zsh-users/zsh-syntax-highlighting

Highlight semicolon that doesn't terminate a sublist

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#569 opened on Oct 24, 2018

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

Description

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.

Contributor guide

Tech stack
shell
Domain
developer experience
Issue type
feature
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.
1-3 hours
Activity statusHow available the issue appears right now: fresh, active, stale, blocked, or waiting on maintainer input.
stale
ClarityHow clearly the issue explains the expected change, acceptance criteria, and next step.
clear
Prerequisites
Understanding of Zsh syntaxFamiliarity with zsh syntax highlighting codebase
Newbie friendlinessA 1-100 score estimating how approachable this issue is for first-time contributors.
65
Research direction
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.