Sorting array of doubles [closed]

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I've array of doubles and I'm sorting it using Arrays.sort problem is that in the output array I've something like this(this is array output after reverse):

0.002385171991295645

...

9.914204103773398E-4

...

1.00139601969068E-4

9.975711760353395E-5

and so on, as we can see shortest number are at the top and longest are at the bottom. Numbers in given range e.g. E-4 are sorted good. I've even created a sample code to test this:

double l1 = 0.002385171991295645;
double l2 = 9.914204103773398E-4;
if(l1 > l2) {
    System.out.println("TRUE");
} else {
    System.out.println("FALSE");
}

And it gives me "TRUE", how to sort this array?

Answer 1

You probably don't know what's the meaning of that E-4 at the end.

9.914204103773398E-4

Is actually

9.914204103773398 * 10^-4

Which is smaller than 0.002385171991295645.

For more details, visit the JLS - 3.10.2. Floating-Point Literals.

Answer 2

You are getting confused by the scientific notation (also known as standard form).

9.914204103773398E-4 means 9.91... x 10^(-4), i.e. 0.0009914.... So this value is indeed smaller than 0.0023...

Answer 3

The "e" or "E" means "exponent," which denotes scientific notation.

Here's a useful example from the primitive data types tutorial on Oracle.com:

double d1 = 123.4;
double d2 = 1.234e2;

Those are both the same number. The "e" means, basically, 10^x (10 raised to the power of x) where the number that follows is x. So in the above, 1.234 * 10^2. 10^2 is 100, so 1.234 * 100 is 1.234.

So looking at one of your numbers, 9.914204103773398E-4, that's 9.914204103773398 * 10^-4. 10^-4 is 0.0001, so 9.914204103773398 * 0.0001, which is 0.0009914103773398.