Re: [PATCH v2 07/20] merge-ort: avoid repeating fill_tree_descriptor() on the same tree
- From
Derrick Stolee <stolee@gmail.com>
- Date
- Nov 11, 2020, 14:51 UTC
- Message-ID
- <c4b9b64c-f0a0-5824-44b9-072370cfc465@gmail.com>
- In-Reply-To
- <20201102204344.342633-8-newren@gmail.com>
On 11/2/2020 3:43 PM, Elijah Newren wrote:
Show 13 quoted lines
> @@ -99,6 +99,15 @@ static int collect_merge_info_callback(int n, > unsigned mbase_null = !(mask & 1); > unsigned side1_null = !(mask & 2); > unsigned side2_null = !(mask & 4); > + unsigned side1_matches_mbase = (!side1_null && !mbase_null && > + names[0].mode == names[1].mode && > + oideq(&names[0].oid, &names[1].oid)); > + unsigned side2_matches_mbase = (!side2_null && !mbase_null && > + names[0].mode == names[2].mode && > + oideq(&names[0].oid, &names[2].oid)); > + unsigned sides_match = (!side1_null && !side2_null && > + names[1].mode == names[2].mode && > + oideq(&names[1].oid, &names[2].oid));
If the *_null values were in an array, instead, then all of these lines could be grouped as a macro:
unsigned null_oid[3] = {
!(mask & 1),
!(mask & 2),
!(mask & 4)
};#define trivial_merge(i,j) (!null_oid[i] && !null_oid[j] && \ names[i].mode == names[j].mode && \ oideq(&names[i].oid, &names[j].oid))
unsigned side1_matches_mbase = trivial_merge(0, 1); unsigned side2_matches_mbase = trivial_merge(0, 2); unsigned sides_match = trivial_merge(1, 2);
I briefly considered making these last three an array, as well, except the loop below doesn't use 'i' in a symmetrical way:
Show 6 quoted lines
> + if (i == 1 && side1_matches_mbase) > + t[1] = t[0]; > + else if (i == 2 && side2_matches_mbase) > + t[2] = t[0]; > + else if (i == 2 && sides_match) > + t[2] = t[1];
Since the 'i == 2' case has two possible options, it wouldn't be possible to just have 'side_matches[i]' here.
Show 7 quoted lines
> + else {
> + const struct object_id *oid = NULL;
> + if (dirmask & 1)
> + oid = &names[i].oid;
> + buf[i] = fill_tree_descriptor(opt->repo,
> + t + i, oid);
> + }I do appreciate the reduced recursion here!
Thanks, -Stolee