zsh-users/zsh-syntax-highlighting

Highlight semicolon that doesn't terminate a sublist

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#569 创建于 2018年10月24日

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Improvementgood first issueresolution:if-and-when-neededsyntax:command-word

描述

false && ; foo does not execute foo because ; is equivalent to \n. Highlight it accordingly, differently to a ; that terminates a sublist.

See also workers/43726.

贡献者指南

技术栈
shell
领域
developer experience
议题类型
feature
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。
1-3 hours
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。
stale
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
Understanding of Zsh syntaxFamiliarity with zsh syntax highlighting codebase
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。
65
研究方向
Investigate how semicolons are currently highlighted in the zsh syntax highlighting codebase. Identify the relevant code that handles command separators and sublist termination. Compare the behavior of `;` as newline vs sublist terminator and implement a new highlighting rule to differentiate them. See the referenced zsh workers email for details.