Re: [Caml-list] float rounding

[Olivier Andrieu <oandrieu@nerim.net>]

 Christophe Raffalli [Tuesday 3 October 2006] :
 >
 > 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 ...).

float_of_string (which uses strtod) already knows how to read a
hexadecimal float and there's a hack to have the C printf do the writing:
,----
| # external format_float: string -> float -> string = "caml_format_float" ;;
| external format_float : string -> float -> string = "caml_format_float"
| # let hex_float = format_float "%a" ;;
| val hex_float : float -> string = <fun>
| # hex_float 1.25 ;;
| - : string = "0x1.4p+0"
| # float_of_string "0x1.4p+0" ;;
| - : float = 1.25
| # hex_float 0.1 ;;
| - : string = "0x1.999999999999ap-4"
`----

-- 
   Olivier