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。

为了补充前面的答案，在处理问题中解决的问题时，除了BigDecimal之外，还可以选择在Java中实现Joda-Money。Java模块名称为org.joda.money。

它需要Java SE 8或更高版本，并且没有依赖关系。

更准确地说，存在编译时依赖关系，但它不是 必需的。

<dependency>
  <groupId>org.joda</groupId>
  <artifactId>joda-money</artifactId>
  <version>1.0.1</version>
</dependency>

使用Joda Money的例子:

  // create a monetary value
  Money money = Money.parse("USD 23.87");
  
  // add another amount with safe double conversion
  CurrencyUnit usd = CurrencyUnit.of("USD");
  money = money.plus(Money.of(usd, 12.43d));
  
  // subtracts an amount in dollars
  money = money.minusMajor(2);
  
  // multiplies by 3.5 with rounding
  money = money.multipliedBy(3.5d, RoundingMode.DOWN);
  
  // compare two amounts
  boolean bigAmount = money.isGreaterThan(dailyWage);
  
  // convert to GBP using a supplied rate
  BigDecimal conversionRate = ...;  // obtained from code outside Joda-Money
  Money moneyGBP = money.convertedTo(CurrencyUnit.GBP, conversionRate, RoundingMode.HALF_UP);
  
  // use a BigMoney for more complex calculations where scale matters
  BigMoney moneyCalc = money.toBigMoney();

文档: http://joda-money.sourceforge.net/apidocs/org/joda/money/Money.html 实现示例: https://www.programcreek.com/java-api-examples/?api=org.joda.money.Money

美国货币可以很容易地用美元和美分来表示。整数是100%精确的，而浮点二进制数并不完全匹配浮点小数。

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。

摘自Bloch, J.， Effective Java，(第二版，第48项。第3版，项目60):

float和double类型是 尤其不适用于货币 因为这是不可能的 表示0.1(或任何其他。 10的负次方)作为浮点数或 完全的两倍。 例如，假设您有1.03美元 你花了42c。多少钱? 你走了? System.out.println(1.03 - .42); 输出0.6100000000000001。 解决这个问题的正确方法是 使用BigDecimal, int或long 用于货币计算。

虽然BigDecimal有一些警告(请参阅当前接受的答案)。

如果你的计算涉及到不同的步骤，任意的精度算法都不能100%覆盖你。

使用完美的结果表示(使用自定义Fraction数据类型，将除法操作批处理到最后一步)并且仅在最后一步转换为十进制的唯一可靠方法。

任意精度不会有帮助，因为总有可能有很多小数点后的数字，或者一些结果，如0.6666666……最后一个例子没有任意的表示法。所以每一步都会有小误差。

这些错误会累积起来，最终可能变得不再容易被忽视。这被称为错误传播。