From: Florian Hassanen <florian.hassanen@gmail.com>
Date: Thu, 11 Dec 2014 23:16:32 GMT
Subject: Re: [PATCH] git-check-ignore.txt: Clarify exit codes
Message-ID: <CAMAJHBYVMy-tc_aCjyeN7=VMG=k7gjWh9evL6bcJLzYfQcf6yg@mail.gmail.com>
In-Reply-To: <xmqqiohhdc5j.fsf@gitster.dls.corp.google.com>

>> check-ignore disregards whether a path is matched by a
>> positive or negative pattern. Thus for a file that is _not_
>> ignored, but is captured by negative pattern in .gitignore,
>> the exit code is 0. The docs suggested otherwise.
>
> I am not sure that is the actual behaviour of the command.  Given
> this .gitignore file:
>
>     $ cat .gitignore; echo END
>     !vendor.o
>     *.o
>     END
>
> which is designed to allow binary-only blob "vendor.o" supplied by
> the vendor to be tracked, but to ignore all the other usual build
> artifacts, you see this:
>
>     $ for o in a.o vendor.o; do git check-ignore $o >/dev/null; echo $?; done
>     0
>     1
>
> Puzzled...

Maybe your global ignore file gets in the way?

on both of my machines (cygwin + git 2.1.1, linux + git @master) I have:

  $ cat > .gitignore << EOF
  !vendor.o
  *.o
  EOF

  for o in a.o vendor.o; do git check-ignore $o >/dev/null; echo $?; done
  0
  0

which corresponds to how I understand the command so far :)

is one of a.o and vendor.o already in your index?
there is a --no-index as well (which enables yet another different behavior)