Question

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Retain precision with Doubles in java

Do you know the difference between these two operations in Java.

final double m1 = 11d / 1e9; // gives 1.1e-8
final double m2 = 11d * 1e-9; // gives 1.1000000000000001e-8

I see in the generated bytecode that the precompiled result of m2 is already not what I expected.
In the output of javap -verbose -c I can see the following value :

const #3 = double   1.1E-8d;
[...]
const #6 = double   1.1000000000000001E-8d;

When I use m1 or m2 in other expressions, I don’t have the same result.

When I try the same thing in C, m1 and m2 is strictly 1.1e-8

I think my problem is in the way java handles double precision computation but I can’t explain myself what I miss.

Answer 1

The difference you have is that 1e9 is exactly representable and 1e-9 which is not exactly representable. A small representation error results in an arithmetic error which you can see.

System.out.println(new BigDecimal(1e9));
System.out.println(new BigDecimal(1e-9));

prints

1000000000
1.0000000000000000622815914577798564188970686927859787829220294952392578125E-9

The difference is not whether you have used * or / but that you have used a whole number in one case and a fraction in the other.

System.out.println(11 * 1e9);
System.out.println(11 * 1e-9);

prints

1.1E10
1.1000000000000001E-8