[PATCH v2 5/5] cache-tree: avoid an unnecessary check

Johannes Schindelin via GitGitGadget
2024-02-07T16:47:50Z

From: Johannes Schindelin <johannes.schindelin@gmx.de>

The first thing the `parse_tree()` function does is to return early if
the tree has already been parsed. Therefore we do not need to guard the
`parse_tree()` call behind a check of that flag.

As of time of writing, there are no other instances of this in Git's
code bases: whenever the `parsed` flag guards a `parse_tree()` call, it
guards more than just that call.

Suggested-by: Patrick Steinhardt <ps@pks.im>
Signed-off-by: Johannes Schindelin <johannes.schindelin@gmx.de>
---
cache-tree.c | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/cache-tree.c b/cache-tree.c
index c6508b64a5c..78d6ba92853 100644
--- a/cache-tree.c
+++ b/cache-tree.c
@@ -779,7 +779,7 @@ static void prime_cache_tree_rec(struct repository *r,
struct cache_tree_sub *sub;
struct tree *subtree = lookup_tree(r, &entry.oid);

- if (!subtree->object.parsed && parse_tree(subtree) < 0)
+ if (parse_tree(subtree) < 0)
exit(128);
sub = cache_tree_sub(it, entry.path);
sub->cache_tree = cache_tree();