merge-ort: avoid repeating fill_tree_descriptor() on the same tree

Three-way merges, by their nature, are going to often have two or more
trees match at a given subdirectory.  We can avoid calling
fill_tree_descriptor() on the same tree by checking when these trees
match.  Noting when various oids match will also be useful in other
calculations and optimizations as well.

Signed-off-by: Elijah Newren <[email protected]>
Signed-off-by: Junio C Hamano <[email protected]>

Author: newren

merge-ort.c

@@ -223,6 +223,15 @@ 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));
 
 	/* n = 3 is a fundamental assumption. */
 	if (n != 3)
@@ -275,10 +284,19 @@ static int collect_merge_info_callback(int n,
 		newinfo.pathlen = st_add3(newinfo.pathlen, p->pathlen, 1);
 
 		for (i = MERGE_BASE; i <= MERGE_SIDE2; i++) {
-			const struct object_id *oid = NULL;
-			if (dirmask & 1)
-				oid = &names[i].oid;
-			buf[i] = fill_tree_descriptor(opt->repo, t + i, oid);
+			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];
+			else {
+				const struct object_id *oid = NULL;
+				if (dirmask & 1)
+					oid = &names[i].oid;
+				buf[i] = fill_tree_descriptor(opt->repo,
+							      t + i, oid);
+			}
 			dirmask >>= 1;
 		}