在选择任何弹出项目之前，Flutter弹出菜单按钮不会关闭（onCanceled不会调用）

Question

一切正常，但是popupmenubutton直到选择了任何弹出项都不会关闭。 我不明白为什么不叫onCanceled。 请帮助我。

我提供了以下源代码。 谢谢。

//从这里打电话

StatefulWidget> CustomScrollView{
child: callAction(
tooltip: "Call Button",
   child: Container(
        height: double.infinity,
        padding: EdgeInsets.all(20),
        child: Icon(Icons.call)
    ),
  ),
}

// PopupMenuButton小部件

enum CallActionType { DataCall, Sim1Call, Sim2Call}

class callAction extends StatelessWidget {
  Widget child;
  String tooltip;
  callAction({@required this.child, @required this.tooltip});

PopupMenuButton<CallActionType>(
      elevation: 8,
      shape: RoundedRectangleBorder(
          borderRadius: BorderRadius.circular(15.0)),
      onSelected: (CallActionType value) {
        setState(() {
          print(value);
        });
      },
//onCanceled didn't call
      onCanceled: () {
        print('You have not chossed anything');
      },
      tooltip: widget.tooltip,
      offset: Offset(0, 100),
      child: widget.child,
      itemBuilder: (BuildContext context) => <PopupMenuEntry<CallActionType>>[
        new PopupMenuItem<CallActionType>(
          value: "Action 1",
          child: Text('Action 1'),
        ),
        new PopupMenuItem<CallActionType>(
          value: "Action 2",
          child: Text('Action 2'),
        ),
        new PopupMenuItem<CallActionType>(
          value: "Action 3",
          child: Text('Action 3'),
        ),
      ],
    );