float rounding
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Date:   (:) 
From:  Olivier Andrieu <oandrieu@n...> 
Subject:  Re: [Camllist] float rounding 
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 0,625e1 0,3125e1 ...). 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 = "0x1.999999999999ap4" `  Olivier