意外类型平均值

Question

由于某种原因,命令提示符继续询问我输入,但我在第12行(sum + one)= sum 上放置一个值.如果你们可以帮我确定什么是错误的,这将是惊人的.

import java.util.Scanner;

public class FunnyAverage {
   public static void main(String[] args){          
   Scanner in = new Scanner(System.in);
   System.out.print("How many values to read? ");
   int top = in.nextInt();
   System.out.print("Enter Value: ");
   int one = in.nextInt();
   int number = 0;
   int sum = 0;
   (sum + one) = sum;

   while (number>top){
       while (one % 6 != 0&&one % 17 != 0) {
           System.out.print("Enter Value: ");
           one = in.nextInt(); 
           number++;
       }
   }

   if (sum/top != 0){
       System.out.print("Average: " + sum/top);
   }
   System.out.print("None Divisible");
   }
}

Answer 1

你的意思是:

sum = sum + one; // or sum += one;

通过命令提示符,我认为您实际上是指编译器(可以将其错误消息写入命令提示符).错误消息将声明结果(sum + one)不是变量.

见15.26节.Java语言规范的赋值运算符,声明:

赋值运算符的第一个操作数的结果必须是变量,否则会发生编译时错误.