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Floating-point difference between ocamlc and ocamlopt #2480

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vicuna opened this issue Apr 18, 2004 · 2 comments
Closed

Floating-point difference between ocamlc and ocamlopt #2480

vicuna opened this issue Apr 18, 2004 · 2 comments
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@vicuna
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vicuna commented Apr 18, 2004

Original bug ID: 2480
Reporter: administrator
Status: closed
Resolution: fixed
Priority: normal
Severity: minor
Category: ~DO NOT USE (was: OCaml general)

Bug description

Full_Name: Shawn Wagner
Version: 3.07
OS: Linux x86
Submission from: dialup-4.242.105.156.dial1.seattle1.level3.net (4.242.105.156)

ocamlopt doesn't seem to handle literal negative 0. values properly.

The following program:

let signbit x = Int64.shift_right_logical (Int64.bits_of_float x) 63 =
Int64.one
and zero = -0.0 in
Printf.printf "zero = %f and signbit zero = %b\n" zero (signbit zero)

will output 'zero = -0.000000 and signbit zero = true' when compiled with ocamlc
but 'zero = 0.000000 and signbit zero = false' when compiled with ocamlopt. The
negative sign's getting dropped somewhere.

peers at the source

It looks like the problem is in asmcomp/i386/emit.mlp in emit_instr in the
section that matches Lop(Iconst_float s). It compares the constant to 0.0 and
emits a fldz instruction. Since -0.0 and 0.0 are equal, it looses the signbit
there.
@vicuna
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vicuna commented Apr 18, 2004

Comment author: administrator

Some more playing around produces this patch, which fixes the problem for
x86. I assume that all the native-code targets have the same issue, though.

*** asmcomp/i386/emit.mlp.orig Sun Apr 18 15:27:32 2004
--- asmcomp/i386/emit.mlp Sun Apr 18 16:05:11 2004

*** 400,408 ****
movl ${emit_nativeint n}, {emit_reg i.res.(0)}\n
| Lop(Iconst_float s) ->
let f = float_of_string s in
! if f = 0.0 then
! fldz\n
! else if f = 1.0 then
fld1\n
else begin
let lbl = new_label() in
--- 400,412 ----
movl ${emit_nativeint n}, {emit_reg i.res.(0)}\n
| Lop(Iconst_float s) ->
let f = float_of_string s in
! if f = 0.0 then begin
! if Int64.shift_right_logical (Int64.bits_of_float f) 63 = Int64.zero
! then (* +0.0 )
! fldz\n
! else ( -0.0 *)
! fldz\n fchs\n
! end else if f = 1.0 then
fld1\n
else begin
let lbl = new_label() in

@vicuna
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vicuna commented May 3, 2004

Comment author: administrator

Fixed by XL 2004-05-03.

@vicuna vicuna closed this as completed May 3, 2004
@vicuna vicuna added the bug label Mar 19, 2019
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@vicuna