[SR-6207] Misleading error for unqualified use of static variable in method · swiftlang/swift#48759

(8 留言) (0 反應) (0 負責人)Swift (69,989 star) (10,719 fork)batch import

bugcompilerdiagnostics qualityexpressionsgood first issuestatic declarationsswift 6.0type checker

描述


Previous ID	SR-6207
Radar	rdar://problem/24291352
Original Reporter	@huonw
Type	Bug
Status	In Progress
Resolution

Swift 4


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

md5: 47c00d1618aecdca414e977e2fbf19c9

Issue Description:

struct HasStatic {
    func foo() {
        print(cvar)
    }
    static let cvar = 123
}

hasstatic.swift:3:11: error: static member 'cvar' cannot be used on instance of type 'HasStatic'
    print(cvar)
          ^~~~
          HasStatic.

The fixit helps, but it still isn't obvious what's going wrong. The compiler is acting like the use of cvar was written self.cvar, and hints at this with "an instance of", but it's easy to gloss over the subtlety. The text could be tweaked, maybe like

static member 'cvar' can only be used on the type 'HasStatic', not the instance 'X'

where the X is the relevant parent value. (e.g. self for this case, or foo.bar for foo.bar.cvar).

貢獻者指南

技術棧: swift
領域: tooling
議題類型: bug
難度: 2
預計時間: 1-3 hours
活動狀態: stale
清晰度: clear
前置要求: 無
新手友善度: 80
研究方向: The issue is about improving a diagnostic message in the Swift compiler. The error occurs when a static member is used unqualified in an instance context. The diagnostic is emitted by the compiler's semantic analysis (Sema). Look at the file lib/Sema/CSSimplify.cpp or similar files where static member access is resolved. The fix involves changing the error message string to be more explicit, e.g., 'static member 'cvar' can only be used on the type 'HasStatic', not the instance 'X'. Since the issue is labeled StarterBug and is in the swift repository, a good starting point is to search for the exact error message in the codebase and modify it accordingly.