Re: bug:git-check-ignore exit status is wrong for negative patterns when -v option used
Junio C Hamano <gitster@pobox.com> wrote:
>Jeremy Faith <jeremy.faith@jci.com> writes:
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>> 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
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>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.
man git-check-ignore states:-
EXIT STATUS
-----------
0::
One or more of the provided paths is ignored.
1::
None of the provided paths are ignored.
128::
A fatal error was encountered.
So my change matches what the manual states.
>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.
I doubt that changing the exit status when -v is added is intended behaviour.