Re: Address of virtual member function from object (*not* class)
James Kanze wrote:
On Sep 2, 4:07 pm, Gabriel de Dietrich <gabriel.dedietr...@gmail.com>
wrote:
On Sep 2, 3:28 pm, Victor Bazarov <v.Abaza...@comAcast.net> wrote:
"Curiosity" is not really a valid reason, I believe. So, really, why
would you want to do that? You're not using the pointer-to-member in
any way, so what would be the point of returning it?
Well, it is for me... :-)
To answer your question: Suppose you have an arbitrarily large
inheritance hierarchy with root class A, and you want to know
if some subclass reimplements f. If you know the class, no
problem. Just test
&A::f == &B::f
or whatever other class, and it's done.
If I read the standard correctly, that test shoud always return
true (supposing that f is virtual, and that B derives from A).
Regardless of whether B overrides the function or not. (I'm not
100% sure about my reading of the standard here, however, since
both of the compilers available to me say that the comparison is
illegal.)
Pointers to members respect virtuality; a pointer to a virtual
function will "point" to the correct function in whatever object
it happens to be used with.
Yes, but, again, how would he get a pointer to member from a pointer to
an object (possibly of the derived type)? Would 'typeid' help?
Probalby not. IOW, what does one need to put on line 6
class B { public: virtual void foo() {} };
class D1 : public B { void foo() {} };
class D2 : public B { }; // no overrider for 'foo'
bool bar(B* pB) {
return ??? ; // (line 6)
}
#include <iostream>
#include <ostream>
#include <cstdlib>
#include <ctime>
using namespace std;
int main() {
srand(unsigned(time(0)));
D1 d1;
D2 d2;
for (int i = 0; i < 10; ++i) {
cout << "attempt " << i;
B* pB = &d2;
if (rand() & 1)
pB = &d1;
cout << (bar(pB) ? " :-)" : "");
cout << endl;
}
}
to see intermittent smileys?
V
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