Re: [libreoffice-users] a little math problem with Calc

Regina, thanks for your answer.

But can you tell me how to call Calc functions inside the macro?
I searched in docs and in the Google and don't found any clue.

By the way, just now I tought the possibility to rewrite math functions
using "string" to encode the number to avoid error because I remembered
BCD.  :-)



Regards

Alex Mitsio Sato



On Thu, Jul 5, 2012 at 8:27 PM, Regina Henschel <rb.henschel@t-online.de>wrote:

> Hi,
>
> alex sato schrieb:
>
>  Hi,
> > I have a problem with functions I made.
> >
> > Let's say I create the function:
> >
> > Function Test (n)
> > dim r(3)
> > r(0)=int(n)
> > r(1)=n-r(0)
> > r(2)=r(1)*100
> > r(3)=int(r(2))
> > Test = r
> > End Function
> >
> > Then, in a empty spreadsheet take the following steps:
> > -select A1:D1
> > -press [F2]
> > -Insert the formula =TEST(E1)
> > -press [CTRL]+[SHIFT]+[ENTER]
> > -insert 11,18 in E1 and see the result in cell D1
> >
> > I think the result should be 18 but I'm getting 17
> > Where is the error?
>
> It is the normal behavior of data type double. 11,18 has indefinite digits
> in dual system. So the stored value is cut somewhere. Because there are
> enough binary digits to round the dual number to 15 decimal digits, you do
> not notice it. But when you subtract the integral part, the others digits
> will shift left and new digits will be generated at the end, which does not
> represent the original value. The calculated difference becomes a little
> bit to small. After multiply with 100 you do not have 18,00 but 17,99...
> And that results in 17. The problem becomes more visible, when the integral
> part is larger.
>
> You should not use Basic functions, but use Calc functions instead. They
> are often tweaked to avoid such problems. You can call Calc functions
> inside your Macro.
>
> To get the first two decimal digits as integer value, it seems to be more
> accurate to use
>         r(3)=(n * 100) mod 100
> in your macro.
>
> Kind regards
> Regina
>
>
>
>
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