java - 数组中的 ParseInt

Question

import java.util.Scanner;

public class NumAverage {

    public static void main (String [] args)
    {
        //int a,b;
        int[] numbers = new int [10];

        Scanner numreader = new Scanner(System.in);

        try
        {
            System.out.println("Enter 10 num");
            for (int i = 0; i < numbers.length; i++)
                numbers[i] = numreader.nextInt();
                //numbers[i] = Integer.parseInt(s)
            //for (int i = 0; i < numbers.length; i++)
            //{
                //sum += numbers[i];
            //}
        }
        catch(NumberFormatException numfo)
        {
            System.out.println(numfo.getMessage() + "cannot be converted to integer");
        }
  }
}

如果用户尝试输入字符串而不是整数，如何获取从键盘输入到数组中的数字以进行 parseInt

Answer 1

尝试:

import java.util.Scanner;

public class NumAverage {

public static void main (String [] args)
{
    //int a,b;
    int[] numbers = new int [10];

    Scanner numreader = new Scanner(System.in);
    System.out.println("Enter 10 num");
    for (int i = 0; i < numbers.length; i++) {
        // Without the try and catch you get the appropriate exception,
        // NumberFormatException in this case
        String str = numreader.next();
        numbers[i] = Integer.parseInt(str);

        //try {
        //    numbers[i] = Integer.parseInt(numreader.next());
        //} catch(NumberFormatException e) {
        //    number[i] = 0; // or whatever you want
        //}
    }
  }
}