comma operator

[Jay Maynard]

In article <564 at cresswell.quintus.UUCP>, ok at quintus.UUCP (Richard A. O'Keefe) writes:
> > In article <2382 at ihuxv.ATT.COM> rck at ihuxv.ATT.COM (R. C. Kukuk) writes:
> > >> In article <3887 at sigi.Colorado.EDU> swarbric at tramp.Colorado.EDU (SWARBRICK FRANCIS JOHN) writes:
> > >> >#define swap(a,b) ((a) = ((b) = ((a) = (a) ^ (b)) ^ (b)) ^ (a))
> > >This oldie fails when a == b.  Why?  Think about it.
> 	swap(a, a);
> sets a to 0, whatever its previous value was.  If you do
> 	swap(*p, *q);
> it can easily happen that p==q, in which case you get this problem.

Let's explode the macro, and see what it does:

p = 5; q = 5;
swap(p,q);

the swap macro becomes:
((p) = ((q) = ((p) = (p) ^ (q)) ^ (q)) ^ (p))

stripping redundant parentheses:
p = (q = (p = p ^ q) ^ q) ^ p

substituting variables, from the inside out:
p = (q = (p = 5 ^ 5) ^ q) ^ p     (assigns 0 to p, temporarily)
p = (q = 0 ^ 5) ^ p               (assigns 5 to q)
p = 5 ^ 0                         (assigns 5 to p)

This case still works. As pointed out before, though, this macro has other
problems.

-- 
Jay Maynard, K5ZC (@WB5BBW)...>splut!< | GEnie: JAYMAYNARD  CI$: 71036,1603
uucp: {uunet!nuchat,academ!uhnix1,{ihnp4,bellcore,killer}!tness1}!splut!jay
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