On Aug 10, 12:10 pm, Stuart Redmann <DerTop...@web.de> wrote:
> nas wrote:
>
> Consider this example program which i hav taken from some site
>
> #include <iostream>
> using namespace std;
> class Base1 {
> public:
> virtual void f() { }};
>
> class Base2 {
> public:
> virtual void f() { }};
>
> class Base3 {
> public:
> virtual void f() { }};
>
> class Drive : public Base1, public Base2, public Base3 {};
>
> // any non zero value because multiply zero with any no is zero
> #define SOME_VALUE 1
> int main() {
> cout << (DWORD)static_cast<Base1*>((Drive*)SOME_VALUE)-SOME_VALUE;
> cout << (DWORD)static_cast<Base2*>((Drive*)SOME_VALUE)-SOME_VALUE;
> cout << (DWORD)static_cast<Base3*>((Drive*)SOME_VALUE)-SOME_VALUE;
> return 0;}
>
> Output of this prgram is:
> 048
>
> Stuart Redmann wrote:
> >> Seehttp://www.phpcompiler.org/doc/virtualinheritance.htmlfor a good
> >>explanation of these compiler specific implementation details (I know no C++
> >>compiler that uses an implementation that really differs at more than minor
> >>pointers from the design explained there).
> nas wrote:
> > But the initially how we got 0,4,8 ??? What might be the reson for
> > this?
>
> Let's take a closer look: If you cast the (integer) value 1 to a Drive* pointer,
> the compiler will not change the value but change the type attached to this
> value (this means that you now have a Drive* pointer pointing to the memory
> location 0x00000001). The memory layout of a Drive object (located at memory
> address 1) would like the following scheme:
>
> 1 5 9 13 (memory addresses)
> _______________________________________________________
> | Base1 (4 Bytes) | Base2 (4 Bytes) | Base3 (4 Bytes) |
> -------------------------------------------------------
> | ' = ptr to VMT ' = ptr to VMT '
> | of Base2 of Base3 (4 Bytes)
> | + members + members (0 Bytes)
> L--: ptr to vtable of Drive (this vtable can be used as VMT for both
> Base1 and Drive).
>
> Now if you upcast a pointer to Drive to Base1, the compiler doesn't need to
> change the address, as the internal Base1 object inside Drive's memory layout
> lays at the internal offset zero. The internal Base2 object has offset 4, so
> you'll get the address 5 for the Base2 object.
>
> I hope this makes it a bit clearer for you.
>
> Regards,
> Stuart

Thank you so much..your answer smashed my questions!
Thanks again