The nonexistent operator (along = v. == lines)

[Darren Morbey]

I've noticed in my writing C code that there is no such operator
as ^^ (which would be to ^ as || is to |).  I feel in this case
I *have* to write a macro for this *nonexistent* operator (you may
recall I wrote #define XOR(a,b) ((a)^(b)) ).  However...

The Pascal xor operator is <>, which follows since it has a Boolean
type on which its and & or operators can (and only will) operate.
Since C uses any non-zero value as 'true', it seems the only reason
to include && and || to avoid trouble with & and |.  The problem is,
there is no ^^ to avoid trouble with ^.

I would like to use one of the following three macros, but each has
its own particular problems.  I would appreciate any advice on which
one should be used.  I would like it to operate like && and ||.

1.    #define XOR(a,b) ( ( !(a) && (b) ) || ( (a) && !(b) ) )
      - Althought this is straight from the definition of XOR
        ( as (NOT A AND B) OR (A AND NOT B) ) it must evaluate
        a and b *twice* (obvious loss of flexibility)

2.    #define XOR(a,b) ( !(a) != !(b) )
      - This is meant to overcome C's acceptance of multiple TRUEs
        ( unlike Pascal's single TRUE value ).  This relies on the
        following statements being true:
        ( 0 == 0L ) (since a and b may yield different sized int's)
        ( !0 == !(0L) )
        ( !(0L) == !('\0') ) (which may happen!)
        etc.  Can such expressions be *guaranteed* true?

3.    #define XOR(a,b) ( (a) ? !(b) : (b) )  /* my favourite. */
      - This relies simply on one thing: in the conditional expression,
        do *both* !(b) and (b) get evaluated, regardless of which is
        returned?  Or is only the value to be returned evaluated.
(Final note: This *is* meant to be *portable*.  Be very careful.)

[ Darren Morby     +1 613 957 9281     {uunet|lsuc|fts1}!revcan!darren ]
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