Re: bug:git-check-ignore exit status is wrong for negative patterns when -v option used

From: Junio C Hamano <gitster@pobox.com>
Date: Apr 28, 2021, 07:13 UTC
Message-ID: <xmqqfszac25x.fsf@gitster.g>
In-Reply-To: <CY4P132MB00885C970ACF5A277F06E40385419@CY4P132MB0088.NAMP132.PROD.OUTLOOK.COM>

Jeremy Faith <jeremy.faith@jci.com> writes:

Show 10 quoted lines

> git version 2.31.1.362.g311531c9de
> git-check-ignore 
> When a negative pattern is the last .gitignore match the -v option causes the exit status to be 0 rather than the expected 1.
> e.g say .gitignore contains  one line: !hello
> git check-ignore hello #outputs nothing
> echo $?  #shows correct exit status=1 i.e None of the provided paths are ignored.
> but
> git check-ignore -v hello #output is next line
> .gitignore:4:!hello	hello
> echo $?  #shows wrong exit status=0 i.e. One or more of the provided paths is ignored

Hmph. This is kind of understandable given the history of the command, which was *not* about programatically ask "is this path ignored?" question at all. Instead, it was invented to answer this question: I am puzzled by the fact that Git considers this path is to be ignored (or "not to be ignored"). Show me which entry in what exclude file made the final decision to ignore (or "not to ignore") it to help me debug my ignore file(s).

And the exit code was to signal "yes, I found a relevant entry", which made sense for the tool's nature as a debugging aid.

So, I suspect that this is working as designed/intended. I agree that it is debatable that the way it was designed to work is a good one, though.