CODEWITHSUNDEEP
javabigdecimal
user2342776
user2342776

Reputation: 23

BigDecimal - unexpected round behaviour

I think I have found a bug:

MathContext mathContext = new MathContext(5, RoundingMode.HALF_UP);
result = BigDecimal.valueOf(0.004798).round(mathContext); // fails
// result is 0.004798!!! (same value)

I had to use the following alternative:

BigDecimal bigDecimal = BigDecimal.valueOf(0.004798);
BigDecimal new_divisor = BigDecimal.valueOf(1, 5);
bigDecimal_array = bigDecimal.divideAndRemainder(new_divisor);
MathContext mathContext = new MathContext(5, RoundingMode.HALF_UP);
result = bigDecimal.subtract(bigDecimal_array[1], mathContext);
result = result.stripTrailingZeros();

This bug (if it is so) is very dangerous in my opinion.

Upvotes: 0

Views: 267

Answers (1)

Federico klez Culloca
Federico klez Culloca

Reputation: 27149

No, there's no bug. You're just misunderstanding what "precision" means.

From BigDecimal's documentation

the total number of digits to return is specified by the MathContext's precision setting; this determines the result's precision. The digit count starts from the leftmost nonzero digit of the exact result.

(emphasis mine).

You have 4 digits in this case. So any precision greater then or equal to 4 will have no effect on rounding.

Compare with

result = BigDecimal.valueOf(0.004798).round(new MathContext(3, RoundingMode.HALF_UP));
result ==> 0.00480

or with

jshell> result = BigDecimal.valueOf(1.004798).round(new MathContext(5, RoundingMode.UP));
result ==> 1.0048

Which behave like you expect.

Upvotes: 5

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