round BigDecimal to nearest 5 cents

Question

Hi,

I'm trying to figure out how to round a monetary amount upwards to the nearest 5 cents. The following shows my expected results

1.03     => 1.05
1.051    => 1.10
1.05     => 1.05
1.900001 => 1.10

I need the result to be have a precision of 2 (as shown above).

Update

Following the advice below, the best I could do is this

    BigDecimal amount = new BigDecimal(990.49)

    // To round to the nearest .05, multiply by 20, round to the nearest integer, then divide by 20
   def result =  new BigDecimal(Math.ceil(amount.doubleValue() * 20) / 20)
   result.setScale(2, RoundingMode.HALF_UP)

I'm not convinced this is 100% kosher - I'm concerned precision could be lost when converting to and from doubles. However, it's the best I've come up with so far and seems to work.

Thanks, Don

Answer 1

I'd try multiplying by 20, rounding to the nearest integer, then dividing by 20. It's a hack, but should get you the right answer.

Answer 2

Tom has the right idea, but you need to use BigDecimal methods, since you ostensibly are using BigDecimal because your values are not amenable to a primitive datatype. Something like:

BigDecimal num = new BigDecimal(0.23);
BigDecimal twenty = new BigDecimal(20);
//Might want to use RoundingMode.UP instead,
//depending on desired behavior for negative values of num.
BigDecimal numTimesTwenty = num.multiply(twenty, new MathContext(0, RoundingMode.CEILING)); 
BigDecimal numRoundedUpToNearestFiveCents
  = numTimesTwenty.divide(twenty, new MathContext(2, RoundingMode.UNNECESSARY));

Answer 3

Based on your edit, another possible solution would be:

BigDecimal twenty = new BigDecimal(20);
BigDecimal amount = new BigDecimal(990.49)

// To round to the nearest .05, multiply by 20, round to the nearest integer, then divide by 20
BigDecimal result =  new BigDecimal(amount.multiply(twenty)
                                          .add(new BigDecimal("0.5"))
                                          .toBigInteger()).divide(twenty);

This has the advantage, of being guaranteed not to lose precision, although it could potentially be slower of course...

And the scala test log:

scala> var twenty = new java.math.BigDecimal(20) 
twenty: java.math.BigDecimal = 20

scala> var amount = new java.math.BigDecimal("990.49");
amount: java.math.BigDecimal = 990.49

scala> new BigDecimal(amount.multiply(twenty).add(new BigDecimal("0.5")).toBigInteger()).divide(twenty)
res31: java.math.BigDecimal = 990.5

Answer 4

You can use plain double to do this.

double amount = 990.49;
double rounded = ((double) (long) (amount * 20 + 0.5)) / 20;

EDIT: for negative numbers you need to subtract 0.5