Java IO "NaN"问题

Question

我有一个名为 Numbers.txt 的文件，其中包含以下内容:

8.5

83.45, 90.2

120.00, 11.05

190.00

我编写了使用文件内容来计算文件中数字的总和和平均值的代码，但是当我运行代码时，平均值结果为“NaN”

代码:

package lab13;

import java.util.Scanner; 

import java.io.*;

public class problem1 {

    public static void main(String[] args) throws IOException
    {   

        double sum = 0;     
        int count = 0;
        double num,total = 0;
        double average = total/count;


        File file = new File("Numbers.txt");
        Scanner scan = new Scanner(file);


        while (scan.hasNext())
        {
            double number = scan.nextDouble();

            sum = sum + number;
        }  

        while (scan.hasNextDouble())
        {
            num = scan.nextDouble();
            System.out.println(num);
            count++;
            total += num;

        }

        scan.close();

        System.out.println("The sum of the numbers in " +
                "Numbers.txt is " + sum );       

        System.out.println("The average of the numbers in " +
                "Numbers.txt is " + average );  


    }
}

输出:

Numbers.txt 中的数字总和为 503.2

Numbers.txt 中数字的平均值为 NaN

Answer 1

你需要做

double average = total/count;

之后您将获得total和count的值

但是还要注意

当 while (scan.hasNext()) 流耗尽时，while (scan.hasNextDouble()) 也将耗尽

这个问题可以克服，但只需循环一次