round towards zero in java

Question

How to round in java towards zero?

So -1.9 becomes -1.0 and -0.2 becomes 0.0, 3.4 becomes 3.0 and so on.

Is Math.round() capable of doing this changing some parameters?

Kris · Accepted Answer · 2012-03-16T14:09:55.743

21

I do not believe that the standard library has such a function.

The problem is that you are asking for very different behavior (mathematically speaking) depending on whether the number is larger or smaller than 0 (i.e. rounding up for negative values, rounding down for positive values)

The following method could be used:

public double myRound(double val) {
    if (val < 0) {
        return Math.ceil(val);
    }
    return Math.floor(val);
}

edited Mar 16 '12 at 14:09

answered Dec 05 '11 at 15:11

Kris

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+1: The only answer so far which does the right thing (AFAICS). – Oliver Charlesworth Dec 05 '11 at 15:18

score 10 · Answer 2 · edited Mar 10 '15 at 13:26

10

cast to long like this:

float x= 1.9;

long y = (long)x;

This rounds both positive and negative numbers towards zero.

edited Mar 10 '15 at 13:26

Spycho

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answered Dec 05 '11 at 15:09

confucius

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Provided the result fits in an `int`, which may be a big if – NPE Dec 05 '11 at 15:11
2

Why would you cast to an `int` and store it in a `long`? Cast to a `long`. – Kevin Dec 05 '11 at 15:12
2

**-1:** this fails in cases where the floating point number is too big or too small to be represented by a 64 bit integer. Whether `int` or `long`, the problem is only shifted. Such numbers may not be common in everyday usage – but that only makes it more difficult to find bugs related to over/underflow. – TheOperator Aug 04 '16 at 12:24

score 6 · Answer 3 · answered Dec 05 '11 at 15:09

6

Use RoundingMode.DOWN, it leads towards zero.

Example :

    BigDecimal value = new BigDecimal("1.4");
    value = value.setScale(0, RoundingMode.DOWN);
    System.out.println(value.doubleValue());
    BigDecimal value1 = new BigDecimal("-1.4");
    value1 = value1.setScale(0, RoundingMode.DOWN);
    System.out.println(value1.doubleValue());

answered Dec 05 '11 at 15:09

mprabhat

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Can a `BigDecimal` represent all the values that a `float` or `double` can represent? – Oliver Charlesworth Dec 05 '11 at 15:12
BigDecimal is the way to represent a number in java, so it makes sense to use BigDecimal – mprabhat Dec 05 '11 at 15:33
@Oli Yes. BigDecimal is an arbitrary precision decimal, it can represent all the values a float and double can and more. – Dunes Dec 05 '11 at 16:08
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@mprabhat `BigDecimal` isn't '*the* way to represent a number in Java'. It is one among several. – user207421 Apr 29 '12 at 10:17
It certainly is overkill, especially when a primitive will do just fine. – demongolem Nov 23 '15 at 01:00

Voo · Answer 4 · 2011-12-05T15:22:01.583

5

Just casting to int will do that for you?

Edit: If you want to retain a double this should work simply enough:

if (val < 0) 
   return -Math.floor(-val);
else
   return Math.floor(val);

And just for the people who want branch free code and feel a bit more clever:

long tmp = Double.doubleToLongBits(val);
tmp >>>= 63;
return Math.floor(val) + tmp;

edited Dec 05 '11 at 15:22

answered Dec 05 '11 at 15:08

Voo

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Provided the result fits in an `int`, which may be a big if. – NPE Dec 05 '11 at 15:10
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@aix True enough, but then you also can't guarantee that there exists a double that can represent the number. But yes that's nitpicking on my side ;) – Voo Dec 05 '11 at 15:12

score 1 · Answer 5 · edited Jul 23 '16 at 18:36

1

y=sign(x)*floor(abs(x))

or

y=sign(x)*round(abs(x)-0.5)

It should be easy to implement in Java.

edited Jul 23 '16 at 18:36

dorukayhan

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answered Jul 23 '16 at 17:03

benfuzius

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score 0 · Answer 6 · answered Dec 05 '11 at 15:08

0

Seems like you want to always round-down? You can use Math.floor instead

public static double floor(double a)

Returns the largest (closest to positive infinity) double value that is not greater than the argument and is equal to a mathematical integer. Special cases:

answered Dec 05 '11 at 15:08

TS-

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`Math.floor()` rounds towards -ve infinity, not zero. – Oliver Charlesworth Dec 05 '11 at 15:09

score 0 · Answer 7 · answered Dec 05 '11 at 15:09

0

BigDecimal offers many rounding options.

answered Dec 05 '11 at 15:09

tobiasbayer

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In particular, that would be DOWN to solve the OP's need which is different from FLOOR in that FLOOR follows the correct mathematical behavior. – demongolem Nov 23 '15 at 00:59

score 0 · Answer 8 · answered Dec 05 '11 at 15:11

0

You also can try this:

public static void main(String[] args) {
        Double myDouble = -3.2;
        System.out.println(myDouble.intValue()); //Prints -3

    }

answered Dec 05 '11 at 15:11

Averroes

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round towards zero in java

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