Question

我是一名初学Java程序员，编写一个程序来计算一组正整数的平均值。首先，它向用户查询要输入的整数量。然后，它从用户收集整数并输出计算的平均值。我在处理例外方面遇到了麻烦。当用户尝试输入要进行平均的负数时，会正确显示异常，但它不能正确地继续for循环以收集适当数量的数字。例如，下面是一个示例输出：

Please enter the number of integers to be averaged: 5
Enter a number: 1
Enter a number: 2
Enter a number: -3
NegativeIntegerException: N must be a positive integer.
Enter a number: 3
Enter a number: 4
The average is: 0.0

它正确抛出异常，但没有继续填充5个整数。只有其他4个有效。

以下是代码：

import java.util.*;

class NegativeIntegerException extends Exception {

    public NegativeIntegerException()
    {
        super("N must be a positive integer.");
    }
}


public class intAverage {

    public static void main(String[] args)
    {
        int N = 0; //number of integers to be averaged
        int[] numbers = null; //array to hold integers
        int sum = 0;
        int newInt;
        double average;
        Scanner keyboard = new Scanner(System.in);
        int x = 1; //for do-while loop #1
        int z = 1; //for do-while loop #2

        do {
        try {
            System.out.print("Please enter the number of integers to be averaged: ");
            N = keyboard.nextInt();
            numbers = new int[N]; //setting size of array
            x = 2;
        } 
        catch (Exception e) {
            System.out.println("N must be a positive integer");
        }
        } while (x == 1);

        do {
            for(int i = 0; i < N; i++) //collecting the integers
            {
                System.out.print("Enter a number: ");
                newInt = keyboard.nextInt();
                if (newInt < 0) {
                    try {
                        throw new NegativeIntegerException();
                    }
                    catch(NegativeIntegerException e) {
                        System.out.println(e);
                    }
                }
                newInt = numbers[i];
            }
            z = 2;
        } while (z == 1);


        for(int y = 0; y < N; y++) //calculate average
        {
            sum = sum + numbers[y];
        }
        average = sum / N;

        System.out.println("The average is: " + average);

    }

}

Answer 1

打印出异常时，从i中减1，然后继续。这基本上会重启迭代。

do {
    for(int i = 0; i < N; i++) //collecting the integers
    {
        System.out.print("Enter a number: ");
        newInt = keyboard.nextInt();
        if (newInt < 0) {
            try {
                throw new NegativeIntegerException();
            }
            catch(NegativeIntegerException e) {
                System.out.println(e);
                // Ignore this input
                i--;
                continue;
            }
        }
        numbers[i] = newInt;
    }
    z = 2;
} while (z == 1);

请注意，您不需要为此工作抛出异常 - 您可以轻松地说：

do {
    for(int i = 0; i < N; i++) //collecting the integers
    {
        System.out.print("Enter a number: ");
        newInt = keyboard.nextInt();
        if (newInt < 0) {
            System.out.println("NegativeIntegerException: N must be a positive integer");
            // Ignore this input
            i--;
            continue;
        }
        numbers[i] = newInt;
    }
    z = 2;
} while (z == 1);

这也更具可读性。

无法正确处理异常

1 个答案: