to "OR" or not to "OR"

[Dale Worley]

Various posters say:

   >) I'm tired of Demorgan's law (a | b= !(a & b)) isnt there a way to get around 

   btw, isn't demorgan's law a|b = !(!a & !b)?

In this case, it's: a|b == ~(~a & ~b) and a||b == !(!a && !b).
Remember that |, &, and ~ are bitwise operators, while ||, &&, and !
are logical operators.

Dale Worley		Compass, Inc.			worley at compass.com
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