对双精度值进行减法时未得到预期的结果

Question

我具有以下值，用作执行减法的测试数据。

  

订单总额= $ 17.99

     

项目小计= $ 17.99

     

折扣= 20

String orderTotal = "17.99"
String itemSubTotal = "17.99"
String discountPercent = "20"

double discountPercent = Double.parseDouble(discount);

// calculate the discount if given in percentage
double discountAmount = (Double.parseDouble(itemSubTotal) * discountPercent) / 100;

// round of the amount
double discountAmount = round(discountAmount, 2);

Double finalOrderTotal = (Double.parseDouble(orderTotal)) - discountAmount;

System.out.println("---------------------"+finalOrderTotal);

这是四舍五入值的方法：

public static double round(double value, int numberOfDigitsAfterDecimalPoint) 
{
    BigDecimal bigDecimal = new BigDecimal(value);
    bigDecimal = bigDecimal.setScale(numberOfDigitsAfterDecimalPoint,
            BigDecimal.ROUND_HALF_UP);
    return bigDecimal.doubleValue();
}

因此，手动计算的预期结果将是17.99 - 3.60 = 14.39 但是我得到的是14.389999999999999

我已经按照相同类型问题中的建议尝试过BigDecimal。但是为此，我必须再次进行finalOrderTotal的回合。是吗？

有人在计算时可以帮助我获取简单的代码（使用BigDecimal），以便获得正确的结果。

Answer 1

    String orderTotal = cart.getOrdertotal();
    String itemSubTotal = cart.getItemPrice();

    String orderTotal1 = orderTotal.replace(currency, "");
    String itemSubTotal1 = itemSubTotal.replace(currency, "");

    double discountPercent = Double.parseDouble(discount);

    // calculate the discount if given in percentage
    double discountAmount =
            (Double.parseDouble(itemSubTotal1) * discountPercent) / 100.0;

    // round of the amount
    double discountAmount = round(discountAmount, 2);

    Double finalOrderTotal = (Double.parseDouble(subTotal1)) - discountAmount;
    System.out.println("---------------------"+finalOrderTotal);