>...
>Loren Wilton wrote:
>>> 3 decimal places, not 3 significant digits.
>>>
>>> ie: 10.001 has 5 significant digits, but 3 decimal places.
>>>
>>> AFAIK there are no SA rules with scores more exact than 3 decimal places.
>>>
>>> So, no.. you would not have any rounding issues at that point.
>>>
>>
>> Yes you would, or at least could. .001 is not an exact binary fraction, so
>> trails out to lots more bits than there are in a double. So you can still
>> get decimal fractions that won't necessarily add up in binary even at 3
>> digits. (And might seem to be even worse if it were displayed at 4 digits.)
>>
>> SA would have to maintain all scores in scaled integers to get exact
>> results.
>>
>Erm.. Loren.. While that may be true of binary fractions, nobody uses
>binary fractions.
>
>In IEEE floating point format (single precision or otherwise), 0.001 has
>an exact binary representation.
>
>Very few things in this world use binary fractions. Standard floating
>point numbers on computers is one of them.
>
>
From an i386 (so you might wish to read the byte order reversed).
float 0x6F 0x12 0x83 0x3A
double 0xFC 0xA9 0xF1 0xD2 0x4D 0x62 0x50 0x3F
float 0.001000000047497451
double 0.00100000000000000002081668
This *is* IEEE 754 (I've designed far too many circuits for this).
1 * 10e-3 does not have any exact representation, maybe you're thinking
about 1 * 2e-X. The error is small, but Loren is completely correct.
Paul Shupak
[EMAIL PROTECTED]
