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Fix "read-tree -m A B" priming the cache-tree

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In 456156d a shortcut to priming the index tree reference was
introduced, but the justification for it was completely bogus.

"read-tree -m A B" is to take the index (and the working tree)
that is largely based on (but does not have to match exactly) A
and update it to B, while carrying the local change that does
not overlap the difference between A and B, so there is no reason
to expect that the resulting index should match the tree B.

Noticed and test provided by Heiko Voigt.

Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Junio C Hamano 2010-07-08 17:27:43 -07:00
parent 29981380d0
commit b1f47514f2
2 changed files with 16 additions and 5 deletions
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5
builtin-read-tree.c
View File

@ -199,14 +199,9 @@ int cmd_read_tree(int argc, const char **argv, const char *unused_prefix)
* "-m ent" or "--reset ent" form), we can obtain a fully
* valid cache-tree because the index must match exactly
* what came from the tree.
*
* The same holds true if we are switching between two trees
* using read-tree -m A B. The index must match B after that.
*/
if (nr_trees == 1 && !opts.prefix)
prime_cache_tree(&active_cache_tree, trees[0]);
else if (nr_trees == 2 && opts.merge)
prime_cache_tree(&active_cache_tree, trees[1]);
if (write_cache(newfd, active_cache, active_nr) ||
commit_locked_index(&lock_file))

16
t/t1001-read-tree-m-2way.sh
View File

@ -392,4 +392,20 @@ test_expect_success \
git ls-files --stage | tee >treeMcheck.out &&
test_cmp treeM.out treeMcheck.out'
test_expect_success '-m references the correct modified tree' '
echo >file-a &&
echo >file-b &&
git add file-a file-b &&
git commit -a -m "test for correct modified tree"
git branch initial-mod &&
echo b >file-b &&
git commit -a -m "B" &&
echo a >file-a &&
git add file-a &&
git ls-tree $(git write-tree) file-a >expect &&
git read-tree -m HEAD initial-mod &&
git ls-tree $(git write-tree) file-a >actual &&
test_cmp expect actual
'
test_done