Java JDK-从double到int的可能有损转换

Question

所以我最近写了下面的代码：

    import java.util.Scanner;

public class TrainTicket
{
      public static void main (String args[])
      {

         Scanner money = new Scanner(System.in);
         System.out.print("Please type in the type of ticket you would like to buy.\nA. Child B. Adult C. Elder.");
         String type = money.next();
         System.out.print("Now please type in the amount of tickets you would like to buy.");
         int much = money.nextInt();
         int price = 0;
         switch (type)
          {
            case "A":
            price = 10;
            break;
            case "B":
            price = 60;
            break;
            case "C":
            price = 35;
            break;
            default:
            price = 0;
            System.out.print("Not a option ;-;");
           }
          if (price!=0)
          {
            int total2 = price* much* 0.7;
            System.out.print("Do you have a coupon code? Enter Y or N");
            String YN = money.next();
            if (YN.equals("Y"))
            {
             System.out.print("Please enter your coupon code.");
             int coupon = money.nextInt();
             if(coupon==21)
             {
              System.out.println("Your total price is " + "$" + total2 + ".");
             }
             else
             {
              System.out.println("Invalid coupon code, your total price is " + "$" + price* much + ".");
             }
            }
            else
            {
            System.out.println("Your total price is " + "$" + price* much + "." ); 
            }
          }

       money.close();
      }
}

但是，它一直显示：

TrainTicket.java:31: error: incompatible types: possible lossy conversion from double to int
            int total2 = price* much* 0.7;

当我尝试使用cmd运行它时。

有人可以帮助并解释我所犯的错误吗？任何帮助表示赞赏：）。谢谢！

Answer 1

当您转换double到int，值的精度损失。例如，当您将4.8657（double）转换为int时，int值将为4.Primitive
int不存储十进制数字，因此您将丢失0.8657。

在您的情况下，0.7是一个双精度值（除非提到float-0.7f，否则浮点默认情况下被视为double）。计算时price*much*0.7，答案是一个双精度值，因此编译器不允许您将其存储在整数类型中，因为这可能会导致精度损失。也就是说possible lossy conversion，您可能会失去精度。

那你能做什么呢？您需要告诉编译器您确实想要这样做。您需要告诉编译器您知道自己在做什么。因此，使用以下代码将double显式转换为int：

int total2= (int) price*much*0.7;
 /*(int) tells compiler that you are aware      of what you are doing.*/
 //also called as type casting

在您的情况下，由于您正在计算成本，因此建议您将变量声明total2为double或float类型。

double total2=price*much*0.7;
 float total2=price*much*0.7;
 //will work