Java JDK-从double到int的可能有损转换
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!
回答:
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
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