How to check if a double value has no decimal part

Question

I have a double value which I have to display at my UI. Now the condition is that the decimal value of double = 0 eg. - 14.0 In that case I have to show only 14 on my UI. Also, the max limit for characters is 5 here.

eg.- 12.34 the integer value can be no bigger than 2 digits and so is the decimal value for our double.

What could be the best way of doing this?

score 290 · Accepted Answer · edited Jul 19 '20 at 14:10

290

You could simply do

d % 1 == 0

to check if double d is a whole.

edited Jul 19 '20 at 14:10

Roshana Pitigala

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answered Apr 12 '13 at 05:46

Lone nebula

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this Double.MIN_VALUE % 1 == 0 gives false – Evgeniy Dorofeev Apr 12 '13 at 06:03
19

@Evgeniy After all, `Double.MIN_VALUE` isn't whole. – Lone nebula Apr 12 '13 at 06:06
1

But modulo does only work with integer, doesn't it? – puncher Mar 30 '21 at 18:47
This solution is not Sonar compliant. Better to use relational operator e.g. less than (<) or greater than (>). For example: d % 1.0 > 0 – Héloïse Chauvel Apr 06 '22 at 22:37

Subhrajyoti Majumder · Answer 2 · 2013-04-12T06:04:50.247

25

double d = 14.4;
if((d-(int)d)!=0)
    System.out.println("decimal value is there");
else
    System.out.println("decimal value is not there");

edited Apr 12 '13 at 06:04

answered Apr 12 '13 at 05:45

Subhrajyoti Majumder

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You should probably use `!=` rather than `>`, since it will return false for negative values `d`. – Lone nebula Apr 12 '13 at 05:59
1

it does not work with -11111111111111.1; – Evgeniy Dorofeev Apr 12 '13 at 06:05
dose not work with `Repeating or Endless Decimals` such as `Pi or 1/7`. – minchaej Jul 24 '21 at 01:54

score 13 · Answer 3 · answered Apr 12 '13 at 05:58

13

All Integers are modulo of 1. So below check must give you the answer.

if(d % 1 == 0)

answered Apr 12 '13 at 05:58

Gokul Nath KP

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Ankit · Answer 4 · 2016-08-04T11:26:28.830

12

either ceil and floor should give the same out out put

Math.ceil(x.y) == Math.floor(x.y)

or simply check for equality with double value

x.y == Math.ceil(x.y)
x.y == Math.floor(x.y)

or

Math.round(x.y) == x.y

edited Aug 04 '16 at 11:26

answered Apr 12 '13 at 05:46

Ankit

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What about ```Math.round(x) == x``` where ```x``` a double typed expression. – Valentin Ruano Jan 21 '16 at 20:19

score 2 · Answer 5 · answered Apr 12 '13 at 05:45

2

Compare two values: the normal double, and the double after flooring it. If they are the same value, there is no decimal component.

answered Apr 12 '13 at 05:45

Patashu

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score 1 · Answer 6 · answered Apr 12 '13 at 06:32

You probably want to round the double to 5 decimals or so before comparing since a double can contain very small decimal parts if you have done some calculations with it.

double d = 10.0;
d /= 3.0; // d should be something like 3.3333333333333333333333...
d *= 3.0; // d is probably something like 9.9999999999999999999999...

// d should be 10.0 again but it is not, so you have to use rounding before comparing

d = myRound(d, 5); // d is something like 10.00000
if (fmod(d, 1.0) == 0)
  // No decimals
else
  // Decimals

If you are using C++ i don't think there is a round-function, so you have to implement it yourself like in: http://www.cplusplus.com/forum/general/4011/

score 1 · Answer 7 · answered May 31 '14 at 23:52

Interesting little problem. It is a bit tricky, since real numbers, not always represent exact integers, even if they are meant to, so it's important to allow a tolerance.

For instance tolerance could be 1E-6, in the unit tests, I kept a rather coarse tolerance to have shorter numbers.

None of the answers that I can read now works in this way, so here is my solution:

public boolean isInteger(double n, double tolerance) {
    double absN = Math.abs(n);
    return Math.abs(absN - Math.round(absN)) <= tolerance;
}

And the unit test, to make sure it works:

@Test
public void checkIsInteger() {
    final double TOLERANCE = 1E-2;
    assertThat(solver.isInteger(1, TOLERANCE), is(true));

    assertThat(solver.isInteger(0.999, TOLERANCE), is(true));
    assertThat(solver.isInteger(0.9, TOLERANCE), is(false));

    assertThat(solver.isInteger(1.001, TOLERANCE), is(true));
    assertThat(solver.isInteger(1.1, TOLERANCE), is(false));

    assertThat(solver.isInteger(-1, TOLERANCE), is(true));

    assertThat(solver.isInteger(-0.999, TOLERANCE), is(true));
    assertThat(solver.isInteger(-0.9, TOLERANCE), is(false));

    assertThat(solver.isInteger(-1.001, TOLERANCE), is(true));        
    assertThat(solver.isInteger(-1.1, TOLERANCE), is(false));
}

`double` can exactly represent up to 53 bits, so there's no need for tolerance for current definition of `int` data type in java, which is strictly 32 bit. If the original value was converted from int to double, then it can be converted back exactly, without any loss of precision. See https://stackoverflow.com/a/43656339/3167374, please. — shitpoet, Nov 26 '19 at 09:04

score 0 · Answer 8 · answered Apr 12 '13 at 05:46

0

Use number formatter to format the value, as required. Please check this.

answered Apr 12 '13 at 05:46

Sudhanshu Umalkar

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How to check if a double value has no decimal part

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