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([2600:1700:e72:80a0:c9d4:abc:8d:1568]) by smtp.gmail.com with UTF8SMTPSA id n41sm2464892qtb.18.2020.11.11.06.51.37 (version=TLS1_3 cipher=TLS_AES_128_GCM_SHA256 bits=128/128); Wed, 11 Nov 2020 06:51:37 -0800 (PST) Subject: Re: [PATCH v2 07/20] merge-ort: avoid repeating fill_tree_descriptor() on the same tree To: Elijah Newren , git@vger.kernel.org References: <20201102204344.342633-1-newren@gmail.com> <20201102204344.342633-8-newren@gmail.com> From: Derrick Stolee Message-ID: Date: Wed, 11 Nov 2020 09:51:35 -0500 User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:83.0) Gecko/20100101 Thunderbird/83.0 MIME-Version: 1.0 In-Reply-To: <20201102204344.342633-8-newren@gmail.com> Content-Type: text/plain; charset=UTF-8 Content-Language: en-US Content-Transfer-Encoding: 7bit Precedence: bulk List-ID: X-Mailing-List: git@vger.kernel.org On 11/2/2020 3:43 PM, Elijah Newren wrote: > @@ -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: > + 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. > + 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