Java 8: DecimalFormat with HALF_UP rounding mode

Question

The following test fails:

@Test
public void testRound() throws Exception {
    double v = 0.015d;
    DecimalFormat format = new DecimalFormat("#.##", new DecimalFormatSymbols(Locale.US));
    format.setRoundingMode(RoundingMode.HALF_UP);
    assertEquals("0.02", format.format(v));
}

I see that there are resolved issues for Java 7 and Java 8, but it still doesn't work for me on "jdk1.8.0_144":

junit.framework.ComparisonFailure: null 
Expected :0.02
Actual   :0.01

0.015 cannot be exatly represented as a floating point number. It happens the nearest possible representation is slightly less and therefore is rounded down. — Henry, Aug 03 '17 at 09:16
Doubles and floats can't represent 0.015 exactly. Don't use double; instead use BigDecimal right from the start. — Klitos Kyriacou, Aug 03 '17 at 09:19
So, is it a normal behaviour? it's not a bug in Java 8, is it? — Liubov E., Aug 03 '17 at 09:25
Yes, quite normal and not a bug. It's the way floating point arithmetic works. — Henry, Aug 03 '17 at 09:38
However why will "double v = 5.5d;" be rounded as "6"? I.e. correctly? — Liubov E., Aug 03 '17 at 09:47
Because 5.5 **can** be exactly represented. There are also unrepresentable examples with nearest values slightly larger, they will also be rounded up. — Henry, Aug 03 '17 at 09:53

score 0 · Answer 1 · answered Aug 03 '17 at 09:14

0

Use this:

DecimalFormat format = new DecimalFormat("#.###", new DecimalFormatSymbols(Locale.US));

answered Aug 03 '17 at 09:14

Whitebear

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I need the correctly rounded format "#.##". – Liubov E. Aug 03 '17 at 09:20
It won't work with 2 deceimal points as you'll get 0.01 which round_up will round to 0.01 that's why it won't work for you correctly. – Whitebear Aug 03 '17 at 09:22

Java 8: DecimalFormat with HALF_UP rounding mode

1 Answers1