From: tlaronde@polynum.com
To: Fans of the OS Plan 9 from Bell Labs <9fans@9fans.net>
Subject: Re: [9fans] BUG!!! in Plan9 compiler!
Date: Thu, 22 Apr 2010 23:15:51 +0200 [thread overview]
Message-ID: <20100422211551.GA987@polynum.com> (raw)
In-Reply-To: <20100422200740.49F7D5B73@mail.bitblocks.com>
Bingo! see below...
On Thu, Apr 22, 2010 at 01:07:40PM -0700, Bakul Shah wrote:
> On Thu, 22 Apr 2010 21:32:36 +0200 tlaronde@polynum.com wrote:
> > On Thu, Apr 22, 2010 at 03:08:40PM -0400, geoff@plan9.bell-labs.com wrote:
> > > What type is `smallnumber'?
> >
> > typedef unsigned char smallnumber;
> ^^^^^^^^
> Aha!
>
> > translated from Pascal:
> >
> > small_number=0..63;
>
> IIRC in C89 integer promotions rules changed. See 6.3.1.8
> (Usual arithmetic conversions)
>
> [Otherwise,]the integer promotions are performed on both
> operands. Then the following rules are applied to the
> promoted operands:
>
> If both operands have the same type, then no
> further conversion is needed.
>
> Otherwise, if both operands have signed integer
> types or both have unsigned integer types, the
> operand with the type of lesser integer conversion
> rank is converted to the type of the operand with
> greater rank.
>
> Otherwise, if the operand that has unsigned
> integer type has rank greater or equal to the rank
> of the type of the other operand, then the operand
> with signed integer type is converted to the type
> of the operand with unsigned integer type.
>
> >>>> Otherwise, if the type of the operand with signed
> integer type can represent all of the values of
> the type of the operand with unsigned integer
> type, then the operand with unsigned integer type
> is converted to the type of the operand with
> signed integer type.
>
> Otherwise, both operands are converted to the
> unsigned integer type corresponding to the type of
> the operand with signed integer type.
>
> Try this on both gcc and 8c (with suitable changes):
>
> #define N(i) atoi(v[i])
> int f(int x, int y, unsigned char z) { return (x + y + z) / 2; }
> int main(int c, char**v) { printf("%d\n", f(N(1), N(2), N(3))); }
I get:
gcc: -201
ken-cc: 2147483447 (2^31 - 201)
This is: signed long + signed long + unsigned char.
Do you mean that there is first promotion :
1) unsigned char is promoted to unsigned int (A6.1).
2) And since there is an arithmetic operator (/ or shift),
unsigned int value may exceeds signed long == signed int, the
signed long is converted to unsigned long (A6.2)?
Is it this?!!!!?
And when I do first assignment, there is only promotion (since no
operator is here). Yielding the correct value in x2, that is then
divided (or shifted) by 2, hence signed, and no problem?
Well, thanks for the lessons, yesterday, today and tomorrow...;)
--
Thierry Laronde <tlaronde +AT+ polynum +dot+ com>
http://www.kergis.com/
Key fingerprint = 0FF7 E906 FBAF FE95 FD89 250D 52B1 AE95 6006 F40C
next prev parent reply other threads:[~2010-04-22 21:15 UTC|newest]
Thread overview: 20+ messages / expand[flat|nested] mbox.gz Atom feed top
2010-04-22 15:29 tlaronde
2010-04-22 17:03 ` Bakul Shah
2010-04-22 17:36 ` tlaronde
2010-04-22 17:50 ` tlaronde
2010-04-22 19:08 ` geoff
2010-04-22 19:32 ` tlaronde
2010-04-22 20:07 ` Bakul Shah
2010-04-22 21:15 ` tlaronde [this message]
2010-04-22 21:26 ` tlaronde
2010-04-22 22:49 ` Bakul Shah
2010-04-23 7:42 ` tlaronde
2010-04-23 18:53 ` C H Forsyth
2010-04-23 18:51 ` tlaronde
2010-04-23 20:08 ` Bakul Shah
2010-04-23 20:46 ` ron minnich
2010-04-23 21:44 ` erik quanstrom
2010-04-23 22:34 ` erik quanstrom
2010-04-24 18:59 ` Bakul Shah
2010-04-24 21:47 ` Charles Forsyth
2010-04-25 0:31 ` erik quanstrom
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