Java的哈利指数外边界的异常 [英] Java Arry Index Out of Bound Exception

问题描述

我一直在努力，当我需要5用户输入的值保存到一个数组，将其发送到的方法，并找到和显示的最低值这个基本的Java程序。

该程序是很简单的，它运行，但是当我进入最后一个号码，我得到的错误：

异常线程mainjava.lang.ArrayIndexOutOfBoundsException：4
        在minNumber.main（minNumber：14）

帮助？

 进口的java.util。*;类minNumber {
    公共静态无效的主要（字符串ARGS []）{    扫描仪输入=新的扫描仪（System.in）;    INT numberArray [] =新INT [4];
    INT findLowest;    的for（int i = 0; I＆LT; = numberArray.length;我++）{
        的System.out.println（插槽输入一个值+（I + 1）+：）;
        numberArray [I] = input.nextInt（）;
    }
    findLowest = getMin（numberArray）;
    displayOutput（findLowest）;
}公共静态INT getMin（INT NUM []）{INT lowestNum = 0;
对于（INT J = 0; J＆LT; = num.length; J ++）{
    如果（NUM [J]＆LT; NUM [J + 1]）{        lowestNum = NUM​​ [J]。
    }
}
返回lowestNum;
}公共静态无效displayOutput（INT最低）{的System.out.println（最小的数字是：+最低）;
}
}
 

解决方案

首先，如果你想在一个数组5个值，然后用5声明它：

  INT numberArray [] =新INT [5];
 

其次，你会关闭阵列的末尾。更改

 的for（int i = 0; I＆LT; = numberArray.length;我++）{
 

到

 的for（int i = 0; I＆LT; numberArray.length;我++）{
 

您将需要更改其他Ĵ 为循环这种方式也。

顺便说一句，你的 getMin 方法需要除了我上面提到的变再变，因为他说 NUM [J + 1] 仍将跑出数组的结尾，即使你对上面的变化。我想你会需要将当前数组元素与比较 lowestNum ，而不是一个数组元素。

I have been working on this basic java program when I need to store 5 user entered values into an array, send it to a method, and find and display the lowest value.

The program is simple enough, and it runs, but when I enter the last number, I get the error:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 4 at minNumber.main(minNumber:14)

Help?

import java.util.*;

class minNumber {


    public static void main(String args[]){

    Scanner input = new Scanner(System.in);

    int numberArray[] = new int[4];
    int findLowest;

    for (int i = 0; i <= numberArray.length; i++){
        System.out.println("Enter a value for slot "+(i+1)+ ":");
        numberArray[i] = input.nextInt();   
    }
    findLowest = getMin(numberArray);
    displayOutput(findLowest);
}

public static int getMin(int num[]){

int lowestNum = 0;
for (int j = 0; j <= num.length; j++){
    if (num[j] < num[j+1]){

        lowestNum = num[j];
    }
}
return lowestNum;
}

public static void displayOutput(int lowest){

System.out.println("The lowest number is: "+lowest);
}
}

解决方案

First, if you want 5 values in an array, then declare it with 5:

int numberArray[] = new int[5];

Second, you are going off the end of the array. Change

for (int i = 0; i <= numberArray.length; i++){

to

for (int i = 0; i < numberArray.length; i++){

You'll need to change your other j for loop this way also.

As an aside, your getMin method needs another change besides the change I mentioned above, because saying num[j+1] will still run off the end of the array even if you make the change above. I think you'll need to compare the current array element versus lowestNum, not the next array element.

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