(100&*)4.57 4.34 4.44
457 434 444
+./(100&*)4.57 4.34 4.44
2.63185e_11
(+./&.(100&*))4.57 4.34 4.44
2.63185e_13
It doesn't work.
Thanks.
Bo.
Den fredag den 15. maj 2020 20.58.47 CEST skrev Hauke Rehr
<hauke.r...@uni-jena.de>:
my 2¢
the result is absolutely correct;
both Henry Rich and Devon McCormick showed why
If you’re dealing with values with two decimal places
everywhere, either apply rounding after each operation,
or (better imho) work on 100-fold values and divide by
100 in the very last step of your computation.
you could also define your own data structure but then
you’re better off using x: in the first place.
Either you want to deal with rationals (x:) or with
whole numbers (algorithm&.(100&*)).
Am 15.05.20 um 17:20 schrieb 'Bo Jacoby' via Programming:
> The expression
> 4.57+.(4.34+.4.44)
>
>
> should evaluate like this
> 4.34+.4.44
> 0.02
>
> 4.57+.0.02
> 0.01
>
>
>
> but it does evaluate to
> 5.39568e_12
>
>
> which is a useless and incorrect result.
> = is tolerant, why isn't +. ?
> Thanks.
> Bo. Den fredag den 15. maj 2020 16.24.10 CEST skrev Henry Rich
> <henryhr...@gmail.com>:
>
> What's wrong with it?
>
> 0j19 ": 4.57 4.34 4.44
> 4.5700000000000002842 4.3399999999999998579 4.4400000000000003908
>
> Henry Rich
>
> On 5/15/2020 10:17 AM, 'Bo Jacoby' via Programming wrote:
>> +./4.57 4.34 4.44
>> 5.39568e_12
>>
>> NB. this was not the correct answer. A bug?
>>
>> +./x:4.57 4.34 4.44
>>
>> 1r100
>>
>> NB. this was a work-around
>> Thanks!
>> Bo.
>>
>>
>>
>>
>> ----------------------------------------------------------------------
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>
>
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