Why java division give some incorrect exponential value

Question

I write below code in java

double scale=1.0;
for(int i=0;i<6;i++){
    scale=scale/10;
}
System.out.println(scale);

Now out put of this is 1.0000000000000002E-6 which seems to be incorrect, but if I loop 5 times then output is correct as below
1.0E-5

Can any body tell me why this is and how can I remove this.

Have a [read about it here](http://stackoverflow.com/q/1661273/2024761). — Rahul, Oct 23 '13 at 10:12
I think the OP is syaing that if the 6 is edited to 5, the output is different? Anyway, it's FP, so neither answer is particularly 'incorrect'. — Martin James, Oct 23 '13 at 10:12
We should seriously think about renaming `double` and `float` to `inexact` and `approximate` — Ingo, Oct 23 '13 at 10:22

score 2 · Accepted Answer · answered Oct 23 '13 at 10:12

2

This is floating point precision problem. Use BigDecimal instead of double.

    BigDecimal scale=BigDecimal.ONE;
    for (int i = 0; i < 6; i++) {
        scale = scale.divide(BigDecimal.valueOf(10));
    }
    System.out.println(scale);

answered Oct 23 '13 at 10:12

Masudul

21,823
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score 0 · Answer 2 · answered Oct 23 '13 at 10:12

0

Because even 0.1 is not representable as an exact double value. These errors sum up and show in the result.

answered Oct 23 '13 at 10:12

Axel

13,939
5
50
79

score -3 · Answer 3 · answered Oct 23 '13 at 10:11

-3

change i initialization.

double scale=1.0;
for(int i=1;i<6;i++){
    scale=scale/10;
}
System.out.println(scale);

now our code will enter the for for values i = [1, 2, 3, 4, 5], its 5 times

answered Oct 23 '13 at 10:11

RamonBoza

8,898
6
36
48

No, that's not correct :( Try it. – Maroun Oct 23 '13 at 10:13
This isn't the issue. The issue is with floating point arithmetic. – Rahul Oct 23 '13 at 10:13

Why java division give some incorrect exponential value

3 Answers3