If the unsigned quantity can be expressed as a signed quantity, then all is well. The only undefined behavior (implementation specific) is if it can't be. However, the conversion from a non-0 quantity to a 0 would be extremely unlikely. You'd get an unexpected signed value, but I can't imagine any compiler turning it into a 0!
On Jan 8, 2014, at 10:57 AM, Yann Ylavic <ylavic....@gmail.com> wrote: > On Wed, Jan 8, 2014 at 2:03 PM, Jim Jagielski <j...@jagunet.com> wrote: > > On Jan 7, 2014, at 3:15 PM, Jeff Trawick <traw...@gmail.com> wrote: > > > > +1 for APR trunk, +0.9 for future APR 1.6.x, -0.9 for APR 1.5.x... > > > > alternate opinions? > > > > As far as I know, C guarantees that > > if (a) > > is the same as > > if (a != 0) > > So I'm unsure of the need for this patch. > > The real question is, for an "unsigned a != 0", can "(int)a" become 0 after > an implicit (undefined) conversion? > > Regards, > Yann.
