为什么Java的RoundingMode HALF_DOWN在1.245到1.25之间？

Question

BigDecimal bd1 = new BigDecimal(1.245);

System.out.println(">> Half up: "
                + bd1.setScale(2, BigDecimal.ROUND_HALF_UP));
System.out.println(">> Half down: "
                + bd1.setScale(2, BigDecimal.ROUND_HALF_DOWN));

Results:

Half up: 1.25

Half down: 1.25

Answer 1

It's the result of the initial double in the constructor. If you do

BigDecimal bd1 = new BigDecimal("1.245");

(ie with quotes) it'll work as expected.

The double value 1.245 cannot accurately be represented, it is actually 1.24500000000000010658141036401502788066864013671875 , hence it is too large for actual rounding "down".

Answer 2

if you print original value, you will see:

1.24500000000000010658141036401502788066864013671875

this is because 1.245 cannot be represented as double number

now read documentation about setScale ROUND_HALF_DOWN

Rounding mode to round towards "nearest neighbor" unless both neighbors are equidistant, in which case round down. Behaves as for ROUND_UP if the discarded fraction is > 0.5 ; otherwise, behaves as for ROUND_DOWN.

as you see discarded fraction is > 0.5, not equal as you expected

if you declare your initial variable as BigDecimal bd1 = new BigDecimal("1.245"); you will get expected result

Answer 3

我认为这是因为你使用了一个带有float的构造函数，它将你的值转换为1.2456345 ，试试这个：

BigDecimal bd1 = new BigDecimal("1.245");

Answer 4

Problem comes from float. You shoud know that using float and double, you will never get precise value. Like 0.1 is not equal to 1/10.

If you are interested in high precision you must use BigDecimal, but unitialize it with a string value, so that you won't loose presicion due to float casting.

Change your code to this:

BigDecimal bd1 = new BigDecimal("1.245");

Answer 5

http://docs.oracle.com/javase/7/docs/api/java/math/BigDecimal.html

You can see that "Note: For values other than float and double NaN and ±Infinity, this constructor is compatible with the values returned by Float.toString(float) and Double.toString(double). This is generally the preferred way to convert a float or double into a BigDecimal, as it doesn't suffer from the unpredictability of the BigDecimal(double) constructor."

So, if you want to set up a BigDecimal by a float number, you should

BigDecimal bd1 = new BigDecimal(Float.toString(1.245f));

You may also check the note of the constructor public BigDecimal(double val)

1. 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.

2. 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.

3. When a double must be used as a source for a BigDecimal, note that this constructor provides an exact conversion; it does not give the same result as converting the double to a String using the Double.toString(double) method and then using the BigDecimal(String) constructor. To get that result, use the static valueOf(double) method.