From: Johannes Schindelin <johannes.schindelin@xxxxxx> 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@xxxxxx> Signed-off-by: Johannes Schindelin <johannes.schindelin@xxxxxx> --- 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(); -- gitgitgadget
