Subject: Re: [PATCH v3 5/5] cache-tree: avoid an unnecessary check

From: Junio C Hamano <[email-masked]>
Date: Thu, 22 Feb 2024 10:00:17 -0800

"Johannes Schindelin via GitGitGadget" <[email-masked]>
writes:

> From: Johannes Schindelin <[email-masked]>
>
> 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 <[email-masked]>
> Signed-off-by: Johannes Schindelin <[email-masked]>
> ---
>  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();

Obviously makes sense.
I see no need for further comments.  Will queue.

Thanks.