Java - Rounding 99999999999999.9999

Question

I am trying rounding and format techniques in Java as I have to keep my values in database with some specific formats.

My database column has the data precision of 18,4 which means 14 integers and 4 decimal places at max.

Now I am trying max possible value case which is 99999999999999.9999. When I execute the below code, I am getting a rounded value of 1000000000000000. However, I want to store the exact value provided.

Can anyone suggest to keep it as it is in java variable?

    DecimalFormat df2 = new DecimalFormat( "#.####" );
    df2.setRoundingMode(RoundingMode.CEILING);
    double number = Double.parseDouble("99999999999999.9999");
    System.out.println("Format: " + df2.format(number));

`precision of 18,2 which means 14 integers and 4 decimal places` that's not possible with double which has precision of only 53 bits ~15 digits — phuclv, Mar 18 '20 at 05:32
Shouldn't the precision of your database column be 18,4? If you want 4 decimal places — jeprubio, Mar 18 '20 at 05:36
Doesn't 18,2 mean one sign, 15 integers and 2 decimal places? — cup, Mar 18 '20 at 05:38
Sorry, it was my typo, thats 18,4 which means 14 integral values and 4 decimal parts. — Jawad-Dev, Mar 18 '20 at 05:47
If you want decimal math, use a decimal number type (`BigDecimal`) instead of a binary number type. — chrylis -cautiouslyoptimistic-, Mar 18 '20 at 05:54
You're losing information on this statement, `double number = Double.parseDouble("99999999999999.9999");` you cannot use a `double`. — matt, Mar 18 '20 at 05:59

score 1 · Answer 1 · answered Mar 18 '20 at 05:54

1

You can use BigDecimal

BigDecimal number = new BigDecimal("99999999999999.9999");

answered Mar 18 '20 at 05:54

Nonika

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Java - Rounding 99999999999999.9999

1 Answers1