无论num1，num2或num3输入的总分总是等于0.0

Question

此报价有何错误？它总是输出0.0，而不是总比分结果

import java.util.Scanner; 

public class d { 

    public static void main(String[] args) { 

     Scanner input = new Scanner(System.in); 

     int num1; 
     int num2; 
     int num3; 
     int percentnum1; 
     int percentnum2; 
     int percentnum3; 
     double totalscore; 

     System.out.print("Enter paper 1 score (out of 90): "); 
     num1 = input.nextInt(); 

     System.out.print("Enter paper 2 score (out of 90): "); 
     num2 = input.nextInt(); 

     System.out.print("Enter IA score (out of 20): "); 
     num3 = input.nextInt(); 

     percentnum1 = (num1/90) * 100; 
     percentnum2 = (num2/90) * 100; 
     percentnum3 = (num3/20) * 80; 

     totalscore = (percentnum1 * 0.4) + (percentnum2 * 0.4) + (percentnum3 * 0.2); 

     System.out.println("The total score is " + totalscore); 


    } 
} 

Answer 1

好像下面的代码引起的问题：如果的num1，num2和num3所有都值低于其分母

percentnum1 = (num1/90) * 100; 
percentnum2 = (num2/90) * 100; 
percentnum3 = (num3/20) * 80; 

结果（totalscore）将为0.0，因为整数除法，即使某些变量是90/90或20/20而其他变量小于分母，您w应该再次得到不正确的结果，并再次由于整数除法（虽然这次不一定是0.0）。

解决方法是简单地使用double类型来保持精度。

double num1; 
double num2; 
double num3 
double percentnum1; 
double percentnum2; 
double percentnum3; 
double totalscore; 

Answer 2

变化int到doubleto avoid rounding down to 0。

Answer 3

如果您喜欢使用int类型，则可以先将数字相乘。希望它有帮助。

percentnum1 = num1 * 100/90; 
percentnum2 = num2 * 90/100; 
percentnum3 = num3 * 80/20;