1. Re: I Need Help -- precision
Reply to Michael Bolin and Lucius Hilley:
>> Hi Michael,
>>
>> Unfortunately, this is not useful to me. If I implement it with:
>>
> >fprint(1, "4.2%f", round(0.745))
>>
>>what do you think I get? Well, it is 0.74! Strange enough, some of the
>>values x.xx5 round up in the correct way, but not all of them! Is this =
a
>>bug in my processor (P75), or in the 'binary floating point' routines?
>>I am puzzled over this for a long time already, as I want to do some
exac=3D
>> t
>>financial calculating in Euphoria. Isn't this possible at all, or shoul=
d
=3D
>> I
>> maybe disable my math coprocessor?
>This is not a bug in my routine, it is another bug resulting from
>binary floating point representation.
>But I have just modified the function to allow a variable amount of
>precision, and it will work in cases such as yours.
>----------------------------------------------------
>function round(atom x,atom precision)
> precision=3Dpower(10,precision)
> return floor(x*precision+.5)/precision
>end function
>----------------------------------------------------
>Just call it with an atom value, and a number of decimal places to
>round to. You can use values less than one:
>If you use 0, it will round to the nearest integer. If you
>use -1, it will round to the nearest 10. If -2, the nearest 100, etc.
>Regards,
> Michael Bolin
Thanks, also to Lucius, who sent me an equal solution.
But still, unfortunately, this are not the right solutions. Imagine that =
in
the binary floating point routines x.xx5 is represented as
x.xx49999.......... (etc) and I want a precision of 2 digits. This means
for instance that .745 should be rounded up to .75. Check your local
'Pocket Japanese'. Multiplied by 100, this gives me 74.49999..., add .5 t=
o
it, you get 74,9999....... Floored, this gives ...... 74!
Exactly this is what happens. But, even more unfortunate, .145 for instan=
ce
is represented as .145000000001 or so.
So, a round() function as you propose it does not always give the right
answers, certainly not the answers I get from my 10$ calculator.
Someone, I think it was Jef Zeitlin, has already pointed this out some ti=
me
ago. But my question remains the same: Is there a way round this, using
Euphoria?
Thanks anyway,
Ad Rienks
