[PATCH v4 5/6] cache-tree: avoid an unnecessary check

From: Johannes Schindelin via GitGitGadget <gitgitgadget@gmail.com>
Date: Feb 23, 2024, 08:34 UTC
Message-ID: <7e5e84a4e7c5733c7f229a7713713d44406ad573.1708677266.git.gitgitgadget@gmail.com>
In-Reply-To: <pull.1651.v4.git.1708677266.gitgitgadget@gmail.com>

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();

-- 
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