Losing accuracy with double plus

Question

I have a simple code here, however it has not any division or complicated multiplying.

for ( double i=.1; i<=1; i+=.1) {
      System.out.println(i);
}

But in some cases value of i is not accurate, e.g. .79999999 or .600000001.

Why did this happen? How can I change this code that becomes accurate?

Possibly the most answer question in programming. In short you must round the result or use integer arithmetic. Try `printf("%.1f%n", i)` instead. Note: `i <= 1` is ideal either. — Peter Lawrey, Aug 12 '13 at 06:35

Bohemian · Answer 1 · 2013-08-12T06:42:21.777

2

double is imprecise - it is designed to hold finite accuracy.

For precise "doubles", use BigDecimal:

BigDecimal d = new BigDecimal(".1");
for ( BigDecimal i= d; ! i.equals(BigDecimal.ONE) ; i = i.add(d) {
    System.out.println(i);
}

edited Aug 12 '13 at 06:42

answered Aug 12 '13 at 06:36

Bohemian

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Thanushan · Accepted Answer · 2013-08-13T06:22:06.920

2

Floating point numbers aren't represented exactly, they lose precision. Try

for ( int i=1; i<=10; i+=1) {
      System.out.println("%.f", (double)i/10);
}

edited Aug 13 '13 at 06:22

answered Aug 12 '13 at 06:39

Thanushan

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This outputs "0 0 0 0 0 0 0 0 0 1" - you need to add a cast to float or double. – resueman Aug 12 '13 at 12:40

Maroun · Answer 3 · 2013-08-13T06:43:51.753

1

Using int inside your loop will fix the problem:

for (int i=1;i<=10;i++) {
    System.out.println(i*0.1);
}

This is happening because decimal fractions might lose precision, they're not represented exactly.
You might want to to know What Every Computer Scientist Should Know About Floating-Point Arithmetic for more information.

edited Aug 13 '13 at 06:43

answered Aug 12 '13 at 06:36

Maroun

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Downvoter, please let me know what I did wrong. We all want to learn from our mistakes. – Maroun Aug 12 '13 at 06:39
The link may theoretically answer the question, but [it would be preferable](http://meta.stackexchange.com/q/8259) to include the essential parts of the answer here, and provide the link for reference. – JJJ Aug 12 '13 at 06:40
@Juhana The link provides the answer. No need to copy paste parts from it IMHO. Reading it will answer OP question. I don't think this is a good reason for downvoting. – Maroun Aug 12 '13 at 06:41
@MarounMaroun: I do. Links are great for pointing to supplementary info or evidence...but for a number of reasons, the post should be able to stand on its own without them. If it can't, it's at best a half-answer. – cHao Aug 13 '13 at 06:32

Alexander · Answer 4 · 2016-02-09T01:39:55.560

1

This is a well known property of double.
Doubles store numbers in the format of 0.x * 2^y (or 1.x * 2^x).
This means not all integers are representable exactly, and, most importantly,
not all decimal fractions can be represented exactly.

I'm guessing you should be using something like BigDecimal here.

edited Feb 09 '16 at 01:39

answered Aug 12 '13 at 06:37

Alexander

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score 0 · Answer 5 · answered Aug 12 '13 at 06:38

0

This is because of floating-point-operations. If you want exact value, there are special classes like BigDecimal

answered Aug 12 '13 at 06:38

MrD

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score 0 · Answer 6 · answered Aug 12 '13 at 06:41

0

for (double i = .1; i <= 1; i += .1) {
    System.out.println(BigDecimal.valueOf(i).setScale(1, RoundingMode.HALF_UP));
}

answered Aug 12 '13 at 06:41

Alex

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Losing accuracy with double plus

6 Answers6