如何在android studio中随机生成随机运算符

Posted 2023-04-18

【中文标题】如何在android studio中随机生成随机运算符

【英文标题】：How to randomly generate random operators in android studio

【发布时间】：2021-09-12 11:38:34

【问题描述】：

公共类 MainActivity 扩展 AppCompatActivity

Button startButton;
ArrayList<Integer> answers=new ArrayList<Integer>();
int LocationOfRightAnswer;
TextView resultTextView;
TextView pointText;
Button button1;
Button button2;
Button button3;
Button button4;
int score=0;
int numberofquestions = 0;
TextView sumText;
TextView timertext;
RelativeLayout game;
Button playAgain;



public void playAgain(View view) 

    score = 0;
    numberofquestions = 0 ;
    timertext.setText("30s");
    pointText.setText("0/0");
    resultTextView.setText("");
    playAgain.setVisibility(View.INVISIBLE);

    generateQuestion();

    new CountDownTimer(30100,1000) 

        @Override
        public void onTick(long millisUntilFinished) 

            timertext.setText(String.valueOf(millisUntilFinished / 1000) + "s");

        

        @Override
        public void onFinish() 

            playAgain.setVisibility(View.VISIBLE);
            timertext.setText("0s");
            resultTextView.setText("Your Score is " + Integer.toString(score) + "/" + Integer.toString(numberofquestions));

        
    .start();





public void generateQuestion() 
    Random random = new Random();

    int a = random.nextInt(21);
    int b = random.nextInt(21);

    char[] ops = '+' , '-' ,'*', '/';
    int l = random.nextInt(3) ;


    sumText.setText(Integer.toString(a) + "+"  + Integer.toString(b) );

    LocationOfRightAnswer = random.nextInt(4);
    answers.clear();

    int incorrectAnswer;

    for (int i=0; i<4; i++) 

        if (i == LocationOfRightAnswer) 

            answers.add(a + b);

        else 

            incorrectAnswer = random.nextInt(41);

            while (incorrectAnswer == a + b) 
                incorrectAnswer = random.nextInt(41);
            
            answers.add(incorrectAnswer);
        
    


    button1.setText(Integer.toString(answers.get(0)));
    button2.setText(Integer.toString(answers.get(1)));
    button3.setText(Integer.toString(answers.get(2)));
    button4.setText(Integer.toString(answers.get(3)));






public void chooseAnswer(View view) 

    if (view.getTag().toString().equals(Integer.toString(LocationOfRightAnswer))) 

        score++;
        resultTextView.setText("Correct");

     else 
        resultTextView.setText("Wrong!");

    

    numberofquestions++;
    pointText.setText(Integer.toString(score) + "/" + Integer.toString(numberofquestions));
    generateQuestion();



public void start (View view) 

    startButton.setVisibility(View.INVISIBLE);
    game.setVisibility(View.VISIBLE);




@Override
protected void onCreate(Bundle savedInstanceState) 
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    startButton =(Button) findViewById(R.id.startButton);
   sumText = (TextView) findViewById(R.id.sumTextView);
  button1 = (Button) findViewById(R.id.button1);
     button2 = (Button) findViewById(R.id.button2);
     button3 = (Button) findViewById(R.id.button3);
     button4 = (Button) findViewById(R.id.button4);
    resultTextView = (TextView) findViewById(R.id.resultTextView);
    pointText = (TextView) findViewById(R.id.pointsTextView);
    timertext = (TextView) findViewById(R.id.timerTextView);
    playAgain = (Button) findViewById(R.id.playAgain);
    game = (RelativeLayout) findViewById(R.id.gamelayout) ;




playAgain(findViewById(R.id.playAgain));

请帮助我调整我的代码，以便我的应用程序可以使用所有运算符，而不仅仅是加法。我试过这样做，但我只设法为运算符创建数组列表，但我找不到更改代码以包含所有运算符的方法。

【参考方案1】：

试试这个：

ArrayList<Double> answers=new ArrayList<Double>();




public void generateQuestion() 
    Random random = new Random();

    int a = random.nextInt(21);
    int b = random.nextInt(21);

    char[] ops = '+' , '-' ,'*', '/';
    int l = random.nextInt(3) ;
    
    char op = ops[random.nextInt(4)];
    double correctAns=0;
    
    switch(op)
        case '+' : correctAns = a+b;
           break;
        case '/' : correctAns = (double)a/b;
           break;
        case '*' : correctAns = a*b;
           break;
        case '-' : correctAns = a-b;
           break;
      


    sumText.setText(Integer.toString(a) + op  + Integer.toString(b) );

    LocationOfRightAnswer = random.nextInt(4);
    answers.clear();

    int incorrectAnswer;

    for (int i=0; i<4; i++) 

        if (i == LocationOfRightAnswer) 

            answers.add(correctAns);

        else 

            incorrectAnswer = random.nextInt(41);

            while (incorrectAnswer != correctAns) 
                incorrectAnswer = random.nextInt(41);
            
            answers.add((double)incorrectAnswer);
        
    


    button1.setText(Double.toString(answers.get(0)));
    button2.setText(Double.toString(answers.get(1)));
    button3.setText(Double.toString(answers.get(2)));
    button4.setText(Double.toString(answers.get(3)));

将ArrayList<Integer> answers=new ArrayList<Integer>(); 更改为ArrayList<Double> answers=new ArrayList<Double>();（int/int 可以是double） 创建一个包含正确答案的double 变量（int/int 可以是double） 随机选择一个操作。 使用switch 定义在a 和b 上需要哪些操作，以防选择每个操作。 将文本设置为Integer.toString(a) + op + Integer.toString(b) 而不是Integer.toString(a) + "+" + Integer.toString(b) 将a+b 交换为correctAns

【讨论】：

答案.add(correctAns);不工作它说它应该被初始化

ArrayList<Double> answers=new ArrayList<Double>(); 用这个代替ArrayList<Integer> answers=new ArrayList<Integer>();

它告诉我将 'default' 分支添加到初始化 'correctAns' 的 'switch' 语句中

@ZuhayrAlarakha 哦。没有必要，因为无论如何都会定义correctAns 。反正我改了。只需将 correctAns 初始化为 0 查看更新