git fetch: Take '-n' to mean '--no-tags'

Prior to commit 8320199 (Rewrite builtin-fetch option parsing to use
parse_options().), we understood '-n' as a short option to mean "don't
fetch tags from the remote". This patch reinstates behaviour similar,
but not identical to the pre commit 8320199 times.

Back then, -n always overrode --tags, so if both --tags and -n was
given on command-line, no tags were fetched regardless of argument
ordering. Now we use a "last entry wins" strategy, so '-n --tags'
means "fetch tags".

Since it's patently absurd to say both --tags and --no-tags, this
shouldn't matter in practice.

Spotted-by: Artem Zolochevskiy <azol@altlinux.org>
Reported-by: Dmitry V. Levin <ldv@altlinux.org>
Signed-off-by: Johannes Schindelin <johannes.schindelin@gmx.de>
Tested-by: Andreas Ericsson <ae@op5.se>
Signed-off-by: Junio C Hamano <gitster@pobox.com>

builtin-fetch.c

@@ -40,6 +40,8 @@ static struct option builtin_fetch_options[] = {
 		    "force overwrite of local branch"),
 	OPT_SET_INT('t', "tags", &tags,
 		    "fetch all tags and associated objects", TAGS_SET),
+	OPT_SET_INT('n', NULL, &tags,
+		    "do not fetch all tags (--no-tags)", TAGS_UNSET),
 	OPT_BOOLEAN('k', "keep", &keep, "keep downloaded pack"),
 	OPT_BOOLEAN('u', "update-head-ok", &update_head_ok,
 		    "allow updating of HEAD ref"),