cache-tree: avoid an unnecessary check
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>
Johannes Schindelin committed
Feb 23, 2024 at 08:34 UTC
5aca024a74e900bd9bc2c14a8e99494063ea4cc5
1 file changed
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-1
cache-tree.c
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-1
@@ -779,7 +779,7 @@ static void prime_cache_tree_rec(struct repository *r,
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struct cache_tree_sub *sub;
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struct tree *subtree = lookup_tree(r, &entry.oid);
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- if (!subtree->object.parsed && parse_tree(subtree) < 0)
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+ if (parse_tree(subtree) < 0)
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exit(128);
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sub = cache_tree_sub(it, entry.path);
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sub->cache_tree = cache_tree();