Why does the double type behave this way?

Asked Jan 28 '15 at 14:00

Active Jan 28 '15 at 14:00

Viewed 56 times

Look at the following code. Why does the 0.01D constant make the output so inaccurate?

double x = 0D;
while (x < 2D) {
    System.out.println(x);
    x = x + 0.01D;
}

0.0
0.01
0.02
0.03
0.04
0.05
0.060000000000000005
0.07
0.08
0.09
0.09999999999999999
0.10999999999999999
0.11999999999999998
0.12999999999999998
0.13999999999999999
0.15

asked Jan 28 '15 at 14:00

spongebob

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What do you believe the *exact* value of `0.01D` to be? (It's not 0.01...) – Jon Skeet Jan 28 '15 at 14:03
Because floating point numbers are not precise: [link](http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html) – Dragondraikk Jan 28 '15 at 14:04
1

It's actually much worse than this, but part of it is hidden by the way the conversion from `double` to `String` works. – harold Jan 28 '15 at 14:05
Duplicate of http://stackoverflow.com/questions/1661273/floating-point-arithmetic-not-producing-exact-results – Ivaylo Slavov Jan 28 '15 at 14:05
@JonSkeet What is it? – spongebob Jan 28 '15 at 14:07
3

@Joiner 0.01000000000000000020816681711721685132943093776702880859375 – harold Jan 28 '15 at 14:08
Now would be a good time to learn more about the `double` type then, via the linked question... – Jon Skeet Jan 28 '15 at 14:13

Why does the double type behave this way?

0 Answers0