Re: accessing subclass members via a base pointer?

From:

Tim H <thockin@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Wed, 31 Oct 2007 05:11:31 -0000

Message-ID:

<1193807491.796542.234780@v23g2000prn.googlegroups.com>

On Oct 30, 4:59 pm, Me <no...@all.com> wrote:

I need to be able to acces non-virtual members of sublcasses via a
base class pointer...and without the need for an explicit type cast.
I thought a pure virtual getPtr() that acts as a type cast would solve
the problem, but it appears not to.

I need this functionality to make object serialization a reality.

I think you're stuck. Why not make a virtual SerializeIt() method
instead?

See below for details

#include <iostream>
#include <iomanip>

struct base {
    virtual base* getPtr()=0;};

struct subclass: public base {
    subclass* getPtr() { return this; }
    const char* isA() { return "subclass"; }

};

getPtr() is not virtual so it is statically bound based on the type,
which is known at compile time.

using namespace std;
int main(int argc, char** argv) {
base* p=new subclass();

p has type "pointer to base"

cout << p->getPtr()->isA() << endl;

Since p is a "base" and getPtr() is not virtual, I would expect the
compiler to generate a call to base::getPtr(). Since base doesn't
have an isA() method in this code, I'd expect it to fail. Does it
actually compile, or does your real code have a base::isA() method?

return 0;

};

I would EXPECT p->getPtr() to do a correct type cast of the pointer so
that ->isA() is called with a pointer of the correct type, but that's not
happening...WTF!