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java rounding problem

Posted on 2002-07-30
10
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Last Modified: 2012-05-04
i want to ask someone,
how to round the decimal like this:
from 10.48 to 10.45
from 289.41 to 289.40 ???
ASAP
0
Comment
Question by:kankan
  • 4
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10 Comments
 
LVL 2

Expert Comment

by:nir2002
ID: 7188036
Hi,

you mean 10.48 => 10.5
If need to round to the first digit after the floating point.

x (double poistive value or positive float)
x = Math.floor(x*10+0.5)/10.0;

for negtive use

x = Math.ceil(x*10-0.5)/10.0;

Hope it help
Best regards
Nir
0
 
LVL 35

Expert Comment

by:girionis
ID: 7188172
 This is a really weird rounding you are asking. Did you actualyl mean rounf the 10.48 to 10.50 instead? If yes below is a quick fix:

static BigDecimal format(BigDecimal n, int prec, int round)
{
    return n.setScale(prec, round);
}

public static void main(String[] args)
{
    BigDecimal bd = format(new BigDecimal("10.48"), 1, BigDecimal.ROUND_HALF_UP);
    System.out.println(format(bd, 2, BigDecimal.ROUND_UNNECESSARY));
     bd = format(new BigDecimal("289.41"), 1, BigDecimal.ROUND_HALF_DOWN);
         System.out.println(format(bd, 2, BigDecimal.ROUND_UNNECESSARY));
}

  If you indeed meant round the 10.48 to 10.45 I will have to put more thought in it. There is also a good tutorial on BigDecimal. It might come in handy:

http://www.javaworld.com/javaworld/jw-06-2001/jw-0601-cents.html

  Hope it helps.
0
 
LVL 35

Expert Comment

by:girionis
ID: 7188178
 The: new BigDecimal("10.48") can also be: new BigDecimal(10.48) as well.
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LVL 26

Expert Comment

by:ksivananth
ID: 7189733
Hope this helps,

    private final double roundValue(double val)
    {
           java.math.BigDecimal bdVal = new java.math.BigDecimal(""+val);

          bdVal = bdVal.setScale(2, java.math.BigDecimal.ROUND_HALF_UP);

          return bdVal.doubleValue();
    }
0
 
LVL 2

Expert Comment

by:nir2002
ID: 7189818
Hi,
 
public static double roundValue(double val,int floating_point_digit_to_round )  {
   double round_result;
   double factor = Math.pow(10.0,(double)floating_point_digit_to_round );
   if(val>0) {
      round_result = Math.floor(val*factor)+0.5)/factor;
   } else {
      round_result = Math.ceil(val*factor)+0.5)/factor;
   }
   return round_result;
}

general function to round floating point
for your case use floating_point_digit_to_round = 1
This solution is more effiective (performance) than the other since you don't need to create temporary objects.

Best regards
Nir
0
 

Author Comment

by:kankan
ID: 7190274
actually, no,
i just want
e.g. 14.48 it will be 14.45
round down to 5
or if 14.41 it will be 14.40
not rounding``
0
 

Author Comment

by:kankan
ID: 7190279
actually, no,
i just want
e.g. 14.48 it will be 14.45
round down to 5
or if 14.41 it will be 14.40
not rounding``
0
 
LVL 2

Expert Comment

by:nir2002
ID: 7190310
you want >0.45 => 0.45 and <0.45 =>0.40 ?
0
 

Author Comment

by:kankan
ID: 7192593
actually,

if 0.45 => 0.45
if 0.41 => 0.40
if 0.48 => 0.45
if 1.59 => 1.55
if 1.53 => 1.50
can u get it ??
ASAP
Thanks
0
 
LVL 2

Accepted Solution

by:
nir2002 earned 100 total points
ID: 7192738
Hi

it is uncommon rounding what you need.

public static double speicalRoundDown(double value,int digit) {
     int newDigit;
     int difference,absDifference;
     int digitToRound;
     int manipulateValue;
     double returnValue;
     manipulateValue = (int)(value*Math.pow(10.0,(double)digit));
     digitToRound = manipulateValue % 10;
     difference = digitToRound - 5;
     absDifference = Math.abs(difference);
     newDigit = (difference + absDifference) / 2 - digitToRound*(difference - absDifference)/2/Math.max(absDifference,1);
     manipulateValue = manipulateValue - newDigit;
     returnValue = ((double)manipulateValue)/Math.pow(10.0,(double)digit);
     return returnValue;          
}

that should do the trick.
for your case:
speicalRoundDown(10.47,2);
speicalRoundDown(289.41,2);

Best regards
Nir
0

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