How to check if a double value has no decimal part

问题:

This question already has an answer here:

• How to test if a double is an integer 12 answers

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?

回答1:

You could simply do this: d % 1 == 0 to see if some double d is whole.

回答2:

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");

回答3:

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

if(d % 1 == 0)

回答4:

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

回答5:

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

回答6:

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

回答7:

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/

回答8:

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));
}