我尝试在两个Future列表上调用Future.await：

Future<void> first() async {
  await Future.delayed(Duration(seconds: 1));
  print('first');
}

Future<void> second() async {
  print('second');
}

void main() async {
  final list1 = [0, 1].map((e) => first()).toList();
  final list2 = [2, 3].map((e) => second()).toList();
  final appended = list1.addAll(list2); // result is not defined
  // await Future.wait(appended); // <= error: This expression has a type of 'void' so its value can't be used.
  await Future.wait([...list1, ...list2]); // <= this works
}

为什么addAll不能像预期的那样工作？