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From: "guihaoc at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug target/103628] ICE: Segmentation fault (in
 gfc_conv_tree_to_mpfr)
Date: Wed, 22 Feb 2023 09:00:47 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D103628

HaoChen Gui <guihaoc at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
           Assignee|unassigned at gcc dot gnu.org      |guihaoc at gcc dot =
gnu.org
                 CC|                            |guihaoc at gcc dot gnu.org

--- Comment #5 from HaoChen Gui <guihaoc at gcc dot gnu.org> ---
The memory representation of IBM long double is not unique. It's actually t=
he
sum of two 64-bit doubles.=20

During decoding, the real variable b can be=20
b =3D =3D {cl =3D 1, decimal =3D 0, sign =3D 0, signalling =3D 0, canonical=
 =3D 0,
  uexp =3D 67108357, sig =3D {0, 0, 9295712554570040320}}
which is sum of following two doubles
u =3D {cl =3D 1, decimal =3D 0, sign =3D 0, signalling =3D 0, canonical =3D=
 0,
  uexp =3D 67108356, sig =3D {0, 0, 9295712899447228416}}
v =3D {cl =3D 1, decimal =3D 0, sign =3D 0, signalling =3D 0, canonical =3D=
 0,
  uexp =3D 67108356, sig =3D {0, 0, 9295712209692852224}}

During encoding, the real variable b can be
b =3D {cl =3D 1, decimal =3D 0, sign =3D 0, signalling =3D 0, canonical =3D=
 0,
  uexp =3D 67108357, sig =3D {0, 0, 9295712554570040320}}
which is splited to following two doubles
u =3D {cl =3D 1, decimal =3D 0, sign =3D 0, signalling =3D 0, canonical =3D=
 0,
  uexp =3D 67108357, sig =3D {0, 0, 9295712554570039296}}
v =3D {cl =3D 1, decimal =3D 0, sign =3D 0, signalling =3D 0, canonical =3D=
 0,
  uexp =3D 67108304, sig =3D {0, 0, 9223372036854775808}}

After decoding and encoding, the memory representation changes. After PR954=
50
added a verification of decoding/encoding check, native_interpret_expr retu=
rns
a NULL tree for this case which causes ICE.

Shall we disable Hollerith constant for IBM long double(-mabi=3Dibmlongdoub=
le)?
Or just throw it to upper layer and let parser report an error? Please advi=
ce.=