Home/ Questions/Q 7667415
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-31T15:05:49+00:00

I know the following behavior is an old problem, but still I don’t understand.

  • 0

I know the following behavior is an old problem, but still I don’t understand.

System.out.println(0.1 + 0.1 + 0.1);

Or even though I use BigDecimal

System.out.println(new BigDecimal(0.1).doubleValue()
    + new BigDecimal(0.1).doubleValue()
    + new BigDecimal(0.1).doubleValue());

Why this result is: 0.30000000000000004 instead of: 0.3?

How can I solve this?

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    What you actually want is

    new BigDecimal("0.1")
 .add(new BigDecimal("0.1"))
 .add(new BigDecimal("0.1"));

    The new BigDecimal(double) constructor gets all the imprecision of the double, so by the time you’ve said 0.1, you’ve already introduced the rounding error. Using the String constructor avoids the rounding error associated with going via the double.

    • 0