CODEWITHSUNDEEP
c++templatesinheritancecasting
abraham_hilbert
abraham_hilbert

Reputation: 2271

Wrapped Type Casting

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.

Upvotes: 0

Views: 52

Answers (1)

Aganju
Aganju

Reputation: 6395

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).

Upvotes: 2

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