[SR-9795] Cannot use `super` in lazy property: 'super' cannot be used outside of class members

Previous ID	SR-9795
Radar	None
Original Reporter	@marcomasser
Type	Bug

Tested with Swift 4.2.1 (Xcode 10.1) and Swift 5 from Xcode 10.2 beta (swiftlang-1001.0.45.7 clang-1001.0.37.7).

Votes	0
Component/s	Compiler
Labels	Bug, StarterBug
Assignee	@theblixguy
Priority	Medium

md5: 75741cacf781b9db21044152b379d272

Issue Description:

I’m not sure this is a bug or if this works as expected:

class Foo {
    var name = "Default Name"
}

class Bar: Foo {
    lazy var fullName: String = {
        return super.name // error: 'super' cannot be used outside of class members
    }()
}

Is there a good reason why super isn’t permitted here? Replacing super with self works fine, as I’d expect because the instance must be fully initialized when lazy properties are accessed.