Question

我试图解决以下简单的编程练习：＆＃34;编写一个程序，读取1到10之间的整数，并计算每个的出现次数。假设输入以0＆＃34;结束。我提出了以下解决方案。我不熟悉调试器并且在过去3个小时内尝试跟踪ArrayIndexOutOfBoundsException。也许有人能够看到ArrayIndexOutOfBoundsException发生的地方？

import java.util.Scanner;

public class Exercise07_03 {
    /** Main method */
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        int[] numbers = new int[10];
        System.out.print("Enter up to 10 integers between 1 and 10" + 
            "inclusive (input ends with zero): ");

        int i = 0;
        do {
            numbers[i] = input.nextInt();
            i++;
        } while (numbers[i] != 0 && i != 9);

        displayCounts(countNumbers(numbers));
    }

    /** Count the occurrences of each number */
    public static int[] countNumbers(int[] numbers) {
        int[] counts = new int[10];

        for (int i = 0; i < counts.length; i++)
            counts[numbers[i] - 1]++;

        return counts;
    }

    /** Display counts */
    public static void displayCounts(int[] counts) {
        for (int i = 1; i < counts.length + 1; i++)
            if (counts[i - 1] != 0)
                System.out.println(i + " occurs " + counts[i - 1] + " time");
    }
}

Answer 1

         counts[numbers[i] - 1]++;

如果输入包含0，

结果为-1。

Answer 2

问题在于：

public static int[] countNumbers(int[] numbers) {
    int[] counts = new int[10];

    for (int i = 0; i < counts.length; i++)
        counts[numbers[i] - 1]++;

    return counts;
}

解决方案是：

 counts[(number - 1 > 0) ? number - 1 : 0]++;

检查（number-1）是否大于零。如果不是，则使用0.如果为true，则使用数字-1。

无法跟踪java.lang.ArrayIndexOutOfBoundsException：

2 个答案: