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SubjectRe: [BUG] Bad #define, nonportable C, missing {}
The great thing about the C standard is that you don't have to know, or guess,
or remember. I respectfully suggest that all of those who are interested in
this topic buy a copy of K&R Second Edition (ISBN 0-13-110370-9), and read
chapter 2, particularly section 2.12 "Precedence and order of Evaluation".

And take this facinating topic off-line.

Sean

On Wed, Nov 21, 2001 at 02:37:38PM +0100, Jan Hudec wrote:
> > > *a++ = byte_rev[*a]
> > It looks perferctly okay to me. Anyway, whenever would you listen to a
> > C++ book talking about good C coding :p
>
> AFAIK the ANSI C specification explicitely claims, that it's not defined.
> The trick is, that the specification explicitly allows the compiler to
> choose wether it does the inc/dec right after/before the fetch, or at the
> begin/end of evaluation. Thus the second reference to a might return the
> original or incremented value at compiler's will.
>
> > Go read up on C operator precedence. Unary ++ comes before %, so if we
> > rewrite the #define to make it more "readable" it would be #define
> > MODINC(x,y) (x = (x+1) % y)
>
> *NO*
> MODINC(x,y) (x = (x+1) % y)
> is correct and beaves as expected. Unfortunately:
> MODINC(x,y) (x = x++ % y)
> is a nonsence, because the evaluation is something like this
> x++ returns x
> x++ % y returns x % y
> x is assigned the result and it's incremented IN UNDEFINED ORDER!!!
> AFAIK the ANSI C spec explicitly undefines the order.
>
> > > *a++ = byte_rev[*a];
> > C std says *always* evaluate from right to left for = operators, so this
> > will always make perfect sense.
> Yes, this should make perfect sense, but I fear the spec talks about
> operand used twice, once with side-efect generaly. So to be ANSI C
> correct, it's not good.
>
> -------------------------------------------------------------------------------
> - Jan Hudec `Bulb' <bulb@ucw.cz>
> -
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