Re: Re: c/7284: incorrectly simplifies leftshift followed by signed power-of-2 division

["Al Grant" <AlGrant at myrealbox dot com>]

The following reply was made to PR c/7284; it has been noted by GNATS.

From: "Al Grant" <AlGrant@myrealbox.com>
To: falk.hueffner@student.uni-tuebingen.de
Cc: nathan@gcc.gnu.org,
	algrant@acm.org,
	gcc-bugs@gcc.gnu.org,
	gcc-prs@gcc.gnu.org,
	nobody@gcc.gnu.org,
	gcc-gnats@gcc.gnu.org
Subject: Re: Re: c/7284: incorrectly simplifies leftshift followed by signed power-of-2 division
Date: Fri, 12 Jul 2002 16:50:21 +0000

 >Right, I just assumed it to be very unlikely that this was changed to
 >be undefined in C99.=20
 
 So did I, otherwise I would have checked it!
 
 >I don't have the C89 standard; could you perhaps
 >cite the passage that shows this was defined >behaviour in C89?
 
 3.3.7 (something else in ISO, maybe 6.3.7)
 
   The result of E1 << E2 is E1 left-shifted E2 bit
   positions; vacated bits are filled with zeros.
 
 It then goes on to say "If E1 has an unsigned type..." and notes that it is=
  equivalent to a multiplication.
 For signed types it says nothing more.
 
 Now if signed left-shift is defined at all, in terms
 of the representation, I don't see there's any lack of definition in "0x000=
 00080 left-shifted 24 bit positions", it is clearly 0x80000000 (of the sa=
 me type).  So it's defined unless the standard says otherwise, which only=
  C99 seems to.
 
 There's no reason given to treat it specially if it happens to represent a =
 negative signed number.  This is not a case of an ideal signed arithmetic=
  operation producing some result in the ideal integers that cannot be rep=
 resented in the particular signed type, so the undefinedness in that case=
  doesn't apply here.