循环内部是异步的，外部如何同步[重复]

Question

例如：

void main() {
  List<String> list = ['1', '2', '3', '4'];
  gogo(list);
  print('main end');
}

void gogo(List list) async {
  list.forEach((element) async {
    await pS(element);
  });

  print('gogo end');
}

Future pS(String s) async {
  Future.delayed(Duration(seconds: 1), () {
    print(s);
  });
}

将输出：

gogo 结束 主端 1 2 3 4

我期待的结果：

1 2 3 4 gogo结束 主端

Answer 1

这是您所期望的，我也在示例中进行了说明。

// Create an async main method to work with Future
void main() async {
  List<String> list = ['1', '2', '3', '4'];
  await gogo(list);
  print('main end');
}

Future gogo(List list) async {
  // Use this method if you want to run all functions at the same time
  // and wait until all functions are completed
  //
  // You can't use List.foreach() because it is not a Future function
  // so you have no way to await it
  await Future.wait([
    for (var element in list) pS(element),
  ]);

  print('gogo end');
}

Future pS(String s) async {
  // You also need to await this function
  await Future.delayed(Duration(seconds: 1), () {
    print(s);
  });
}