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Re: changing "configure" to default to "gcc -g -O2 -fwrapv ..."
- From: "Seongbae Park" <seongbae dot park at gmail dot com>
- To: "Gabriel Dos Reis" <gdr at integrable-solutions dot net>
- Cc: "Andrew Pinski" <pinskia at physics dot uc dot edu>, "Marc Espie" <espie at quatramaran dot ens dot fr>, "Richard Kenner" <kenner at vlsi1 dot ultra dot nyu dot edu>, gcc at gcc dot gnu dot org, richard dot guenther at gmail dot com
- Date: Fri, 29 Dec 2006 19:59:43 -0800
- Subject: Re: changing "configure" to default to "gcc -g -O2 -fwrapv ..."
- References: <200612300147.kBU1lODl026375@localhost.localdomain> <m31wmi3z1g.fsf@zeus.integrable-solutions.net>
On 30 Dec 2006 03:20:11 +0100, Gabriel Dos Reis
<gdr@integrable-solutions.net> wrote:
...
The C standard, in effect, has an appendix (Annex H) that was not
there in the C89 edition, and that talks about the very specific issue
at hand
H.2.2 Integer types
[#1] The signed C integer types int, long int, long long
int, and the corresponding unsigned types are compatible
with LIA-1. If an implementation adds support for the LIA-1
exceptional values ``integer_overflow'' and ``undefined'',
then those types are LIA-1 conformant types. C's unsigned
integer types are ``modulo'' in the LIA-1 sense in that
overflows or out-of-bounds results silently wrap. An
implementation that defines signed integer types as also
being modulo need not detect integer overflow, in which
case, only integer divide-by-zero need be detected.
which clearly says LIA-1 isn't a requirement - notice "if" in the
second setence.
H.1 makes it clear that the entire Annex H doesn't add any extra rule
to the language but merely describes what C is in regard to LIA-1.
H.2 doubly makes it clear that C as it defined isn't LIA-1
- again, notice "if" in H.2p1.
The second sentence of H.3p1 confirms this again:
C's operations are compatible with LIA−1 in that C
allows an implementation to cause a notification to occur
when any arithmetic operation
returns an exceptional value as defined in LIA−1 clause 5.
i.e. "compatible" means C's definition doesn't prevent
a LIA-1 conformant implementation.
In other words, all LIA-1 comformant compiler is conformant to C99
in terms of arithmetic and types.
However, not all C99 conformant compiler aren't LIA-1 conformant.
C isn't conformant to LIA-1 but merely compatible,
exactly because of the undefined aspect.
That's enough playing a language laywer for me in a day.
--
#pragma ident "Seongbae Park, compiler, http://seongbae.blogspot.com";