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Re: "git rm" seems to do recursive removal even without "-r"




On Sun, 8 Oct 2017, Junio C Hamano wrote:

> "Robert P. J. Day" <rpjday@xxxxxxxxxxxxxx> writes:
>
> > ... so if, in the kernel source
> > tree, i ran:
> >
> >   $ git rm \*.c
> >
> > i would end up removing *all* 25,569 "*.c" files in the kernel
> > source repository.
>
> Yes, as that is exactly what the command line asks Git to do.

  ok, i truly want to understand this, so let me dig through this
carefully. i can now see (from the man page and the recent
explanations) that "git rm" will accept *escaped* fileglobs to remove
and that, further, "File globbing matches across directory
boundaries." which is why, in the linux kernel source tree, if i run
one of:

  $ git rm \*.c
  $ git rm '*.c'

the "git rm" command will internally process the fileglob and apply it
across directory boundaries. and that's why, when i try a dry run, i
can see the effect it would have on the kernel source:

  $ git rm -n '*.c' | wc -l
  25569
  $

> If you said
>
>     $ git rm *.c
>
> then the shell expands the glob and all Git sees is that you want to
> remove a.c b.c d.c ...; if you said "git rm -r *.c", unless b.c is
> not a directory, these and only these files are removed.

  right, that's just regular shell fileglob processing, no surprise
there. (let's stick to just file removal for now.)

> >   however, let's say i wanted to remove, recursively, all files with a
> > *precise* (non-globbed) name, such as "Makefile". so i, naively, run:
> >
> >   $ git rm Makefile
> >
> > guess what ... the lack of globbing means i remove only the single
> > Makefile at the top of the working directory.
>
> Again, that is exactly what you asked Git to do.

  yes, now i get it -- a lack of fileglob arguments disallows
traversing directory boundaries, so one gets the "normal" behaviour.

>     $ git rm $(find . -name Makefile -print)
>
> would of course one way to remove all Makefiles.  If you let POSIX
> shell glob, i.e.
>
>     $ git rm */Makefile
>
> the asterisk would not expand nothing but a single level, so it may
> remove fs/Makefile, but not fs/ext4/Makefile (some shells allow
> "wildmatch" expansion so "git rm **/Makefile" may catch the latter
> with such a shell).

  sure, all regular shell fileglob processing.

> By letting Git see the glob, i.e.
>
>     $ git rm Makefile \*/Makefile
>
> you would let Git to go over the paths it knows/cares about to find
> ones that match the pathspec pattern and remove them (but not
> recursively, even if you had a directory whose name is Makefile; for
> that, you would use "-r").

  right ... i can now see that '*/Makefile' would pick up all
Makefiles *below* the current directory, so you need that initial
reference to 'Makefile' to catch the top one. this just seems ...
awkward.

  but as i asked in my earlier post, if i wanted to remove *all* files
with names of "Makefile*", why can't i use:

  $ git rm 'Makefile*'

just as i used:

  $ git rm '*.c'

are those not both acceptable fileglobs? why does the former clearly
only match the top-level Makefile, and refuse to cross directory
boundaries?

  $ git rm -n 'Makefile*'
  rm 'Makefile'
  $

rday

-- 

========================================================================
Robert P. J. Day                                 Ottawa, Ontario, CANADA
                        http://crashcourse.ca

Twitter:                                       http://twitter.com/rpjday
LinkedIn:                               http://ca.linkedin.com/in/rpjday
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