swiftlang/swift

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

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#52,220 创建于 2019年1月29日

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 (12 评论) (0 反应) (1 负责人)Swift (69,989 star) (10,719 fork)batch import
bugcompilergood first issue

描述
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.

贡献者指南

技术栈
swift
领域
tooling
议题类型
bug
难度面向新贡献者的预计实现难度，1 表示很小改动，5 表示专家级工作。
3
预计时间有经验贡献者完成调查、实现、测试并准备 pull request 的粗略时间范围。
1-2 days
活动状态议题当前的可参与程度：新鲜、活跃、陈旧、阻塞或等待维护者输入。
blocked
清晰度议题是否清楚说明期望改动、验收标准和下一步。
clear
前置要求
Swift language basicsUnderstanding of lazy properties
新手友好度1-100 的估计分数，表示该议题对首次贡献者的友好程度。
35
研究方向
The issue is a compiler bug where `super` is incorrectly disallowed in lazy property closures. The fix likely involves modifying the Swift type checker in the `lib/Sema` directory to permit `super` in that context. Review the existing comments (12 comments) for any proposed solutions or constraints. The issue is assigned, so coordination with the assignee may be needed.