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cache-tree: avoid an unnecessary check

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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>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
This commit is contained in:
Johannes Schindelin 2024-02-23 08:34:24 +00:00 committed by Junio C Hamano
parent aa9f618909
commit 5aca024a74
1 changed files with 1 additions and 1 deletions
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2
cache-tree.c
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@ -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();