Re: Convert Derived** to Base**

danil52@mail.ru wrote:


Consider this example (which will fail to compile, but is illustrative).
Ignore the memory leaks, they're not relevant to the point being made.

#include <iostream>

struct B {};

struct D1 : public B
{
    void f() const
    {
        std::cout << "D1::f()" << std::endl;
    }
};

struct D2 : public B {};

void f(B *arr[])
{
    arr[0] = new D2;
}

int main()
{
    D1 *arr[2];
    arr[0] = new D1;
    arr[1] = new D1;
    f(arr);
    arr[0]->f(); // BOOM
    return 0;
}

If the conversion from D1** to B** were allowed, f() could change arr[0]
to point to a D2 (since a D2 is a B, a D2* can be converted to a B*).
Back in main(), arr[0] is expected to point to a D1. The call
arr[0]->f() would thus be valid. The only problem is...arr[0] no longer
points to a D1! Result: danil52 0, Computer 1.

So it's really just as well that this conversion isn't allowed. The same
thing goes for containers in general. A container of D1s is *not* a
container of Bs, because you can add a D2 to a container of Bs, and you
can't add a D2 to a container of D1s.

Regards,
Stu