Java more precision in arithmetic

Question

I am building a web app in Java that does math and shows steps to the user. When doing basic arithmetic with decimals I often get the messy in accurate outputs.

Here is my problem:

double a = 0.15;
double b = 0.01;
System.out.println(a - b);
// outputs 0.13999999999999999

float a = 0.15;
float b = 0.01;
System.out.println(a - b);
// outputs 0.14

float a = 0.16f;
float b = 0.01f;
System.out.println(a - b);
// outputs 0.14999999

double a = 0.16;
double b = 0.01;
System.out.println(a - b);
// outputs 0.15

Neither is reliable for complete accuracy. Is there a numeric class that is more precise or should I just round the values off?

You can use strictfp as well instead of BigDecimal. Explained at http://stackoverflow.com/questions/517915/when-should-i-use-the-strictfp-keyword-in-java — ch4nd4n, Aug 29 '12 at 07:21
strictfp is still floating-point, so that won't help (it will just ensure that all his users get the exact same messy results). — Thilo, Aug 29 '12 at 07:22
The Sun/Oracle uses strictfp almost all the time so its unlikely to make any difference. — Peter Lawrey, Aug 29 '12 at 07:37
But the major point is that `strictfp` **reduces** your precision, if anything! — Marko Topolnik, Aug 29 '12 at 08:05
http://docs.oracle.com/javase/1.5.0/docs/api/java/math/BigDecimal.html — arun, Aug 29 '12 at 07:16
Can you use a link from Java 7 and provide an example of how this is done in BigDecimal? — Peter Lawrey, Aug 29 '12 at 07:30

jqno · Accepted Answer · 2012-08-29T07:26:11.787

6

You can use BigDecimal for this. It's ugly, but it works:

BigDecimal a = new BigDecimal("0.15");
BigDecimal b = new BigDecimal("0.01");
System.out.println(a.subtract(b));

Be sure to construct them either with a String parameter, or with the valueOf method, like this:

BigDecimal x = new BigDecimal("0.15");   // This is ok
BigDecimal x = BigDecimal.valueOf(0.15); // This is also ok

And not with a double parameter, like this:

BigDecimal x = new BigDecimal(0.15); // DON'T DO THIS

Because if you pass in a double, you will also pass in double's inaccuracy into the new BigDecimal instance. If you pass in a String, BigDecimal will know™ and do the right thing™.

edited Aug 29 '12 at 07:26

answered Aug 29 '12 at 07:16

jqno

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Instead of `new BigDecimal(0.15)` you should use `BigDecimal.valueOf(0.15)` – Peter Lawrey Aug 29 '12 at 07:23
@PeterLawrey You're right, I missed that one. I've updated my answer. – jqno Aug 29 '12 at 07:26
Won't work too well for "1/3", though. For that you need rational numbers. – Thilo Aug 29 '12 at 23:54
@Thilo True, but OP asked for "basic arithmetic with decimals", not for "basic arithmetic with rationals" :). – jqno Aug 30 '12 at 07:53

score 2 · Answer 2 · answered Aug 29 '12 at 07:23

You need to apply a sensible round when using double. If you do you can get 15 digits of accuracy which is more than enough in most cases.

double a = 0.15;
double b = 0.01;
System.out.printf("%.2f%n", a - b);

double a2 = 0.16;
double b2 = 0.01;
System.out.printf("%.2f%n", a2 - b2);

prints

0.14
0.15

score 1 · Answer 3 · edited May 23 '17 at 12:14

1

How about BigDecimal?

Will not be completely accurate for all divisions, though (such as 1/3). For that you'd need fractions, which is not part of the JDK, but easy to implement, for example as two integers (or BigIntegers).

edited May 23 '17 at 12:14

Community

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answered Aug 29 '12 at 07:17

Thilo

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score 0 · Answer 4 · answered Aug 29 '12 at 07:23

0

  BigDecimal  a = new BigDecimal(".15");
  BigDecimal  b = new BigDecimal(".01"),
  BigDecimal  c = new BigDecimal(0);

  c = a.subtract(b);      
  System.out.println(c);

answered Aug 29 '12 at 07:23

Sumit Singh

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Java more precision in arithmetic

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