rounding decimal points

Question

Possible Duplicate:
Round a double to 2 significant figures after decimal point

I am trying to work with converting a decimal degree (23.1248) into a minutes style degree(23 7'29.3"). this is what I have so far:

   double a=23.1248;
   int deg=(int)a;//gives me the degree
   float b=(float) (a-deg);
   int min=(int) (b*60);//gives me the minutes
   double sec= (double) ((c*60)-min);//gives me my seconds

everything works fine, but I would like to round the seconds up to either the nearest tenth or hundrenth. I have looked at decimal formatting, but would prefer not to cast it to a string. I have also looked at bigdecimal but do not think that would be helpful,

I thought decimal points were already round. And minus signs, plus signs and division were all flat :) To "round up to 100's", just use "Math.round()": http://docs.oracle.com/javase/1.4.2/docs/api/java/lang/Math.html double x = round(y/100.0) * 100.0; — paulsm4, Jan 11 '12 at 19:14
BigDecimal works fine, you can create a round(number,decimals) function like the one in this answer. Try it, I don't see why it wouldn't work. http://stackoverflow.com/a/8911683/744859 — Jav_Rock, Jan 18 '12 at 15:46

score 16 · Accepted Answer · answered Jan 11 '12 at 19:21

Try using Math.round(double) on the number after scaling it up, then scaling it back down.

double x = 1.234;
double y = Math.round(x * 100.0) / 100.0; // => 1.23

You can also use BigDecimal if you want to get really heavyweight:

BigDecimal a = new BigDecimal("1.234");
BigDecimal b = a.setScale(2, RoundingMode.DOWN); // => BigDecimal("1.23")

score 0 · Answer 2 · answered Jan 11 '12 at 19:19

First off, there are library functions to do this, so why not just use those? See Math.round(). No need to reinvent the wheel. If you wanted to, though, you could try what follows. To round a double to the hundredth's place:

x = 0.01 * floor(x * 100.0)

To round a double to the tenth's place:

x = 0.1 * floor(x * 10.0)

To round a double to the 10^k place:

x = 10^k * floor(x / 10^k)

The implementation in any language - including Java - should be straightforward. A problem with this is that it doesn't really round, but truncates, to your position. To fix this, you can simply add 0.5 * 10^k to your number before rounding. If you just want to round up, use the versions above, and add 10^k before or after the computation.

Wouldn't you want to use round() instead of floor()? I'd expect 0.193 to be rounded to 0.2 if to 1dp. — Dan Hardiker, Jan 11 '12 at 19:22
@brainzzy I discuss how to achieve the same thing as round() and ceiling() using the floor() primitive for base-10 numbers. But essentially, yes, if he wants rounding to the nearest (rather than always up or always down), round() should work. — Patrick87, Jan 11 '12 at 19:24

rounding decimal points

2 Answers2