Re: [Caml-list] string_of_float (0.1 +. 0.2)

[Andreas Rossberg <rossberg@mpi-sws.org>]

You can use sprintf, which allows you to specify a precision:

# Printf.sprintf "%g" (0.1 +. 0.2);;
- : string = "0.3"
# Printf.sprintf "%.16g" (0.1 +. 0.2);;
- : string = "0.3"
# Printf.sprintf "%.18g" (0.1 +. 0.2);;
- : string = "0.300000000000000044"
# Printf.sprintf "%.24g" (0.1 +. 0.2);;
- : string = “0.300000000000000044408921”

The OCaml manual does not say what the default is, but it appears to be .6 for printf (like in C), while string_of_float is equivalent to (sprintf "%.12g”) and the REPL uses .18 (which is the maximum meaningful decimal precision for 64 bit floats):

# Printf.sprintf "%g" 0.1234567890123456789;;
- : string = "0.123457"
# string_of_float 0.123456789012345678;;
- : string = “0.123456789012”
# 0.1234567890123456789;;
- : float = 0.123456789012345677

/Andreas


> On 15. 6. 2022, at 03:59, Kenichi Asai <asai@is.ocha.ac.jp> wrote:
> 
> On OCaml 4.12.0 on M1 mac, I got:
> 
> # 0.1 +. 0.2;;
> - : float = 0.300000000000000044
> # string_of_float (0.1 +. 0.2);;
> - : string = "0.3"
> 
> Why don't I obtain "0.300000000000000044" here?
> 
> Here is some background.  I am writing an OCaml interpreter that
> mimics most part of the original OCaml interpreter.  In the OCaml
> interpreter, 0.1 and 0.2 are represented as
> 
> Pexp_constant (Pconst_float ("0.1", None))
> Pexp_constant (Pconst_float ("0.2", None))
> 
> When I add these two numbers, I would have to execute
> 
> let a = float_of_string "0.1"
> let b = float_of_string "0.2"
> let c = a +. b
> let d = string_of_float c
> 
> and then return
> 
> Pexp_constant (Pconst_float (d, None))
> 
> At this point, however, since d is "0.3" instead of
> "0.300000000000000044" (even though c is 0.300000000000000044), I
> cannot return 0.300000000000000044 as a result.  How can I mimic the
> OCaml behavior?
> 
> Sincerely,
> 
> -- 
> Kenichi Asai