On 2017-04-09 19:20:27 -0400, Tom Lane wrote:
> Andres Freund <andres@anarazel.de> writes:
> > For a while I've been getting warnings like
> > /home/andres/src/postgresql/src/backend/utils/adt/inet_cidr_ntop.c: In function ‘inet_cidr_ntop_ipv6’:
> > /home/andres/src/postgresql/src/backend/utils/adt/inet_cidr_ntop.c:205:11: warning: left shift of negative value
[-Wshift-negative-value]
> > m = ~0 << (8 - b);
> > ^~
>
> I imagine forcing the LHS to unsigned would silence that, though you'd
> have to be careful that the sign extension (widening) happened before
> you changed the value to unsigned, in the int64 cases. It's a bit odd
> though that it seems to think ~0 is signed.
Hm, why's that odd? By default integers are signed, and ~ doesn't change
signedness?
> > If I understand C99 correctly, the behaviour of a left-shift of a
> > negative number is undefined (6.5.7 4.).
>
> As I read that, it's only "undefined" if overflow would occur (ie
> the sign bit would change). Your compiler is being a useless annoying
> nanny, but that seems to be the in thing for compiler authors these
> days.
"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 × 2 E2 , reduced modulo
one more than the maximum value representable in the result type. If E1 has a signed
type and nonnegative value, and E1 × 2 E2 is representable in the result type, then that is
the resulting value; otherwise, the behavior is undefined."
As I read this it's defined iff E1 is signed, nonnegative *and* the the
result of the shift is representable in the relevant type. That seems,
uh, a bit restrictive, but that seems to be the only reading? Note that
the otherwise is preceded by a semicolon...
- Andres