Re: [CLOSED] git checkout <branch> allowed with uncommitted changes

From: Jakub Narebski <jnareb@gmail.com>
Date: Oct 13, 2011, 19:01 UTC
Message-ID: <m3vcrs7m3v.fsf@localhost.localdomain>
In-Reply-To: <loom.20111013T181801-923@post.gmane.org>

arQon <arqon@gmx.com> writes:

> (Though if someone can come up with a script / hook / whatever that improves
> the "visibility" of stash, that would be awesome. Or one that makes the
> refusal to switch branches consistent).

Well, if you use __git_ps1 from contrib/completion/git-completion.bash
(installed with git-core package for some time), there is an option to
add '$' to branch name if stash is non-empty (though it doesn't actually
check if stash was on said branch).

Show 9 quoted lines

> Looking at the manpage for checkout in the hope that there might be a "--safe"
> switch, I don't understand why
>
>   "-f  Proceed even if the index *or the working tree* differs from HEAD."
>
> even exists, since it proceeds under those conditions anyway.
> "--safe" appears to be exactly what the behavior should be if you DON'T
> specify -f, except that -f nukes the working tree outright rather than just
> bleeding it across. Hopefully it'll be clearer after some sleep.  :)

 
Without '-f' git-checkout would switch branches only if uncomitted
changes (which do not belong to any branch) could be "floated" on top
of new branch.

If branch you are switching to has differences from current branch that conflict with uncomitted changes, git would refuse switching branches. Now '-f' would get rid of your uncomitted changes, and '-m' try to merge it with changes brought by new branch.

HTH

-- 
Jakub Narębski