Re: overflow in snprintf() when printing INT64_MIN

Hi,

On 2018-09-27 20:18:12 -0400, Tom Lane wrote:
> Andres Freund <andres(at)anarazel(dot)de> writes:
> > I just noticed, while reviewing a patch that corrects overflow handing
> > in snprintf, that we don't correctly handle INT64_MIN in snprintf.c:
>
> Well, you still get the right answer, even if the "-value" is
> nominally undefined.

Right.

> > I suspect the best way to fix this, would be to instead do:
>
> > /* Handle +/- */
> > if (dosign && adjust_sign((value < 0), forcesign, &signvalue);
> > uvalue = -(uint64) value;
> > else
> > uvalue = (uint64) value;
>
> Hm, what does -x mean for an unsigned value? I'm not really
> convinced this is conceptually better.

6.2.5 (9): "... A computation involving unsigned operands can never
overflow, because a result that cannot be represented by the resulting
unsigned integer type is reduced modulo the number that is one greater
than the largest value that can be represented by the resulting type."

(unsigned)((int)-1) == 4294967295
-(unsigned)4294967295 == 1

I think that's well defined.

Greetings,

Andres Freund