The float and double types are designed primarily for scientific and engineering calculations. They perform binary floating-point arithmetic, which was carefully designed to furnish accurate approximations quickly over a broad range of magnitudes. They do not, however, provide exact results and should not be used where exact results are required. The float and double types are particularly ill-suited for monetary calculations because it is impossible to represent 0.1 (or any other negative power of ten) as a float or double exactly.
float 和 double 类型主要用于科学计算和工程计算。它们执行二进制浮点运算，该算法经过精心设计，能够在很大范围内快速提供精确的近似值。但是，它们不能提供准确的结果，也不应该在需要精确结果的地方使用。float 和 double 类型特别不适合进行货币计算，因为不可能将 0.1（或 10 的任意负次幂）精确地表示为 float 或 double。
For example, suppose you have $1.03 in your pocket, and you spend 42¢. How much money do you have left? Here’s a naive program fragment that attempts to answer this question:
例如，假设你口袋里有 1.03 美元，你消费了 42 美分。你还剩下多少钱？下面是一个简单的程序片段，试图回答这个问题：
System.out.println(1.03 - 0.42);Unfortunately, it prints out 0.6100000000000001. This is not an isolated case. Suppose you have a dollar in your pocket, and you buy nine washers priced at ten cents each. How much change do you get?
不幸的是，它输出了 0.6100000000000001。这不是一个特例。假设你口袋里有一美元，你买了 9 台洗衣机，每台 10 美分。你能得到多少零钱？
System.out.println(1.00 - 9 * 0.10);According to this program fragment, you get $0.09999999999999998.
根据这个程序片段，可以得到 0.0999999999999999998 美元。
You might think that the problem could be solved merely by rounding results prior to printing, but unfortunately this does not always work. For example, suppose you have a dollar in your pocket, and you see a shelf with a row of delicious candies priced at 10¢, 20¢, 30¢, and so forth, up to a dollar. You buy one of each candy, starting with the one that costs 10¢, until you can’t afford to buy the next candy on the shelf. How many candies do you buy, and how much change do you get? Here’s a naive program designed to solve this problem:
你可能认为，只需在打印之前将结果四舍五入就可以解决这个问题，但不幸的是，这种方法并不总是有效。例如，假设你口袋里有一美元，你看到一个架子上有一排好吃的糖果，它们的价格仅仅是 10 美分，20 美分，30 美分，以此类推，直到 1 美元。你每买一颗糖，从 10 美分的那颗开始，直到你买不起货架上的下一颗糖。你买了多少糖果，换了多少零钱？这里有一个简单的程序来解决这个问题：
// Broken - uses floating point for monetary calculation!
public static void main(String[] args) {
double funds = 1.00;
int itemsBought = 0;
for (double price = 0.10; funds >= price; price += 0.10) {
funds -= price;
itemsBought++;
}
System.out.println(itemsBought +"items bought.");
System.out.println("Change: $" + funds);
}If you run the program, you’ll find that you can afford three pieces of candy, and you have $0.3999999999999999 left. This is the wrong answer! The right way to solve this problem is to use BigDecimal, int, or long for monetary calculations.
如果你运行这个程序，你会发现你可以买得起三块糖，你还有 0.399999999999999999 美元。这是错误的答案！解决这个问题的正确方法是 使用 BigDecimal、int 或 long 进行货币计算。
Here’s a straightforward transformation of the previous program to use the BigDecimal type in place of double. Note that BigDecimal’s String constructor is used rather than its double constructor. This is required in order to avoid introducing inaccurate values into the computation [Bloch05, Puzzle 2]:
这里是前一个程序的一个简单改版，使用 BigDecimal 类型代替 double。注意，使用 BigDecimal 的 String 构造函数而不是它的 double 构造函数。这是为了避免在计算中引入不准确的值 [Bloch05, Puzzle 2]：
public static void main(String[] args) {
final BigDecimal TEN_CENTS = new BigDecimal(".10");
int itemsBought = 0;
BigDecimal funds = new BigDecimal("1.00");
for (BigDecimal price = TEN_CENTS;funds.compareTo(price) >= 0;price = price.add(TEN_CENTS)) {
funds = funds.subtract(price);
itemsBought++;
}
System.out.println(itemsBought +"items bought.");
System.out.println("Money left over: $" + funds);
}If you run the revised program, you’ll find that you can afford four pieces of candy, with $0.00 left over. This is the correct answer.
如果你运行修改后的程序，你会发现你可以买四颗糖，最终剩下 0 美元。这是正确答案。
There are, however, two disadvantages to using BigDecimal: it’s a lot less convenient than using a primitive arithmetic type, and it’s a lot slower. The latter disadvantage is irrelevant if you’re solving a single short problem, but the former may annoy you.
然而，使用 BigDecimal 有两个缺点：它与原始算术类型相比很不方便，而且速度要慢得多。如果你只解决一个简单的问题，后一种缺点是无关紧要的，但前者可能会让你烦恼。
An alternative to using BigDecimal is to use int or long, depending on the amounts involved, and to keep track of the decimal point yourself. In this example, the obvious approach is to do all computation in cents instead of dollars. Here’s a straightforward transformation that takes this approach:
除了使用 BigDecimal，另一种方法是使用 int 或 long，这取决于涉及的数值大小，还要自己处理十进制小数点。在这个例子中，最明显的方法是用美分而不是美元来计算。下面是一个采用这种方法的简单改版：
public static void main(String[] args) {
int itemsBought = 0;
int funds = 100;
for (int price = 10; funds >= price; price += 10) {
funds -= price;
itemsBought++;
}
System.out.println(itemsBought +"items bought.");
System.out.println("Cash left over: " + funds + " cents");
}In summary, don’t use float or double for any calculations that require an exact answer. Use BigDecimal if you want the system to keep track of the decimal point and you don’t mind the inconvenience and cost of not using a primitive type. Using BigDecimal has the added advantage that it gives you full control over rounding, letting you select from eight rounding modes whenever an operation that entails rounding is performed. This comes in handy if you’re performing business calculations with legally mandated rounding behavior. If performance is of the essence, you don’t mind keeping track of the decimal point yourself, and the quantities aren’t too big, use int or long. If the quantities don’t exceed nine decimal digits, you can use int; if they don’t exceed eighteen digits, you can use long. If the quantities might exceed eighteen digits, use BigDecimal.
总之，对于任何需要精确答案的计算，不要使用 float 或 double 类型。如果希望系统来处理十进制小数点，并且不介意不使用基本类型带来的不便和成本，请使用 BigDecimal。使用 BigDecimal 的另一个好处是，它可以完全控制舍入，当执行需要舍入的操作时，可以从八种舍入模式中进行选择。如果你使用合法的舍入行为执行业务计算，这将非常方便。如果性能是最重要的，那么你不介意自己处理十进制小数点，而且数值不是太大，可以使用 int 或 long。如果数值不超过 9 位小数，可以使用 int；如果不超过 18 位，可以使用 long。如果数量可能超过 18 位，则使用 BigDecimal。
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