Standard extensions (logical exclusive or)

[Joe Orost]

<>
>Joseph Orost suggests a new operator ^^ such that
>
>	a^^b
>
>would be equivalent to
>
>	a? (b?0:1): (b?1:0)
>
>What is wrong with writing
>
>	a!=b
>
>for this purpose?

Sorry, this is not the same.  Take, for instance, a=3 and b=4.

	a^b  = 7
	a!=b = 1
	a^^b = 0

Next take a=3 and b=0:

	a^b  = 3
	a!=b = 1
	a^^b = 1

Another expression for a^^b is (a==0)^(b==0).
Yet another one is (a!=0)^(b!=0).
And yes, you have to evaluate both a and b no matter what.

					regards,
					joe

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