Rounding Bigdecimal to *Always* Have Two Decimal Places

Rounding Bigdecimal values with 2 Decimal Places

I think that the RoundingMode you are looking for is ROUND_HALF_EVEN. From the javadoc:

Rounding mode to round towards the "nearest neighbor" unless both neighbors are equidistant, in which case, round towards the even neighbor. Behaves as for ROUND_HALF_UP if the digit to the left of the discarded fraction is odd; behaves as for ROUND_HALF_DOWN if it's even. Note that this is the rounding mode that minimizes cumulative error when applied repeatedly over a sequence of calculations.

Here is a quick test case:

BigDecimal a = new BigDecimal("10.12345");
BigDecimal b = new BigDecimal("10.12556");

a = a.setScale(2, BigDecimal.ROUND_HALF_EVEN);
b = b.setScale(2, BigDecimal.ROUND_HALF_EVEN);

System.out.println(a);
System.out.println(b);

Correctly prints:

10.12
10.13

UPDATE:

setScale(int, int) has not been recommended since Java 1.5, when enums were first introduced, and was finally deprecated in Java 9. You should now use setScale(int, RoundingMode) e.g:

setScale(2, RoundingMode.HALF_EVEN)

Rounding BigDecimal to *always* have two decimal places

value = value.setScale(2, RoundingMode.CEILING)

round up to 2 decimal places in java?

Well this one works...

double roundOff = Math.round(a * 100.0) / 100.0;

Output is

123.14

Or as @Rufein said

 double roundOff = (double) Math.round(a * 100) / 100;

this will do it for you as well.

BigDecimal with at least two decimals

Actually, you still can use setScale, you just have to check if the current scale if greater than 2 or not:

public static void main(String[] args) {
    BigDecimal bd = new BigDecimal("3.001");

    bd = bd.setScale(Math.max(2, bd.scale()));
    System.out.println(bd);
}

With this code, a BigDecimal that has a scale lower than 2 (like "3") will have its scale set to 2, and if it's not, it will keep its current scale.

Discrepancy in decimal rounding results

According to the JavaDoc of RoundingMode.HALF_EVEN:

Rounding mode to round towards the {@literal "nearest neighbor"} unless both neighbors are equidistant, in which case, round towards the even neighbor.

That means, with a scale of 0 that you set in the .setScale() (meaning you want 0 decimals):

• 3084.5 is equidistant from 3084 and 3085, so it will be the even neighbor (hence 3084 that is even, not 3085 that is odd).
• 3084.51 is not equidistant, it is 0.01 closer to 3085 than 3084, hence it will be the nearest neighbor.

Java BigDecimal Rounding while having pointless 0's in the number

Providing you have already performed your rounding, you can simply use DecimalFormat("0.#").

final double value = 5.1000;
DecimalFormat format = new DecimalFormat("0.#");
System.out.println(format.format(value));

The result here will be 5.1 without the trailing zeroes.

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