Michael Haggerty <[email protected]> writes:
> 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).
> -
> - 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;
> }
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