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sparse-index: silently return when cache tree fails

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If cache_tree_update() returns a non-zero value, then it could not
create the cache tree. This is likely due to a path having a merge
conflict. Since we are already returning early, let's return silently to
avoid making it seem like we failed to write the index at all.

If we remove our dependence on the cache tree within
convert_to_sparse(), then we could still recover from this scenario and
have a sparse index.

Signed-off-by: Derrick Stolee <dstolee@microsoft.com>
Reviewed-by: Elijah Newren <newren@gmail.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Derrick Stolee 2021-09-08 01:42:28 +00:00 committed by Junio C Hamano
parent 72d84ea347
commit 5dc16756b2
1 changed files with 6 additions and 4 deletions
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10
sparse-index.c
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@ -178,10 +178,12 @@ int convert_to_sparse(struct index_state *istate)
/* Clear and recompute the cache-tree */
cache_tree_free(&istate->cache_tree);
if (cache_tree_update(istate, 0)) {
warning(_("unable to update cache-tree, staying full"));
return -1;
}
/*
* Silently return if there is a problem with the cache tree update,
* which might just be due to a conflict state in some entry.
*/
if (cache_tree_update(istate, 0))
return 0;
remove_fsmonitor(istate);