Dave Cramer wrote:
> Oliver,
>
> I don't believe you will lose precision if the number is below MAX_LONG
> ? When I tested it on my system, I was able to retrieve a double that
> was equal to MAX_LONG without losing precision.
The attached testcase says otherwise. It produces this output:
> Mismatch: 9223372036854775806 => 9.223372036854776E18 => 9223372036854775807
> Mismatch: 9223372036854775805 => 9.223372036854776E18 => 9223372036854775807
> Mismatch: 9223372036854775804 => 9.223372036854776E18 => 9223372036854775807
> Mismatch: 9223372036854775803 => 9.223372036854776E18 => 9223372036854775807
> Mismatch: 9223372036854775802 => 9.223372036854776E18 => 9223372036854775807
> Mismatch: 9223372036854775801 => 9.223372036854776E18 => 9223372036854775807
> Mismatch: 9223372036854775800 => 9.223372036854776E18 => 9223372036854775807
> Mismatch: 9223372036854775799 => 9.223372036854776E18 => 9223372036854775807
[...]
> Mismatch: 9223372036854775296 => 9.223372036854776E18 => 9223372036854775807
> Mismatch: 9223372036854775295 => 9.2233720368547748E18 => 9223372036854774784
> Mismatch: 9223372036854775294 => 9.2233720368547748E18 => 9223372036854774784
and so on.
The problem is that near MAX_LONG you need almost 64 bits of mantissa to
exactly represent the value -- but a double is only a 64-bit value
including space for the exponent. I can't remember the exact split but
from the above it looks like there is around 10 bits of exponent so you
only have ~54 bits for the mantissa -- so you only get a precision of
about +/- 512 when you're dealing with numbers of a magnitude around
MAX_LONG.
-O
public class TestDoublePrecision {
public static void main(String[] args) {
for (long l = Long.MAX_VALUE; l != Long.MIN_VALUE; --l) {
double d = (double)l;
long check = (long)d;
if (check != l)
System.out.println("Mismatch: " + l + " => " + d + " => " + check);
}
}
}