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author | Jeff King <peff@peff.net> | 2020-10-27 03:36:02 -0400 |
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committer | Junio C Hamano <gitster@pobox.com> | 2020-10-27 12:41:54 -0700 |
commit | 0b809c82480a870f4a46fa66f0eddc1c1b04cf6b (patch) | |
tree | ac766b1f5e226b95d804eb911a3b7533e35d4ce4 /t/t5324-split-commit-graph.sh | |
parent | Git 2.29.1 (diff) | |
download | tgif-0b809c82480a870f4a46fa66f0eddc1c1b04cf6b.tar.xz |
checkout-index: drop error message from empty --stage=all
If checkout-index is given --stage=all for a specific path, it will try
to write stages 1-3 (if present) for that path to temporary files.
However, if the file is present only at stage 0, it writes nothing but
gives a confusing message:
$ git checkout-index --stage=all -- Makefile
git checkout-index: Makefile does not exist at stage 4
This is nonsense. There is no stage 4 (it's just an internal enum value
we use for "all"), and the documentation clearly states:
Paths which only have a stage 0 entry will always be omitted from the
output.
Here it's talking about the list of tempfiles written to stdout, but it
seems clear that this case was not meant to be an error. We even have a
test which covers it, but it only checks that the command reports an
exit code of 0, not its stderr. And it reports 0 only because of another
bug which fails to propagate errors (which will be fixed in a subsequent
patch).
So let's make the test more thorough. We'll also cover the case that we
found _no_ entry, not even a stage zero, which should still be an error.
However, because of the other bug, we'll have to mark this as expecting
failure for the moment.
Signed-off-by: Jeff King <peff@peff.net>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
Diffstat (limited to 't/t5324-split-commit-graph.sh')
0 files changed, 0 insertions, 0 deletions