class B extends A<Foo> {}
rather than the current approximation:
class B {
A<Foo> b;
equals(that@B) = b.equals(that);
// other wrappers for pass through functions to make B act like an A<Foo>
}
Yet B exists primarily so that functions can require a B rather than allow in any kind of A<Foo>
-- RohanHart - 08 Aug 2003
Yes,
class B extends A<Foo>would be possible. In such cases the compiler only have to remember that class B has a hidden type parameter Foo. For complex cases and subclasses with more type parameters as the superclass, i don't know if it will be possible maybe can Daniel tell more about that. -- ArjanB
| Topic ChangeNumberOfTypeParametersInSubclasses . { Edit | Attach | Ref-By | Printable | Diffs | r1.2 | > | r1.1 | More } |
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Revision r1.2 - 08 Aug 2003 - 16:13 GMT - ArjanB Parents: WebHome > NiceCompiler > TodoList |
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