Re: [PATCH 14/33] refs: extract a function peel_entry()

From: Junio C Hamano <gitster@pobox.com>
Date: Apr 15, 2013, 17:38 UTC
Message-ID: <7v38urwv6o.fsf@alter.siamese.dyndns.org>
In-Reply-To: <1365944088-10588-15-git-send-email-mhagger@alum.mit.edu>

Michael Haggerty <mhagger@alum.mit.edu> writes:

Show 14 quoted lines

>  	if (read_ref_full(refname, base, 1, &flag))
>  		return -1;
>  
> -	if ((flag & REF_ISPACKED)) {
> +	/*
> +	 * If the reference is packed, read its ref_entry from the
> +	 * cache in the hope that we already know its peeled value.
> +	 * We only try this optimization on packed references because
> +	 * (a) forcing the filling of the loose reference cache could
> +	 * be expensive and (b) loose references anyway usually do not
> +	 * have REF_KNOWS_PEELED.
> +	 */
> +	if (flag & REF_ISPACKED) {
>  		struct ref_entry *r = get_packed_ref(refname);

This code makes the reader wonder what happens when a new loose ref masks a stale packed ref, but the worry is unfounded because the read_ref_full() wouldn't have gave us REF_ISPACKED in the flag in such a case.

But somehow the calling sequence looks like such a mistake waiting to happen. It would be much more clear if a function that returns a "struct ref_entry *" is used instead of read_ref_full() above, and we checked (r->flag & REF_ISPACKED) in the conditional, without a separate get_packed_ref(refname).

Show 10 quoted lines

> -
> -		if (r && (r->flag & REF_KNOWS_PEELED)) {
> -			if (is_null_sha1(r->u.value.peeled))
> -			    return -1;
> +		if (r) {
> +			if (peel_entry(r))
> +				return -1;
>  			hashcpy(sha1, r->u.value.peeled);
>  			return 0;
>  		}