Re: Delegation through pure virtual
On May 1, 7:25 pm, aksinghdce <aksingh...@gmail.com> wrote:
Please consider this example from C++ FAQ
{http://www.parashift.com/c++-faq-lite}
class Base {
public:
virtual void foo() = 0;
virtual void bar() = 0;
};
class Der1 : public virtual Base {
public:
virtual void foo();
};
void Der1::foo()
{ bar(); }
class Der2 : public virtual Base {
public:
virtual void bar();
};
class Join : public Der1, public Der2 {
public:
...
};
int main()
{
Join* p1 = new Join();
Der1* p2 = p1;
Base* p3 = p1;
p1->foo();
p2->foo();
p3->foo();
}
Could you please explain how the compiler ( in general ) would
create virtual tables for Der1, Der2 and Join classed.
The same way it creates any virtual table. It determs,
according to the rules of the language, which function should be
called, and puts the address of that function in the vtable.
(Often, it also puts additional information, such as any
corrections needed for the this pointer, in the vtable.)
I really need to understand how this->bar() in function
Der1::foo() gets translated to a call to Der2::bar() call.
The same way any virtual function call is translated.
As per the explanation in 20.4, the Der1's vtable should have a
pointer &Base::bar() because bar() is not implemented in Der1; by what
mechanism this pointer points to &Der2:bar()?
The vptr always points to a vtable corresponding to the most
derived class (once constructors have finished, and before
destructors have started). The vptr in Der1 will point to a
vtable_Der1InJoin. (The compiler may optimize this, if, for
example, vtable_Der1InJoin is identical to the start of
vtable_Join.)
--
James Kanze