Re: git rebase regression: cannot pass a shell expression directly to --exec

From: Jeff King <peff@peff.net>
Date: May 16, 2017, 17:41 UTC
Message-ID: <20170516174106.l2kmlprrimxku5vr@sigill.intra.peff.net>
In-Reply-To: <CA+zRj8WFuqaNV_6U5=VUrEmeQ_fNaRs8mKxdSd=Z0QMyGk4pHw@mail.gmail.com>

On Tue, May 16, 2017 at 10:29:31AM -0700, Eric Rannaud wrote:

Show 16 quoted lines

> > It does not really work that way. Git runs in a separate process that
> > does not have access to your current shell. That's why you need to do
> > 'export -f foo'.
> >
> > If you want git to be able to ecute the foo shell function, git needs to
> > start a _new_ shell process, which reads the environment, recognize the
> > exported function and run that.
> >
> > This is not the same as git executing the command in your shell. Not
> > exported variables would not be available in this function (as it would
> > be in your equivalent).
> 
> I'm sorry, I didn't mean (or say) "my shell process". Indeed, it
> doesn't work that way. And to be clear, there is no problem with
> having to "export -f foo". The only question is how should git run the
> <cmd> passed to --exec: should it run directly or using a shell?

It's definitely "using a shell". Most things Git runs on your behalf behave the same, but there are some exceptions due to historical warts (that would break backwards compatibility if we switched them). E.g., $GIT_SSH does not use the shell, but we introduced GIT_SSH_COMMAND as an alternative which does use the shell.

But note that "a shell" may not be necessarily be your login shell. We always execute "/bin/sh -c" (and you can override the shell path at build time). So your "export -f" trick only works because your /bin/sh is a symlink to bash (or because you specifically built with the SHELL_PATH knob pointed at bash).

-Peff