swiftlang/swift

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

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#52,220 opened on Jan 29, 2019

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 (12 comments) (0 reactions) (1 assignee)Swift (10,719 forks)batch import
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Description
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.

Contributor guide

Research direction
Investigate the Swift compiler's handling of `super` expressions in lazy property initializers. Trace the semantic analysis pass that rejects `super` outside of class members and modify it to allow `super` in lazy properties, since lazy properties are guaranteed to be accessed only on an initialized instance.
Tech stack
swift
Domain
compilers
Issue type
Bug
DifficultyEstimated implementation difficulty for a new contributor, from 1 for very small changes to 5 for expert-level work.
3
Estimated timeA rough time range for an experienced contributor to investigate, implement, test, and prepare a pull request.
1-2 days
Activity statusHow available the issue appears right now: fresh, active, stale, blocked, or waiting on maintainer input.
Active
ClarityHow clearly the issue explains the expected change, acceptance criteria, and next step.
Clear
Prerequisites
Swiftcompiler internals
Newbie friendlinessA 1-100 score estimating how approachable this issue is for first-time contributors.
60