Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
727 views
in Technique[技术] by (71.8m points)

rounding - how does java.math.RoundingMode work?

I'm having trouble with rounding. Specifically, after reading all the javadoc, I was expecting the following code:

int n = (integer between 0 and 9, included)
new BigDecimal(n + 0.555d).setScale(2, RoundingMode.HALF_UP).doubleValue()

to return n + 0.56. Instead, these are the return values for n from 0 to 4:

 new BigDecimal(0.555d).setScale(2, RoundingMode.HALF_UP).doubleValue()
 0.56
 new BigDecimal(1.555d).setScale(2, RoundingMode.HALF_UP).doubleValue()
 1.55
 new BigDecimal(2.555d).setScale(2, RoundingMode.HALF_UP).doubleValue()
 2.56
 new BigDecimal(3.555d).setScale(2, RoundingMode.HALF_UP).doubleValue()
 3.56
 new BigDecimal(4.555d).setScale(2, RoundingMode.HALF_UP).doubleValue()
 4.55

I have also tried to change the rounding mode:

int n = (integer between 0 and 9, included)
new BigDecimal(n + 0.555d).setScale(2, RoundingMode.HALF_DOWN).doubleValue()

expecting n + 0.55 as a result for each and every n. Instead, the return values are exactly the same as the previous example:

 new BigDecimal(0.555d).setScale(2, RoundingMode.HALF_DOWN).doubleValue()
 0.56
 new BigDecimal(1.555d).setScale(2, RoundingMode.HALF_DOWN).doubleValue()
 1.55
 new BigDecimal(2.555d).setScale(2, RoundingMode.HALF_DOWN).doubleValue()
 2.56
 new BigDecimal(3.555d).setScale(2, RoundingMode.HALF_DOWN).doubleValue()
 3.56
 new BigDecimal(4.555d).setScale(2, RoundingMode.HALF_DOWN).doubleValue()
 4.55

Am I missing something?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The problem you have is that double is not a precise representation and you are round based on this imprecise number.

BigDecimal bd = new BigDecimal(1.555d);
System.out.println("bd=" + bd);
bd = bd.setScale(2, RoundingMode.HALF_UP);
System.out.println("after rounding bd=" + bd);
double d = bd.doubleValue();
System.out.println("after rounding d=" + d);

prints

bd=1.5549999999999999378275106209912337362766265869140625
after rounding bd=1.55
after rounding d=1.55

however

BigDecimal bd = BigDecimal.valueOf(1.555d);
System.out.println("bd=" + bd);
bd = bd.setScale(2, RoundingMode.HALF_UP);
System.out.println("after rounding bd=" + bd);
double d = bd.doubleValue();
System.out.println("after rounding d=" + d);

prints

bd=1.555
after rounding bd=1.56
after rounding d=1.56

This works because BigDecimal.valueOf does some extra rounding based on how double would appear if you printed it.


However I wouldn't use BigDecimal unless performance/simplicity is not an issue.

double d = 1.555d;
System.out.println("d=" + d);
d = roundToTwoPlaces(d);
System.out.println("after rounding d=" + d);

public static double roundToTwoPlaces(double d) {
    return ((long) (d < 0 ? d * 100 - 0.5 : d * 100 + 0.5)) / 100.0;
}

prints

d=1.555
after rounding d=1.56

For more details Double your money again compares the performance of different ways of rounding.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...