How to check if a double value has no decimal part

2019-01-17 05:42发布

This question already has an answer here:

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?

标签: java double
8条回答
我命由我不由天
2楼-- · 2019-01-17 05:57

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

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beautiful°
3楼-- · 2019-01-17 05:59

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

if(d % 1 == 0)
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Fickle 薄情
4楼-- · 2019-01-17 06:00

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/

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冷血范
5楼-- · 2019-01-17 06:05

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

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Viruses.
6楼-- · 2019-01-17 06:08

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));
}
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该账号已被封号
7楼-- · 2019-01-17 06:12

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
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