Compile Time decision static or member

[Anees]

Dear Fellows:

I have following scenario:

class A {
public:
void me_print () { cout << "A" << endl; }
static void print() { cout << "static A" << endl; }
};

class B {
};

class C : public A, public B {
public:
void foo() { PRINT; }
};

class D: public B {
public:
void foo() { PRINT; }
};

int _tmain(int argc, _TCHAR* argv[])
{
C c;
D d;

c.foo();
d.foo();

return 0;
}

I want to define a macro (#define) such that when instance of A's
derived class (i.e. C) call PRINT; it gets A's member method me_print
(), and when instance of NON A's derived class (i.e. D) call PRINT, it
should be resolved to A's static subroutine print().

Thank you.

Regards,

Anees Haider

[Maxim Yegorushkin]

You don't need a macro for that, you can do that in pure C++ (for
clarity, I changed class to struct and removed all public keywords):

struct A {
void me_print () { std::cout << "A\n"; }
static void print() { std::cout << "static A\n"; }
};

void do_print(A* a) { a->me_print(); }
void do_print(void*) { A::print(); }

struct B {};

struct C : A, B {
void foo() { do_print(this); }
};

struct D : B {
void foo() { do_print(this); }
};

int main()

{
C c;
D d;
c.foo();
d.foo();
}

The idea here is that there are two overloads of do_print(). One takes
A*, the other one void*. The first one is chosen when you pass A* or
its publicly derived sub-class pointer. The void* overload is chosen
for all other pointers.

--
Max

[Anees]

On Jun 1, 6:42 pm, Maxim Yegorushkin <maxim.yegorush...@gmail.com>
wrote:

> Max- Hide quoted text -
>
> - Show quoted text -

Thanks Max, it works :)