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float rounding
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| Date: | -- (:) |
| From: | Christophe Raffalli <Christophe.Raffalli@u...> |
| Subject: | Re: [Caml-list] float rounding |
Sean McLaughlin a écrit :
> Hello,
>
> I'm using Ocaml for an interval arithmetic application. I"m curious
> about
> what the Ocaml parser/compiler does to float constants. May I assume
> that for any constant I enter, eg. 3.1415... (for N digits of pi), that
> the compiler will give me a closest machine representable number?
> i.e., if I bound a constant by the previous and next floating point
> value to
> that given me by the compiler,
> will it always be the case that my original (mathematical) constant
> lies in that interval?
>
By the way, float constants need to be written in hexadecimal and this
is missing to
the printf/scanf functions (it is what man printf says at least) ...
just compute how many decimals you need to write the exact value of
2^{-n} as a decimal float constant (0,5 0,25 0,125 0,625e-1 0,3125e-1 ...).
Then, I am not sure the algorithm used to parse the constant will give
you the correct value and you will never know how many decimals you need
to write your constant.
The only problem with float and hexadecimal is to learn the hexadecimals
of pi, e, ln 2, sqrt 2, ... ;-)
Christophe Raffalli