I think there are specs like strict math in Java, which allows to use
more
than 64 bits of precision for intermediate calculations - you know
Intel's
x86 FPU has 80 bits. There was a discussion about utilizing those bits
in Java.
Jacob Nikom
Juergen Kreileder wrote:
>
> >>>>> hilbrink writes:
>
> >> This most likely has nothing to do with the Java
> >> implementation, but of the floating point characteristics of
> >> the underlying machine. If you set a double (in C) to 0.3,
> >> it's value is not necessarily 0.3 but something very very
> >> close:
>
> hilbrink> well yes that is what I would expect, only number ^2
> hilbrink> could be represented properly, but what I think happens
> hilbrink> is that the 'new' double is not properly initialized, ie
> hilbrink> all bytes are not set to '0'. Apparently the exponent
> hilbrink> part is transferred correctly, but the mantissa part is
> hilbrink> not, hence the error in the mantissa.
>
> No. You already lose precision when you store 0.3 in a float. There
> is no way to get higher precision by converting this value to a
> double.
>
> 0.3f:
> 0 01111101 00110011001100110011010
>
> (double) 0.3f:
> 0 01111111101 0011001100110011001101000000000000000000000000000000
>
> 0.3d:
> 0 01111111101 0011001100110011001100110011001100110011001100110011
>
> hilbrink> But what I would like to know is if other people
> hilbrink> experience the same, how does a JVM under windows
> hilbrink> behave.
>
> The result will be the same.
>
> Juergen
>
> --
> Juergen Kreileder, Blackdown Java-Linux Porting Team
> http://www.blackdown.org/java-linux.html
>
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