问题描述

我在显示平均小数时遇到问题.它一直显示 0.00 或 0.1.我试着把它加倍，但我仍然得到相同的结果.另外，我想将我输入的第一个整数包含在总和中，但我不知道如何.请帮助:

I have a problem in showing the decimals on the average. It keeps showing .00 or .1. I tried to put it in double, but I still get the same results.Also, I want to include the first integer that I input to the sum, but I have no idea how.Please help:

import java.io.*;
import java.util.*;

public class WhileSentinelSum 
{
static final int SENTINEL= -999;

public static void main(String[] args)
{
    // Keyboard Initialization
        Scanner kbin = new Scanner(System.in);

    //Variable Initialization and Declaration
        int number, counter;
        int sum =0;
        int average;

    //Input
        System.out.print("Enter an integer number: " );
        number = kbin.nextInt();
        counter=0;


    //Processing
    //Output
        while(number != SENTINEL)
        {
            counter++;
            System.out.print("Enter an  integer number: ");
            number = kbin.nextInt();
            sum += number;
        }

        if (counter !=0)
            System.out.println("
Counter is: " + counter);
        else
            System.out.println("no input");

        average= sum/counter;
        System.out.println("The sum is " + sum);
        System.out.println("The average is " + average);


        System.out.println();

    }//end main
}//end class

推荐答案

即使你声明 average 为 double，sum 和counter 变量是 int，所以 Java 仍然使用 整数除法 在分配给 average 之前，例如5/10 结果是 0 而不是 0.5.

Even if you declared average to be double, the sum and counter variables are int, so Java still uses integer division before assigning it to average, e.g. 5 / 10 results in 0 instead of 0.5.

在除法之前将变量之一转换为 double 以强制进行浮点运算:

Cast one of the variables to double before the division to force a floating point operation:

double average;

...
average = (double) sum / counter;

您可以通过在进入 while 循环之前对其进行处理来将第一个数字包含在计算中(如果它不是标记值).

You can include the first number in your calculation by processing it before you go into the while loop (if it's not the sentinel value).

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