Question

我在使用此代码时遇到了困难。当我输入95.95作为购买价格和100作为支付价格时，我不断得到一个舍入错误，例如它返回4美元，4便士而不是4美元和5便士。所有其他值都可以正常工作，但由于某种原因，当小数位为.95或.05时，它很难返回正确的便士......

public static void main(String[] args){

        //Read price per pack
        Scanner in = new Scanner(System.in);

        System.out.println("Enter purchase price: ");
        double purchasePrice = in.nextDouble();
        System.out.println("Enter amount paid: ");
        double amountPaid = in.nextDouble();

        //Compute change due

        if (amountPaid < purchasePrice)
        {System.out.printf("Error, you need more money!");
        }

        else {
        int diff = (int)(( amountPaid - purchasePrice)*100);


        int dollars = diff/100;
            diff = diff%100;

        int quarters = diff/25;
            diff = diff%25;

        int dimes = diff/10;
            diff = diff%10;

        int nickels = diff/5;
            diff = diff%5;

        int pennies = diff/1;
            diff = diff%1;

        System.out.println("--------------------");    
        System.out.println("Total Change due: " + (amountPaid - purchasePrice));
        System.out.println("--------------------");
        System.out.println("Dollars: " + dollars);
        System.out.println("Quarters: " + quarters);
        System.out.println("Dimes: " + dimes);
        System.out.println("Nickels: " + nickels);
        System.out.println("Pennies: " + pennies);

        }
    }
}

Answer 1

你的问题就在这一行

int diff = (int)(( amountPaid - purchasePrice)*100);

您将向下的数量向下舍入到下一个int 以下价格的100倍。由于double并不一定要精确地表示十进制数，因此可能存在浮点错误，当您进行乘法时，该错误会被放大100倍。该产品可能略高于或低于整数分数，因此您的舍入可以导致金额向下舍入，几乎整整一分。

您实际想要做的是将分数的double值舍入最接近的 int，而不是下面的int。您可以在回合前添加0.5来做到这一点。

int diff = (int)(( amountPaid - purchasePrice)*100 + 0.5);

一般来说，如果你想用它做精确的计算，你绝不应该使用double存储一笔钱。我建议学习使用BigDecimal类，其中像这样的问题就会消失。

java收银机的舍入错误

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