屏幕未在Flutter中的statefulWidget中更新

问题描述

我在Flutter中有一个std::array<int,3> arr = {1,2,3}; std::array<int,3> *ptr = &arr; // arr is equivalent to an array of 1 element int temp = ptr[0][2]; // same as (*ptr).operator[](2) ,如下所示:

  .approvalContainer {
    align-self: flex-end;
    width: 24px;
    height: 24px;
    background-color: var(--color-background-success-dark);
    vertical-align: flex-end;
  }

  .approvalIcon {
    width: 100px;
    height: 100px;
  }

问题是,当我第一次使用此小部件启动应用程序时,我可以在屏幕上看到<div style=approvalContainer> <svg style=approvalIcon/> </div> ,但是如果我继续并将statefulWidget更改为{{ 1}},即使打开了热重装功能,它也不会更新屏幕,并且仍然显示class GameScreen extends StatefulWidget { @override GameScreenState createState() => GameScreenState(); } class GameScreenState extends State<GameScreen> { List<String> selectedWord = ['h','e','l','o']; Widget _letterinput() { return Center( child: Row( mainAxisAlignment: MainAxisAlignment.spaceAround,crossAxisAlignment: CrossAxisAlignment.center,children: <Widget>[ for (var letter in selectedWord) LetterInput(letter: letter) ],),); } @override Widget build(BuildContext context) { return Scaffold( body: SafeArea( child: Column( mainAxisAlignment: MainAxisAlignment.spaceEvenly,children: <Widget>[ _letterinput(),],)),); } } class LetterInput extends StatelessWidget { LetterInput({this.letter}); final String letter; @override Widget build(BuildContext context) { return Container( padding: EdgeInsets.fromLTRB(5,5,0),decoration: Boxdecoration( border: BorderDirectional( bottom: BorderSide(width: 6.0,color: Colors.green))),child: Text(letter,textAlign: TextAlign.center,style: GoogleFonts.acme(fontSize: 28.0,fontWeight: FontWeight.w400))); } } 。我必须去重新启动应用程序,以便它显示hello。我该如何解决

解决方法

根据我的经验,热重装可以保持状态。尝试热重启吗?

关于您的评论,如果您想继续使用热重载,建议您将变量拉出到小部件本身(如果您可以选择的话),如下所示:

class GameScreen extends StatefulWidget {
  final List<String> selectedWord = ['h','e','l','o'];

  @override
  GameScreenState createState() => GameScreenState();
}

class GameScreenState extends State<GameScreen> {
  Widget _letterInput() {
    return Center(
      child: Row(
        mainAxisAlignment: MainAxisAlignment.spaceAround,crossAxisAlignment: CrossAxisAlignment.center,children: <Widget>[
          for (var letter in widget.selectedWord) LetterInput(letter: letter)
        ],),);
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      body: SafeArea(
          child: Column(
            mainAxisAlignment: MainAxisAlignment.spaceEvenly,children: <Widget>[
              _letterInput(),],)),);
  }
}

相关问答

Selenium Web驱动程序和Java。元素在(x,y)点处不可单击。其...
Python-如何使用点“。” 访问字典成员?
Java 字符串是不可变的。到底是什么意思?
Java中的“ final”关键字如何工作?(我仍然可以修改对象。...
“loop:”在Java代码中。这是什么,为什么要编译?
java.lang.ClassNotFoundException:sun.jdbc.odbc.JdbcOdbc...