Question

我不太确定这里发生了什么。

代码应该根据固定ID号数组验证输入。但每当我故意输入错误的数字时，就会说＆＃34;数组超出范围＆＃34;。

对导致问题的原因不太确定，也许有人可以指出我的错误？

import java.util.Scanner;
 public class isValid
{
    static int accNum[] = {11111, 22222, 33333, 44444, 55555, 66666, 77777, 88888, 99999, 10101, 20202, 30303, 40404, 50505, 60606, 70707, 80808, 90909};

public static void main(String[] args)
{
    Scanner keyboard = new Scanner(System.in);

    int conti = -99;
    int search = 0;
    do
    {
        System.out.print("Enter 5-digit account number you want to validate: ");

        search = keyboard.nextInt();

        sequentialSearch(accNum, search);
        System.out.println("");
        System.out.print("Enter -9 to exit program, or any other number to validate another ID: ");
        conti = keyboard.nextInt();

    } while (conti != -9);
}

public static void sequentialSearch(int[] array,int value)
{
    int index = 0;        
    int element = -1;      
    boolean found = false;    

    while (!found && index < array.length)
    {
        if (array[index] == value)
        {
            found = true;
            element = index;
            break; //prevent index addition if value found
        }
        index++;
    }

    if (array[index] == value)
    {
        System.out.println("Account " + value + " is valid.");
    }
    else 
    {
        System.out.println("Account " + value + " is invalid.");
    }
}
}

问题：http://imgur.com/39caZxD 错误消息：http://imgur.com/GEr95Wb

Answer 1

我修改了代码，它应该是这样的：

    import java.util.Scanner;
    public class isValid
    {
        static int accNum[] = {11111, 22222, 33333, 44444, 55555, 66666, 77777, 88888, 99999, 10101, 20202, 30303, 40404, 50505, 60606, 70707, 80808, 90909};

        public static void main(String[] args)
        {
            Scanner keyboard = new Scanner(System.in);

            int conti = -99;
            int search = 0;
            do
            {
                System.out.print("Enter 5-digit account number you want to validate: ");

                search = keyboard.nextInt();

                sequentialSearch(accNum, search);
                System.out.println("");
                System.out.print("Enter -9 to exit program, or any other number to validate another ID: ");
                conti = keyboard.nextInt();

        } while (conti != -9);
    }

    public static void sequentialSearch(int[] array,int value)
    {
        int index = 0;        
        int element = -1;      
        boolean found = false;    

        while (!found && index < array.length)
        {
            if (array[index] == value)
            {
                found = true;
                element = index;
                break; //prevent index addition if value found
            }
            index++;
        }

        if (found)
        {
            System.out.println("Account " + value + " is valid.");
        }
        else 
        {
            System.out.println("Account " + value + " is invalid.");
        }
    }
  }

注意在sequentialSearch函数的while循环之后if条件的变化。

为什么？由于索引等于超出数组索引的值，以防数组中不存在该值。

Answer 2

好的，我真的很抱歉我以前的解决方案有点愚蠢，但现在我可以建议你解决方案的第二种方式就是这样做，

while (!found && index < accNum.length)
{
    if (accNum[index] == value)
    {
        found = true;
        element = index;
        System.out.println("Account " + value + " is valid.");
    }
    else
    {
        System.out.println("Account " + value + " is invalid.");
        break;
    }
    index++;
}

描述： - 您的值将匹配一次并打印valid消息。如果它与值不匹配，那么它将打印无效消息并且循环将中断。

无法执行数组验证（超出范围） - java

2 个答案: