Java 中的 BigDecimal 行为

Question

public class Test {
    public static void main(String[] args){
         System.out.println(new BigDecimal(58.34));
    }
}

如果我运行在 Java 中给定的程序上面，它给我的输出如下： 58.340000000000003410605131648480892181396484375

这是为什么？

Answer 1

在构造 BigDecimal 时传递了 double，因此精度已经丢失。最简单的修复可能是

System.out.println(new BigDecimal("58.34"));

输出是

58.34

Answer 2

这将导致由于 double 值

 System.out.println(new BigDecimal(58.34));
                                    ^^^

要避免这种情况，您可以使用字符串值

 System.out.println(new BigDecimal("58.34"));

Answer 3

我认为 BigDecimal.java 最能说明此功能。

API 说的

Translates a double into a BigDecimal which is the exact decimal representation of the double's binary floating-point value. The scale of the returned BigDecimal is the smallest value such that (10scale × val) is an integer.

备注：

The results of this constructor can be somewhat unpredictable. One might assume that writing new BigDecimal(0.1) in Java creates a BigDecimal which is exactly equal to 0.1 (an unscaled value of 1, with a scale of 1), but it is actually equal to 0.1000000000000000055511151231257827021181583404541015625. This is because 0.1 cannot be represented exactly as a double (or, for that matter, as a binary fraction of any finite length). Thus, the value that is being passed in to the constructor is not exactly equal to 0.1, appearances notwithstanding.

The String constructor, on the other hand, is perfectly predictable: writing new BigDecimal("0.1") creates a BigDecimal which is exactly equal to 0.1, as one would expect. Therefore, it is generally recommended that the String constructor be used in preference to this one.

When a double must be used as a source for a BigDecimal, note that this constructor provides an exact conversion

Java 中的 BigDecimal 行为

BigDecimal behavior in Java

java

bigdecimal