> On Aug 17, 2016, at 6:57 PM, John McCall via swift-evolution
> <[email protected]> wrote:
>
> Being able to bypass another class's overrides and jump to a specific
> superclass implementation on an arbitrary method call is badly
> encapsulation-breaking, and I can't think of any OO language with first-class
> support for it besides C++.
Perl 5 does, although of course its object system is a little bit...different.
What these languages have in common is multiple inheritance. Calling a specific
superclass's implementation is necessary when you have more than one
superclass. Swift doesn't have that problem with superclasses, but it *does*
have it with protocol extension members.
My suggestion would be to allow you to call a particular protocol extension's
implementation with:
super(ProtocolName).method()
`super(Foo)` would always use the appropriate member on `Foo`, which must be a
protocol (not a class name), and must be conformed to by this type (not by a
superclass). Unqualified `super` would only be valid in classes and would only
permit calls to members of the superclass (including protocols it conforms to).
That would permit access to default implementations without permitting
encapsulation-breaking shenanigans, while leaving plain `super`'s meaning
unambiguous.
--
Brent Royal-Gordon
Architechies
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