Java double precision with constant multiplication/division [duplicate]

Question

Possible Duplicate:
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

There is no difference between Java and C, both respect the IEEE floating-point standard. The only difference can be the way you print it. The exact number 1.1e-8 is not representable as a double. The difference in output may be attributed to the details on how the error of representation was accumulated, and to what rounding is used to print the result.