flutter - 按下后退按钮时，将indexStack设置为0

Question

我有一个 IndexStack 作为主要的导航方式，但是当用户点击“后退”按钮时，“导航”会返回到先前的支架。
目前我正在使用WillPopScope()，但是我一直收到一条错误消息，说...

type '() => void' is not a subtype of type '() => Future'

这是代码。
导航页

class NavPages extends StatefulWidget {
  @override
  _NavPagesState createState() => _NavPagesState();
}

List<Widget> screens = new List();
List<String> site = new List();

class _NavPagesState extends State<NavPages> {
  int _selectedIndex = 0;

  @override
  Widget build(BuildContext context) {
    site.add('..............');
    return Consumer<User>(
      builder: (context, value, child) {
        screens.add(GotKeysSites(
          user: value,
          departmentClicked: (title) {
            if (title.isNotEmpty) {
              var pos = site.indexWhere((element) => element == title);
              setState(() {
                _selectedIndex = pos;  //Here the index is updated however, i receive the error commented above.
              });
            }
          },
          departmentCallback: (title, departmentsAssigned, startTime, endTime) {
            if (title.isNotEmpty && departmentsAssigned.isNotEmpty) {
              site.add(title);
            }
          },
        ));
        return Scaffold(
          body: IndexedStack(
            index: _selectedIndex,
            children: screens,
          ),
        );
      },
    );
  }
}

typedef void ReturnIndex();

以及我称之为 ReturnIndex函数的类

return WillPopScope(
      onWillPop: widget.returnIndex,
      child: Scaffold(
      ));

Answer 1

OnWillPop范围希望您为其提供适当的功能，因此请尝试执行以下2个更改中的任何一项
更改1

return WillPopScope(
      onWillPop:(){ 
         widget.returnIndex;
      },
      child: Scaffold(
      ));

要么
更改2

void returnIndex(){//Future<void> if it is a future object
}