Question

import java.util.*;

public class GuessNumber{
    public static void main(String[]args){
        Scanner in = new Scanner(System.in);
        int x = 0;
        String num = "65854"; // This is the secret code
        String s;
        System.out.println("This program asks you to guess the code of 5 digits. ");
        System.out.println("You have 5 attempts. ");

        for(int i=0; i<5; i++){
            do{
                s = in.nextLine(); 


                for(int c=0; c<s.length(); c++){
                    if(s.charAt(c)==num.charAt(c)) //if digit in 's' equals the digit in the same position in 'num', increment variable x
                        x++;
                }

                System.out.println("Number of correct digits in right position: " + x);  // here the execution goes out of bounds
            }
            while(!s.equals(num));
            System.out.println("Congrats! You guessed the secret code. ");
            System.exit(0);

        }
    }
}

我试图创建一个简单的java程序，它应该允许用户猜测一个五位数的前缀代码（只有五次尝试）。 do-while循环仅显示前两次尝试的正确值，然后超出界限（显示值> 5，对于仅5位数的代码是不可能的）。有人能解释一下原因吗？

Answer 1

删除do-while循环。它将运行直到用户猜出正确的代码插入以下代码以检查字符串的长度

if(s.length()!=5){
    System.out.println("code should be of length 5");
    continue;
}

您可以为输入字符串中的任何字符添加更多限制。还要在每次外循环开始时重置x。

同时检查每个外部循环中的输入字符串是否正确

if(s.equals(num)){
    System.out.println("Congrats! You guessed the secret code. ");
    System.exit(0);
}

Answer 2

当前代码 -

for (int i = 0; i < 5; i++) {
    do {
         s = in.nextLine();
         for (int c = 0; c < s.length(); c++) { // the length of s can exceed 5 as of **num = "65854"**
             if (s.charAt(c) == num.charAt(c)) //if digit in 's' equals the digit in the same position in 'num', increment variable x
             x++;
             System.out.println("Number of correct digits in right position: " + x);
         }
      }
      while (!s.equals(num)); // this ensures currently N number of attempts NOT JUST 5 as stated otherwise.
      System.out.println("Congrats! You guessed the secret code. ");
      System.exit(0); // with this line the for loop would execute just once
}

建议代码 -

String s;
int count = 0;
int x;
do {
    x = 0;
    if (count >= 5) { break; } // 5 attempts
    s = in.nextLine();
    String formattedNumber = String.format("%05d",
           Integer.parseInt(s)); // make sure the input is of 5 digit integer (padding here with 0's)
    for (int c = 0; c < formattedNumber.length(); c++) {
        if (formattedNumber.charAt(c) == num.charAt(c)) {
            x++;
        }
        if (x == 5) { // correct guess if all chars are equal
            System.out.println("Congrats! You guessed the secret code.");
            break; // break if guessed correct
        } else {
            System.out.println("Number of correct digits in right position: " + x);
        }
    }
    count++; //next attempt
} while (!s.equals(num)); //input not equals the desired, next attempt

Answer 3

此代码将运行 EVER ，因为您输入了真实代码，因为您的!s.equals(num)条件为do-while，因此您必须删除{{1首先，这不是必要的，当您的预测代码等于num时，变量x必须等于5，因此您使用{{1}终止程序打印成功后的语句。要注意return的值，它在每次迭代时必须等于零，我的意思是x！

x=0

循环无法正常工作

3 个答案: