Re: Member offsets in multiple inheritance

[Sam Varshavchik <mrsam at courier-mta dot com>]

pieniek writes:

> Hello,
>
> I have always though, that given a pointer to an object, members are at  
> (this) + (member-offset). But this cannot be true for multiple inheritance,  
> no?
>
> Example:
>
> struct A {
>    int x,y;
> }
>
> struct B {
>    int a, b, c;
> }
>
> struct C : public A, public B {}
>
> int func (B* b, B* actuallyCinstance) {	
>     return b->b + actuallyCinstance->b;
> }
>
> How is the B::b member accessed, especially b->b when b is B? What is the  
> cost of that (x86_64)?

Both pointers received by func() are pointers to an instance of the B class.  
The offset is the same.

If the caller invokes func() with the second argument being a pointer to an  
instance of C, the caller adjust the pointer to point to C's superclass, and  
invokes func() with two pointers to an instance of B.

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