我将冒着被否决的风险，但我认为浮点数在货币计算中的不适用性被高估了。只要确保正确地进行了舍入，并且有足够的有效数字来处理zneak解释的二进制十进制表示不匹配，就不会有问题。

在Excel中使用货币计算的人总是使用双精度浮点数(Excel中没有货币类型)，我还没有看到有人抱怨舍入错误。

当然，你必须在合理范围内;例如，一个简单的网络商店可能永远不会遇到双精度浮点数的任何问题，但如果你做会计或其他需要添加大量(无限制)数字的事情，你不会想要用十英尺的杆子触摸浮点数。

因为浮点数和双精度数不能准确地表示我们用来表示金钱的以10为底的倍数。这个问题不仅适用于Java，还适用于任何使用2进制浮点类型的编程语言。

以10为基数，可以将10.25写成1025 * 10-2(整数乘以10的幂)。IEEE-754浮点数是不同的，但是考虑它们的一个非常简单的方法是乘以2的幂。例如，您可以看到164 * 2-4(整数乘以2的幂)，也等于10.25。这不是数字在内存中的表示方式，但数学含义是相同的。

即使以10为基数，这个符号也不能准确地表示大多数简单的分数。例如，你不能表示1/3:十进制表示是重复的(0.3333…)，所以没有一个有限整数可以乘以10的幂得到1/3。你可以设定一个长序列的3和一个小指数，如333333333 * 10-10，但它是不准确的:如果你乘以3，你不会得到1。

然而，为了数钱，至少对于那些货币价值在美元数量级内的国家，通常你所需要的只是能够存储10-2的倍数，所以1/3不能表示并没有什么关系。

The problem with floats and doubles is that the vast majority of money-like numbers don't have an exact representation as an integer times a power of 2. In fact, the only multiples of 0.01 between 0 and 1 (which are significant when dealing with money because they're integer cents) that can be represented exactly as an IEEE-754 binary floating-point number are 0, 0.25, 0.5, 0.75 and 1. All the others are off by a small amount. As an analogy to the 0.333333 example, if you take the floating-point value for 0.01 and you multiply it by 10, you won't get 0.1. Instead you will get something like 0.099999999786...

把钱表示成双位数或浮点数一开始可能看起来不错，因为软件会消除微小的错误，但当你对不精确的数字进行更多的加减乘除运算时，错误就会加剧，最终你会得到明显不准确的数值。这使得浮点数和双精度数不适用于处理货币，因为货币需要精确计算以10为底数的倍数。

一种适用于任何语言的解决方案是使用整数，并计算美分。例如，1025就是10.25美元。一些语言也有内置的类型来处理钱。其中，Java有BigDecimal类，Rust有rust_decimal板条箱，c#有decimal类型。

The result of floating point number is not exact, which makes them unsuitable for any financial calculation which requires exact result and not approximation. float and double are designed for engineering and scientific calculation and many times doesn’t produce exact result also result of floating point calculation may vary from JVM to JVM. Look at below example of BigDecimal and double primitive which is used to represent money value, its quite clear that floating point calculation may not be exact and one should use BigDecimal for financial calculations.

    // floating point calculation
    final double amount1 = 2.0;
    final double amount2 = 1.1;
    System.out.println("difference between 2.0 and 1.1 using double is: " + (amount1 - amount2));

    // Use BigDecimal for financial calculation
    final BigDecimal amount3 = new BigDecimal("2.0");
    final BigDecimal amount4 = new BigDecimal("1.1");
    System.out.println("difference between 2.0 and 1.1 using BigDecimal is: " + (amount3.subtract(amount4)));

输出:

difference between 2.0 and 1.1 using double is: 0.8999999999999999
difference between 2.0 and 1.1 using BigDecimal is: 0.9

这不是精确与否的问题，也不是精确与否的问题。这是一个满足以10为底而不是以2为底计算的人的期望的问题。例如，在财务计算中使用双精度值不会产生数学意义上的“错误”答案，但它可以产生财务意义上不期望的答案。

即使您在输出前的最后一分钟舍入结果，您仍然可以偶尔使用与期望不匹配的双精度结果。

Using a calculator, or calculating results by hand, 1.40 * 165 = 231 exactly. However, internally using doubles, on my compiler / operating system environment, it is stored as a binary number close to 230.99999... so if you truncate the number, you get 230 instead of 231. You may reason that rounding instead of truncating would have given the desired result of 231. That is true, but rounding always involves truncation. Whatever rounding technique you use, there are still boundary conditions like this one that will round down when you expect it to round up. They are rare enough that they often will not be found through casual testing or observation. You may have to write some code to search for examples that illustrate outcomes that do not behave as expected.

Assume you want to round something to the nearest penny. So you take your final result, multiply by 100, add 0.5, truncate, then divide the result by 100 to get back to pennies. If the internal number you stored was 3.46499999.... instead of 3.465, you are going to get 3.46 instead 3.47 when you round the number to the nearest penny. But your base 10 calculations may have indicated that the answer should be 3.465 exactly, which clearly should round up to 3.47, not down to 3.46. These kinds of things happen occasionally in real life when you use doubles for financial calculations. It is rare, so it often goes unnoticed as an issue, but it happens.

如果您使用以10为基数进行内部计算，而不是使用双数，则如果您的代码中没有其他错误，那么结果总是完全符合人类的预期。

Float is binary form of Decimal with different design; they are two different things. There are little errors between two types when converted to each other. Also, float is designed to represent infinite large number of values for scientific. That means it is designed to lost precision to extreme small and extreme large number with that fixed number of bytes. Decimal can't represent infinite number of values, it bounds to just that number of decimal digits. So Float and Decimal are for different purpose.

有一些方法可以管理货币值的错误:

使用长整数，以分计算。 使用双精度，保持你的有效数字为15，这样小数可以精确模拟。在显示值之前舍入;做计算时经常四舍五入。 使用像Java BigDecimal这样的十进制库，这样就不需要使用double来模拟十进制。

附注:有趣的是，大多数品牌的手持科学计算器工作在十进制而不是浮点数。所以没有人抱怨浮点数转换错误。

虽然浮点类型确实只能表示近似的十进制数据，但如果在表示数字之前将数字舍入到必要的精度，则可以获得正确的结果。通常。

通常是因为双排精度小于16位。如果你要求更高的精度，这不是一个合适的类型。近似也可以累积。

必须指出的是，即使您使用定点算术，您仍然必须对数字进行四舍五入，如果不是因为BigInteger和BigDecimal在获得周期性小数时会给出错误。所以这里也有一个近似。

例如，历史上用于财务计算的COBOL的最大精度为18位数字。所以通常会有一个隐含的舍入。

总之，在我看来，双精度主要不适合它的16位精度，这可能是不够的，而不是因为它是近似值。

考虑以下后续程序的输出。它表明，在舍入double后，得到与BigDecimal相同的结果，精度为16。

Precision 14
------------------------------------------------------
BigDecimalNoRound             : 56789.012345 / 1111111111 = Non-terminating decimal expansion; no exact representable decimal result.
DoubleNoRound                 : 56789.012345 / 1111111111 = 5.111011111561101E-5
BigDecimal                    : 56789.012345 / 1111111111 = 0.000051110111115611
Double                        : 56789.012345 / 1111111111 = 0.000051110111115611

Precision 15
------------------------------------------------------
BigDecimalNoRound             : 56789.012345 / 1111111111 = Non-terminating decimal expansion; no exact representable decimal result.
DoubleNoRound                 : 56789.012345 / 1111111111 = 5.111011111561101E-5
BigDecimal                    : 56789.012345 / 1111111111 = 0.0000511101111156110
Double                        : 56789.012345 / 1111111111 = 0.0000511101111156110

Precision 16
------------------------------------------------------
BigDecimalNoRound             : 56789.012345 / 1111111111 = Non-terminating decimal expansion; no exact representable decimal result.
DoubleNoRound                 : 56789.012345 / 1111111111 = 5.111011111561101E-5
BigDecimal                    : 56789.012345 / 1111111111 = 0.00005111011111561101
Double                        : 56789.012345 / 1111111111 = 0.00005111011111561101

Precision 17
------------------------------------------------------
BigDecimalNoRound             : 56789.012345 / 1111111111 = Non-terminating decimal expansion; no exact representable decimal result.
DoubleNoRound                 : 56789.012345 / 1111111111 = 5.111011111561101E-5
BigDecimal                    : 56789.012345 / 1111111111 = 0.000051110111115611011
Double                        : 56789.012345 / 1111111111 = 0.000051110111115611013

Precision 18
------------------------------------------------------
BigDecimalNoRound             : 56789.012345 / 1111111111 = Non-terminating decimal expansion; no exact representable decimal result.
DoubleNoRound                 : 56789.012345 / 1111111111 = 5.111011111561101E-5
BigDecimal                    : 56789.012345 / 1111111111 = 0.0000511101111156110111
Double                        : 56789.012345 / 1111111111 = 0.0000511101111156110125

Precision 19
------------------------------------------------------
BigDecimalNoRound             : 56789.012345 / 1111111111 = Non-terminating decimal expansion; no exact representable decimal result.
DoubleNoRound                 : 56789.012345 / 1111111111 = 5.111011111561101E-5
BigDecimal                    : 56789.012345 / 1111111111 = 0.00005111011111561101111
Double                        : 56789.012345 / 1111111111 = 0.00005111011111561101252

import java.lang.reflect.InvocationTargetException;
import java.lang.reflect.Method;
import java.math.BigDecimal;
import java.math.MathContext;

public class Exercise {
    public static void main(String[] args) throws IllegalArgumentException,
            SecurityException, IllegalAccessException,
            InvocationTargetException, NoSuchMethodException {
        String amount = "56789.012345";
        String quantity = "1111111111";
        int [] precisions = new int [] {14, 15, 16, 17, 18, 19};
        for (int i = 0; i < precisions.length; i++) {
            int precision = precisions[i];
            System.out.println(String.format("Precision %d", precision));
            System.out.println("------------------------------------------------------");
            execute("BigDecimalNoRound", amount, quantity, precision);
            execute("DoubleNoRound", amount, quantity, precision);
            execute("BigDecimal", amount, quantity, precision);
            execute("Double", amount, quantity, precision);
            System.out.println();
        }
    }

    private static void execute(String test, String amount, String quantity,
            int precision) throws IllegalArgumentException, SecurityException,
            IllegalAccessException, InvocationTargetException,
            NoSuchMethodException {
        Method impl = Exercise.class.getMethod("divideUsing" + test, String.class,
                String.class, int.class);
        String price;
        try {
            price = (String) impl.invoke(null, amount, quantity, precision);
        } catch (InvocationTargetException e) {
            price = e.getTargetException().getMessage();
        }
        System.out.println(String.format("%-30s: %s / %s = %s", test, amount,
                quantity, price));
    }

    public static String divideUsingDoubleNoRound(String amount,
            String quantity, int precision) {
        // acceptance
        double amount0 = Double.parseDouble(amount);
        double quantity0 = Double.parseDouble(quantity);

        //calculation
        double price0 = amount0 / quantity0;

        // presentation
        String price = Double.toString(price0);
        return price;
    }

    public static String divideUsingDouble(String amount, String quantity,
            int precision) {
        // acceptance
        double amount0 = Double.parseDouble(amount);
        double quantity0 = Double.parseDouble(quantity);

        //calculation
        double price0 = amount0 / quantity0;

        // presentation
        MathContext precision0 = new MathContext(precision);
        String price = new BigDecimal(price0, precision0)
                .toString();
        return price;
    }

    public static String divideUsingBigDecimal(String amount, String quantity,
            int precision) {
        // acceptance
        BigDecimal amount0 = new BigDecimal(amount);
        BigDecimal quantity0 = new BigDecimal(quantity);
        MathContext precision0 = new MathContext(precision);

        //calculation
        BigDecimal price0 = amount0.divide(quantity0, precision0);

        // presentation
        String price = price0.toString();
        return price;
    }

    public static String divideUsingBigDecimalNoRound(String amount, String quantity,
            int precision) {
        // acceptance
        BigDecimal amount0 = new BigDecimal(amount);
        BigDecimal quantity0 = new BigDecimal(quantity);

        //calculation
        BigDecimal price0 = amount0.divide(quantity0);

        // presentation
        String price = price0.toString();
        return price;
    }
}