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>
- BitAnd[True,False]