java - 在java代码中添加一个条件，用户只能尝试4次

Question

所以我有这样的代码，用户输入一个从 0 到 9 的随机数，并且必须猜测这个数字，但我想将用户的尝试限制为 4 次，一旦他猜错 4 次，系统就会停止并显示一条消息“you超出了您允许的试验”这是我到目前为止的代码:

public static void main(String[] args) {
    // TODO Auto-generated method stub
    int nb,x;
    int count= 0;
    Scanner inp= new Scanner(System.in);
    nb= (int)(Math.random()*10);
    System.out.print("Guess the number between 0 and 9");
    x=inp.nextInt();

    while(x!= nb) {
        if(x>nb) System.out.println("Lesser");
        else if (x<nb) System.out.println("Bigger");
        System.out.println("Try another Number");
        x=inp.nextInt();
        count++;

    }

    System.out.println("Found!!");
    System.out.println("Number of Guesses:" +count);


    inp.close();

}

Answer 1

还可以用你count变量来跟踪尝试，一旦用户以限制结束，您可以通过打破 while 退出程序使用 break 循环陈述。

import java.util.Scanner;
public class Main
{
    public static void main(String[] args) {
    // TODO Auto-generated method stub
    int nb,x;
    int count= 1;
    Scanner inp= new Scanner(System.in);
    nb= (int)(Math.random()*10);
    System.out.print(nb);
    System.out.print("Guess the number between 0 and 9:");
    x=inp.nextInt();

    while(x!= nb) {
        if (count <= 3){
        if(x>nb) System.out.println("Lesser");
        else if (x<nb) System.out.println("Bigger");
        System.out.println("Try another Number");
        x=inp.nextInt();
        }
        else{
            System.out.print("you exceeded your allowed trials");
            break;

        }
        count++;

    }

    System.out.println("Found!!");
    System.out.println("Number of Guesses:" +count);


    inp.close();

   }
}