Initialization of virtual bases

[Dave Theese]

Hello all,

The code example below has proper behavior (of course), but I'm trying to
understand how the behavior is brought about. Specifically, what happens to
B's and C's initialization of A??? Is it just ignored even though it
explicitly appears in the code?

Thanks,
Dave

#include <iostream>

using namespace std;

class A
{
public:
A(int d): data(d)
{
cout << "data initialized to " << d << endl;
}

private:
int data;
};

class B: virtual public A
{
public:
B(int d): A(d)
{
}
};

class C: virtual public A
{
public:
C(int d): A(d)
{
}
};

class D: public B, public C
{
public:
D(int d): A(d), B(d + 1), C(d + 2)
{
}
};

int main(void)
{
// Displays "data initialized to 1"
D foo(1);

return 0;
}

[John Harrison]

Re: Initialization of virtual bases


"Dave Theese" <cheeser_1998@y ahoo.com> wrote in message
news:QSX1b.7781 $QT5.1646@fed1r ead02...[color=blue]
> Hello all,
>
> The code example below has proper behavior (of course), but I'm trying to
> understand how the behavior is brought about. Specifically, what happens[/color]
to[color=blue]
> B's and C's initialization of A??? Is it just ignored even though it
> explicitly appears in the code?
>
> Thanks,
> Dave
>
> #include <iostream>
>
> using namespace std;
>
> class A
> {
> public:
> A(int d): data(d)
> {
> cout << "data initialized to " << d << endl;
> }
>
> private:
> int data;
> };
>
> class B: virtual public A
> {
> public:
> B(int d): A(d)
> {
> }
> };
>
> class C: virtual public A
> {
> public:
> C(int d): A(d)
> {
> }
> };
>
> class D: public B, public C
> {
> public:
> D(int d): A(d), B(d + 1), C(d + 2)
> {
> }
> };
>
> int main(void)
> {
> // Displays "data initialized to 1"
> D foo(1);
>
> return 0;
> }
>[/color]

Yes I believe so, virtual base classes are initialised by the more derived
object (or something like that). So if you created a B or a C object then B
or C would be responsible for initialising A.

john