[SR-2119] Protocol methods of superclass should be treated as superclass API in subclasses

[an0]

Previous ID	SR-2119
Radar	None
Original Reporter	@an0
Type	Bug
Status	Resolved
Resolution	Duplicate

Environment

Swift 3 in Xcode 8.0 beta 3

Additional Detail from JIRA

Votes	0
Component/s	Compiler
Labels	Bug
Assignee	None
Priority	Medium

md5: 6ccb97ed31171e7abf704a2cb03475e4

duplicates:

• SR-103 Protocol Extension: function's implementation cannot be overridden by a subclass

Issue Description:

Compiler reports this error for `Bar.f` overriding `Foo.f`(offered by default implementation in `P` extension):

error: method does not override any method from its superclass

protocol P {
    func f()
}

extension P {
    func f() {}
}

class Foo: P {
//    func f() {}
}

class Bar: Foo {
    // Why no `override`? From what `Bar` sees its superclass `Foo` implements `P`'s interface, whether with explicit implementation of `f` in itself or borrowed implementation of `f` from `P`'s extension. Uncomment Foo.f() to see the contradictory behavior.
    override func f() {}
}