Query about mod operator
Mick Collins
mickclns at mac.com
Tue Apr 22 10:11:48 EDT 2008
Thanks Jim and Colin, especially the cautions about integer functions
used for non-integer operations. Turns out I snipped out (after
missing in my scan-over) the examples that would have proved your
points, namely:
0.25, 0, 0
0.5, 0, 0
0.75, 0.5, 0
These are the only exact divisors of 8 for which the mods come out
"correctly", that is, as 0.
.25 decimal = .01 binary, exactly
.5 decimal = .1 binary, exactly
.75 decimal = .11 binary, exactly
That said, 4 things:
1) Other languages have an fmod for floating point, could Rev? (or
does it have one that is hidden?).
2) Because I don't know of an fmod, when I need to use such
functionality, I write my own function that bypasses the non-
integer / binary problems.
3) That usually involves converting to integer, using the built in
function, and converting back. I have never been unable to produce
a function this way that gives the correct values (except extreme
fringe-of-precision numbers).
4) A related joke that hard-core programmers probably know already:
Why do programmers have trouble distinguishing between Christmas and
Halloween?
Because 25 dec = 31 oct.
(for those not understanding, octal is base 8, and is a direct
conversion from base 2, so used some in nitty gritty low level and
hardware programming (still used? I'm not sure, I don't). In octal,
"31" represents three 8s and one 1, 24+1= 25 in decimal. )
nerdy, nerdy, nerdy,
- Mick
On Apr 22, 2008, at 6:48 AM, use-revolution-request at lists.runrev.com
wrote:
>
> Message: 3
> Date: Mon, 21 Apr 2008 11:43:20 -0700
> From: Jim Ault <JimAultWins at yahoo.com>
>
> Mick,
>
>> It seems that d divides 8 if and only if the mod is NOT
>> correct (until we get to 1, of course. I got similar results using 7
>> instead of 8.. Seems like it should work, I think it's a bug.
>
> You need to use caution when working with fractions in computer
> operations.
> Binary means that math involving fractions in base 10 is not
> precise as you
> would normally think about it. This has been covered on the list
> in the
> last couple months.
>
> I think that 'mod' would exhibit the same apparent irregularities
> that all
> other math functions do when dealing with fractions.
>
> If you want to be more accurate in this case:
>
> put 100 into mult
> put ((8 * mult) mod (div * mult))/ mult into theDiff
>
> but again, you need to be aware of the binary nature of computers
> expressing
> fractions when calculating theDiff. All computer languages have
> the same
> limitation. This is why floating point libraries are added to
> scientific
> processing to achieve accuracy for using decimal values.
>
> Hope this helps shed light on the subject.
>
> Jim Ault
> Las Vegas
>
> On 4/21/08 11:18 AM, "Mick Collins" <mickclns at mac.com> wrote:
>
>> Interesting, I don't think it has to do with dividing by zero
>>
>> I ran this script:
>>
>> ON tst
>> REPEAT WITH i = 1 to 100
>> put i * .01 into iDiv
>> put 8 mod iDiv into modiDiv
>> put 8 div iDiv into intQuotient
>> put 8 - (intQuotient * iDiv + modiDiv) into theDiff
>> put iDiv & ", " & modiDiv & ", " & theDiff into line i of
>> field 1
>> END repeat
>> END tst
>>
>>
>> The (partial) results are:
>> 0.01, 0.01, -0.01
>> 0.02, 0.02, -0.02
<snip>
>> 0.98, 0.16, 0
>> 0.99, 0.08, 0
>> 1, 0, 0
>>
>>
>> It seems that d divides 8 if and only if the mod is NOT
>> correct (until we get to 1, of course. I got similar results using 7
>> instead of 8.. Seems like it should work, I think it's a bug.
>>
>> - Mick
>>>
>>> Message: 15
>>> Date: Mon, 21 Apr 2008 10:39:33 -0400
>>> From: Colin Holgate <coiin at rcn.com>
>>> On Apr 21, 2008, at 10:11 AM, Paul Williams wrote:
>>>
>>>> why does 8 mod 0.05 return 0.05 ?
>>>
>>> Mod is normally an integer operation, and as you wouldn't want to
>>> risk
>>> a divide by zero issue, perhaps it divides by at least 1? That would
>>> give a remainder that was equal to the fraction.
>>
>> _______________________________________________
>
> Message: 4
> Date: Mon, 21 Apr 2008 13:30:47 -0700
> From: "Mark Wieder" <mwieder at ahsoftware.net>
>
> ...and from the docs...
> Note: Using non-integer number and divisor usually produces sensible
> results. However, mathematically, modulus is generally defined as a
> function
> over the integers, and the results using non-integers may not
> consistently
> be what you expect.
>
>
> --
> Mark Wieder
> mwieder at ahsoftware.net
>
>
>
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