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>
diff --git a/builtin-fetch.c b/builtin-fetch.c
index 320e235..9a6ddce 100644
--- a/builtin-fetch.c
+++ b/builtin-fetch.c
@@ -40,6 +40,8 @@
"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"),
