R.E.Wolff@BitWizard.nl (Rogier Wolff) writes:
> > > Oh, by the way, ANSI defines the behaviour as undefined, as a compiler
> > > is allowed to recognize that (var <<1) >> 1 evaluates to "var" if done
> > > with arbitrary precision.
> >
> > No. For any unsigned type as wide or wider than `unsigned int', that
> > will clear the top bits. For narrower unsigned types it won't. For
> > signed types it is implementation defined.
>
> Hmm. I'm not too familiar with the exact wording of ANSI. However,
> they usually take the road where "more accuracy in intermediates is
> allowed", as that can normally only reduce the amount of trouble that
> you get into when things near the limit of the size you're using.
>From ANSI 6.3.7 "Bitwise shift operators":
$ The result of E1 << E2 is E1 left-shifted E2 bit positions;
$ vacated bits are filled with zeros. If E1 has an unsigned type,
$ the value of the result is E1 multiplied by the quantity, 2
$ raised to the power E2, reduce modulo ULONG_MAX+1 if E1 has type
$ unsigned long, UINT_MAX+1 if otherwise.
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