Flutter showDialog在颤振中关于在其内部调用的方法有错误？

我使用的是showDialog方法，该对话框被称为showDialog，它将出现在地图上。showDialog在proDialog上有一些带有语句的红线

这是我正在使用的对话框小部件：

WidgetBuilder  proDialog = (BuildContext context)  => Dialog(
  backgroundColor: Colors.white,
  child: Padding(
    padding: EdgeInsets.all(17.0),
    child: Column(
      mainAxisAlignment: MainAxisAlignment.start,
      mainAxisSize: MainAxisSize.min,
      children: <Widget>[

        Padding(
          padding: EdgeInsets.all(15.0),
          child: Text(
            'Your Profile',
            style: TextStyle(
              color: Colors.blue,
              fontSize: 26.0,

            ),

          ),
        ),


        Container(color: Colors.black, height: 2),
        Row(mainAxisAlignment: MainAxisAlignment.start, children: <Widget>[
          Padding(
            padding: EdgeInsets.all(15.0),
            child: Text(
              'Take a Photo',
              style: TextStyle(
                fontSize: 26.0,
                color: Colors.black,
              ),
            ),
          ),
        ]),


        Container(color: Colors.black, height: 2),
        SizedBox(height: 8),
        RaisedButton(
          padding: EdgeInsets.symmetric(horizontal: 40.0, vertical: 15.0),
          textColor: Colors.white,
          color: Colors.redAccent[800],
          child: Text('Back', style: TextStyle(fontSize: 16)),
          shape:
              RoundedRectangleBorder(borderRadius: BorderRadius.circular(30.0)),
          onPressed: () {
            Navigator.pop(context);
          },
        )
      ],
    ),
  ),
);

这就是我打电话的地方，在proDialog上出现了一些错误

改变

Widget proDialog = (BuildContext context)  => Dialog(

到

---

您应将对话框作为生成器传递或调用生成器：

showDialog(
  context: context,
  builder: proDialog,
);

或

---

否这是一个错误，说明无法将Dialog类型的值分配给WidgetBuilderbuild context->Widget.类型的变量。已更改为Widget

showDialog(
  context: context,
  builder: proDialog,
);

showDialog(
    context: context,
    builder: (BuildContext context) => proDialog(context),
);