derived : public Base {
public:
    virtual void foo (int i) {cout << "Derived " << i << endl; }
};

/*
output:
Base
Derived 100
*/
int main()
{
    Derived d;
    Base* pb = &d;
    Derived* pd = &d;
    pb -> foo();
    // pb->foo (100); // error C2660: 'Base::foo' : function does not take 1
arguments
    pd -> foo(100);
    // pd->foo(); // error C2660: 'Derived::foo' : function does not take 0
arguments
    return 0;
}
[/Code]

regards,
George

"Igor Tandetnik" wrote: