Re: [HACKERS] "left shift of negative value" warnings

[Markus Nullmeier]

On 10/04/17 22:19, Andres Freund wrote:
> I guess the motivation is that it's not entirely clear what happens with
> the sign bit, when shifting.

Indeed, certain one's complement CPUs even "outlived" C99 by a small
margin, as it were:
http://mainframe.typepad.com/blog/2012/10/sad-day-unisys-abandons-its-mainframe-processors.html

And the respective ABI will be around for some time to come, apparently:
https://www.theregister.co.uk/2016/03/31/free_x86_mainframes_for_all_virtual_mainframes_that_is

:-)

-- 
Markus Nullmeierhttp://www.g-vo.org
German Astrophysical Virtual Observatory (GAVO)



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Re: [HACKERS] "left shift of negative value" warnings

[Andres Freund]

On 2017-04-10 15:25:57 -0400, Tom Lane wrote:
> Andres Freund  writes:
> > On 2017-04-09 19:20:27 -0400, Tom Lane wrote:
> >> 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?
> 
> Oh --- I misread the "nonnegative" as applying to the shift count, but
> you're right, it's talking about the LHS.  That's weird --- the E1 × 2^E2
> definition works fine as long as there's no overflow, so why didn't they
> define it like that?  It seems just arbitrarily broken this way.

I guess the motivation is that it's not entirely clear what happens with
the sign bit, when shifting.  Why they made that UB instead of
implementation defined, is a complete mystery to me, however.

We should do *something* about this?  The warnings are a bit annoying :(

- Andres


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Re: [HACKERS] "left shift of negative value" warnings

[Tom Lane]

Andres Freund  writes:
> On 2017-04-09 19:20:27 -0400, Tom Lane wrote:
>> 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?

Oh --- I misread the "nonnegative" as applying to the shift count, but
you're right, it's talking about the LHS.  That's weird --- the E1 × 2^E2
definition works fine as long as there's no overflow, so why didn't they
define it like that?  It seems just arbitrarily broken this way.

regards, tom lane


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Re: [HACKERS] "left shift of negative value" warnings

[Andres Freund]

On 2017-04-09 19:20:27 -0400, Tom Lane wrote:
> Andres Freund  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


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Re: [HACKERS] "left shift of negative value" warnings

[Tom Lane]

Andres Freund  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.

> 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.

regards, tom lane


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