Re: [PATCH 4/4] Always check `parse_tree*()`'s return value

From: Patrick Steinhardt <ps@pks.im>
Date: Feb 7, 2024, 07:42 UTC
Message-ID: <ZcM0blqMPppLn2oB@tanuki>
In-Reply-To: <93abd7000b81bbcd7f5422715b4bbb60c69a7cde.1707212981.git.gitgitgadget@gmail.com>

On Tue, Feb 06, 2024 at 09:49:41AM +0000, Johannes Schindelin via GitGitGadget wrote:

> From: Johannes Schindelin <johannes.schindelin@gmx.de>

[snip]

Show 12 quoted lines

> diff --git a/cache-tree.c b/cache-tree.c
> index 641427ed410..c6508b64a5c 100644
> --- a/cache-tree.c
> +++ b/cache-tree.c
> @@ -779,8 +779,8 @@ 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);
> +			if (!subtree->object.parsed && parse_tree(subtree) < 0)
> +				exit(128);

It's somewhat weird that we have the `subtree->object.parsed` check in the first place, because the first thing that `parse_tree_gently()` does is to check that flag and immediately return when the tree is parsed already. It's a preexisting issue though, so this is fine.

Patrick