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RE: warning: right shift count >= width of type
- From: "Dave Korn" <dk at artimi dot com>
- To: "'Andrew Haley'" <aph at redhat dot com>
- Cc: "'Dale Johannesen'" <dalej at apple dot com>,<gcc at gcc dot gnu dot org>,"'Nathan Sidwell'" <nathan at codesourcery dot com>
- Date: Mon, 29 Nov 2004 18:33:51 -0000
- Subject: RE: warning: right shift count >= width of type
> -----Original Message-----
> From: gcc-owner On Behalf Of Andrew Haley
> Sent: 29 November 2004 17:11
> Dave Korn writes:
> > So my question is really "Given that it's undefined,
> which means that
> > whatever the compiler does is correct, and given that
> there's already code
> > in there to detect the situation and issue a warning,
> which probably means
> > that it would be very easy at such a point to replace the
> offending RTL with
> > (const_int 0), is there any specific reason why not to?"
>
> I think the idea is that
>
> a << n /* n == 32 */
>
> and
>
> a << 32
>
> should do the same thing. This seems IMO more helpful than
> optimizing away the shift.
Ah, well I can see that as a desirable goal (although who ever said
undefined behaviour had to produce the same results consistently across
different methods of invoking said undefined behaviour?) I suppose.
> No, not at all. The x86 processors interpret this as
>
> a << (n % 32)
>
> > but it's surely only an issue of bugward-compatibility:
> > mathematically, there's really no problem with right-shifting more
> > than the width of the integer, all that happens is that _all_ the
> > bits drop out the right-hand side and you're left with nothing.
>
> That's not what all hardware actually does with shift instructions.
>
> > ISTM reasonable that the result of a right-shift by 32 bits could
> > be assumed to be the same thing you get if you right-shift by 1 bit
> > 32 times....
>
> The chip designers don't agree.
I would argue that the x86 simply does not _provide_ a shift-by-32 bits
instruction, owing to that implicit modulo, any more than a RISC cpu with a
5-bit-wide field in the opcode to encode a shift amount does so.
I myself would want "(n >> 32)" to produce the same result as "((n >> 16)
>> 16)" and indeed "for (int i = 32; i > 0; i--, n >>= 1) ;", and it seems
to be generally agreed that the compiler would be at liberty to so do if it
wants to.
cheers,
DaveK
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