On Sun, Dec 05, 2010 at 05:14:20AM -0600, Peter Samuelson wrote:
>
> [Stefan Sperling]
> > > - is_zeros |= (*checksum)->digest[i];
> > > + is_nonzero |= ((char *)(*checksum)->digest)[i] = x1 << 4 | x2;
> >
> > At the very least, this needs parenthesis unless you want everyone
> > to pull out their copy of K&R to check operator precedence rules.
>
> Can do. x1 << 4 | x2 didn't seem too ambiguous to me, but I can see
> how it might be.
>
> > Can we do one assignment per line instead? Maybe that's easier to parse.
>
> Hmmm.
>
> ((char *)(*checksum)->digest)[i] = x1 << 4 | x2;
> is_nonzero |= ((char *)(*checksum)->digest)[i];
>
> ...The repeated expression is complex, so you have to visually verify
> that it is indeed identical. That seems harder, to me. Or did you
> mean:
>
> is_nonzero |=
> ((char *)(*checksum)->digest)[i] = x1 << 4 | x2;
>
> ...Which doesn't seem much easier to parse either.
What about something like this (pseudo code):
x = x1 << 4 | x2;
((char *)(*checksum)->digest)[i] = x;
is_nonzero |= x;
Stefan
