Wrapped Type Casting

Question

Can someone please help me to understand why this code compiles:

class A {};

class B : public A {};

void foo( A ) {}


int main() {
  B b;

  foo( b );
}

but this not:

class A {};

class B : public A {};

template< typename T >
class wrapper {};

void foo( wrapper<A> ) {}

int main() {
  wrapper<B> b;

  foo( b );
}

The only difference in the second code (i.e., the code that does not compile) is that the classes A and B are wrapped as template parameters in the class wrapper; the wrapping, surprisingly, seems to hinder the conversion from B to A.

It would also be great if someone can help me to fix the second code.

`wrapper` and `wrapper` have no direct inheritance relation. — user0042, Oct 03 '17 at 23:18
For what it's worth, for less trivial examples your first version will exhibit [object slicing](https://stackoverflow.com/questions/274626/what-is-object-slicing) and [will not act polymorphically](https://stackoverflow.com/questions/15188894/why-doesnt-polymorphism-work-without-pointers-references) — Cory Kramer, Oct 03 '17 at 23:18

score 2 · Accepted Answer · answered Oct 03 '17 at 23:22

class B is derived from class A, so each B is an A and can be used in its place. However, wrapper<A> has no relationship with wrapper<B> - why would it?

Consider this example: if B is the son of A, would B’s wife be the son of A’s wife (obviously not - they have no relation).

To accomplish such a relation, you would need to derive wrapper<B> from wrapper<A>, which is not possible with wrapper<> being a template class. Consider other approaches, where you send B into the methods of the wrapper<A> class or similar (works, as it is a A).

I suppose we should all just be glad that A is not a son of a B. — user4581301, Oct 03 '17 at 23:38

Wrapped Type Casting

1 Answers1