RE: warning: right shift count >= width of type

["Dave Korn" <dk@artimi.com>]

> -----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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