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Jordi Castilla
Asked: 2020-12-17 07:20:59 +0800 CST 2020-12-17 07:20:59 +0800 CST

Even using BigDecimal I have problems with rounding

  • 772

Why if I use BigDecimalin this case, I don't get the expected result?

double value1 = 5.68;
double value2 = 2.45;
System.out.println(BigDecimal.valueOf(value1 + value2));

EXIT:

8.129999999999999

EXPECTED:

8.13
java

3 Answers

  1. Best Answer

    What happens is that you are doing the sum of the doublenot of theBigDecimal

    I recommend this instead

    double value1 = 5.68;
double value2 = 2.45;
System.out.println(BigDecimal.valueOf(value1).add(BigDecimal.valueOf(value2)));

    In this way each number is first converted into a BigDecimalbefore performing the addition.

    Since the operator +does not work with BigDecimalthe method must be used .add()instead

    • 15

  2. When using BigDecimal, it's a good idea to always use constructors and methods that don't use the type double, since, as I assume you already know, the type doubledoesn't have exact representations for many numbers that have an exact representation in base 10.

    For your example, you could use:

    /* Usa valueOf(long valorSinEscalar, int escala) */
long value1 = 568;
long value2 = 245;
System.out.println(BigDecimal.valueOf(value1 + value2, 2));

    or maybe:

    /* Usa el constructor BigDecimal(long valorSinEscalar, int escala) */
BigDecimal value1 = BigDecimal.valueOf(568, 2);
BigDecimal value2 = BigDecimal.valueOf(245, 2);
System.out.println(value1.add(value2));

    or this other:

    /* Usa el constructor BigDecimal(String valor) */
BigDecimal value1 = new BigDecimal("5.68");
BigDecimal value2 = new BigDecimal("2.45");
System.out.println(value1.add(value2));

    Note that there is a version of these constructors with an additional parameter MathContext, which allows you to select the rounding mode.

    • 5

  3. Since the question is about rounding using BigDecimal, you should only use the BigDecimal#setScale(int, RoundingMode):

    System.out.println(BigDecimal.valueOf(value1 + value2).setScale(2, RoundingMode.HALF_EVEN));

    The result will be:

    8.13

    You can see an example of this running on ideone .

    Obviously, it doesn't mean that this is the way to proceed with decimal operations. It would be best to follow the example given in @ninjalj's answer , where if you are going to use decimal numbers, initialize BigDecimalusing strings instead of double:

    /* Usa el constructor BigDecimal(String valor) */
BigDecimal value1 = new BigDecimal("5.68");
BigDecimal value2 = new BigDecimal("2.45");
System.out.println(value1.add(value2));

    However, if the above operations must have a fixed number of decimal places, use setScale.

    • 1