使用 % 找零时给出错误的数字

Question

我正在创建一个回馈变化的系统。当我运行程序时，我得到： （假设我给了 100 美元）

You have given 100.0 dollars.
Your change is 14.670000000000002 dollars.
You will recieve:
0.6700000000000017 dollars
0.1700000000000017 quarters
0.0700000000000017 dimes
0.020000000000001697 nickels
1.6965595595053173E-15 pennies

代码

import java.util.Scanner;
public class RunGivingChange {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        double given;
        double cost = 85.33;
        double change;
        int dollars;
        int quarters;
        int dimes;
        int nickels;
        int pennies;

        Scanner input = new Scanner(System.in);

        System.out.println("Your total is " + cost + " dollars.");
        System.out.println("How much money are you giving?");
        given = input.nextDouble();
        if (given >= cost)
        {
            System.out.println("You have given " + given + " dollars.");
            change = given - cost;
            System.out.println("Your change is " + change + " dollars.");
            System.out.println("You will recieve:");

            //dollars
            change = change % 1;
            System.out.println(change + " dollars");

            //quarters
            change = change % 0.25;
            System.out.println(change + " quarters");

            //dimes
            change = change % 0.10;
            System.out.println(change + " dimes");

            //nickels
            change = change % 0.05;
            System.out.println(change + " nickels");

            //pennies
            change = change % 0.01;
            System.out.println(change + " pennies");
        }
        else if (given < cost)
        {
            System.out.println("Not enough money.");
        }
        else
        {
            System.out.println("Error.");
        }

    }

    }

Answer 1

您正在使用双精度浮点值来表示货币。

不要那样做。

浮点数为您提供大约 15 个有效数字的准确度，但具有惊人的范围（和计算性能），使其可用于科学用途。

使用定点类型或 BigDecimal 作为货币值。

Answer 2

Modulo (%) 无法按预期工作（除了精度问题）：

2.67 % 1 不会导致2，而是0.67。

（更多信息请看这里：How do I use modulus for float/double?）

您可能想要计算Math.floor( 2.67 / 1) 等并从change 中减去四舍五入的值。

或者，如果change 始终为正（应该是），您可以将其强制转换（并因此削减）为整数：

int dollars = (int)(change / 1);
change -= dollars;

int quarters = (int)(change / 0.25);
change -= 0.25 * quarters;

...

要克服精确问题，请使用 BigDecimal 或将所有值转换为便士，即零钱 = 1467 便士、1 美元 = 100 便士等。

Answer 3

您可以用最小的单位（便士）来表示成本，而不是用美元表示：

double cost = 85.33;// dollars
you could express it as :
int cost = 8533: .. //pennies

然后您可以使用/ 和% 操作序列来计算每个货币除法。 例如：

int dollars = change / 100; // how many dollars fit in that pennies amount
change = change % 100; // how much pennies are left

//repeat for the other subdivisions

Answer 4

如果您使用双精度，请确保四舍五入到小数位。

change = (double) Math.round(change *100) / 100;

其次，mod (%) 返回语句的剩余部分

14.67 % 1.0 = 0.67

你首先要做的是找出 1.0 将分成多少次 14。

dollars = Math.floor(change / 1);
change = change % 1.0;

这将向下舍入到最接近的整数，并告诉您 1.0 可以进行多少次更改。

这里使用 double 的问题就像 Bathsheba 所说的那样。 要解决此问题，您必须在每次修改更改后进行舍入。

这应该会给你正确的输出。

import java.util.Scanner;

public class RunGivingChange{


    public static void main(String[] args) {


        double given;
        double cost = 85.33;
        double change;
        int dollars;
        int quarters;
        int dimes;
        int nickels;
        int pennies;

        Scanner input = new Scanner(System.in);

        System.out.println("Your total is " + cost + " dollars.");
        System.out.println("How much money are you giving?");
        given = input.nextDouble();
        if (given >= cost) {
            System.out.println("You have given " + given + " dollars.");
            change = given - cost;
            // round the change
            change = (double) Math.round(change *100) / 100;
            System.out.println("Your change is " + change + " dollars.");
            System.out.println("You will recieve:");

            //dollars     

            System.out.println(Math.floor(change / 1) + " dollars");
            change = change % 1;
            change = (double) Math.round(change *100) / 100;
            //quarters

            System.out.println(Math.floor(change / 0.25) + " quarters");
            change = change % 0.25;
            change = (double) Math.round(change *100) / 100;
            //dimes

            System.out.println(Math.floor(change / 0.1) + " dimes");
            change = change % 0.10;
            change = (double) Math.round(change *100) / 100;
            //nickels

            System.out.println(Math.floor(change / 0.05) + " nickels");
            change = change % 0.05;
            change = (double) Math.round(change *100) / 100;
            //pennies

            System.out.println(Math.floor(change / 0.01) + " pennies");

        } else if (given < cost) {
            System.out.println("Not enough money.");
        } else {
            System.out.println("Error.");
        }

    }

}

有更好的方法可以做到这一点，下次您应该搜索该网站。这个话题有多个答案。

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