Re: BitAnd[True,False]

*To*: mathgroup at smc.vnet.net*Subject*: [mg131740] Re: BitAnd[True,False]*From*: "Ernst H.K. Stelzer" <ernst.stelzer at physikalischebiologie.de>*Date*: Thu, 26 Sep 2013 03:44:24 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <20130925063645.E297C6A74@smc.vnet.net>

BooleanTable[{p, q, And[p, q]}, {p, q}] // TableForm True True True True False False False True False False False False BitAnd operates on the bits in an integer {#1, #2, BitAnd[#1, #2]} & @@@ Tuples[{0, 1}, {2}] // TableForm 0 0 0 0 1 0 1 0 0 1 1 1 Greetings Ernst -----Original Message----- From: Alan [mailto:alan.isaac at gmail.com] Sent: Wednesday, 25 September, 2013 08:37 AM To: mathgroup at smc.vnet.net Subject: [mg131740] BitAnd[True,False] I'd hoped BitAnd would work on Boolean Lists so I gave it a try. If it had simply failed completely I'd be disappointed but would understand. But oddly, it half succeeds. Why? BitOr fails the same way, and BitNot always fails. I just want to understand these results. I know there are ways to get the output I want. Thanks, Alan Isaac (using Mathematica 9) In[77]:= BooleanTable[{p,q,BitAnd[p,q]},{p,q}]//TableForm Out[77]//TableForm= True True True True False BitAnd[False,True] False True BitAnd[False,True] False False False

**References**:**BitAnd[True,False]***From:*Alan <alan.isaac@gmail.com>