revision: note the lack of free() in simplify_merges()

Among the three similar-looking loops that walk singly linked
commit_list, the first one is only peeking and the same list is
later used for real work.  Leave a comment not to mistakenly
free its elements there.

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

Author: gitster

revision.c

@@ -2015,23 +2015,31 @@ static struct commit_list **simplify_one(struct rev_info *revs, struct commit *c
 
 static void simplify_merges(struct rev_info *revs)
 {
-	struct commit_list *list;
+	struct commit_list *list, *next;
 	struct commit_list *yet_to_do, **tail;
+	struct commit *commit;
 
 	if (!revs->prune)
 		return;
 
 	/* feed the list reversed */
 	yet_to_do = NULL;
-	for (list = revs->commits; list; list = list->next)
-		commit_list_insert(list->item, &yet_to_do);
+	for (list = revs->commits; list; list = next) {
+		commit = list->item;
+		next = list->next;
+		/*
+		 * Do not free(list) here yet; the original list
+		 * is used later in this function.
+		 */
+		commit_list_insert(commit, &yet_to_do);
+	}
 	while (yet_to_do) {
 		list = yet_to_do;
 		yet_to_do = NULL;
 		tail = &yet_to_do;
 		while (list) {
-			struct commit *commit = list->item;
-			struct commit_list *next = list->next;
+			commit = list->item;
+			next = list->next;
 			free(list);
 			list = next;
 			tail = simplify_one(revs, commit, tail);
@@ -2043,9 +2051,10 @@ static void simplify_merges(struct rev_info *revs)
 	revs->commits = NULL;
 	tail = &revs->commits;
 	while (list) {
-		struct commit *commit = list->item;
-		struct commit_list *next = list->next;
 		struct merge_simplify_state *st;
+
+		commit = list->item;
+		next = list->next;
 		free(list);
 		list = next;
 		st = locate_simplify_state(revs, commit);