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<  Erlang  ~  bnot and Bit String Comprehension
sexmith
Posted: Mon Jan 26, 2009 4:11 am Reply with quote
Joined: 03 Jan 2008 Posts: 5 Location: Ohio USA
I am trying to use bnot within a Bit String Comprehension; but my logic fails every time.

First I am assuming that a bnot on 2#101 would yield
2#010 or 2. No it yields a -6. I am assuming the default is unsigned; yet is this a sign issue or most likely I have no clue as to what bnot does.

bxor works as expected. 2#101 bxor 2#111 yield 2#010.

bxor works as expected within Bit String Comprehension
see below:

io:format("~3.2.0B ~3.2.0B ~3.2.0B ~3.2.0B ~3.2.0B~n",[A,B,C,D,E] = binary_to_list(<< << (2#111 bxor Y) >> || <<Y:3>> <= <<0:3,1:3,2:3,3:3,4:3>> >>)).

111 110 101 100 011
ok

What am I missing about bnot.

R Sexmith
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mhenoch
Posted: Mon Jan 26, 2009 11:10 am Reply with quote
User Joined: 06 Nov 2008 Posts: 29
There is no difference between 2#101 and 2#00000101 (and so on), so bnot can't know which part of the number you want to invert. Try:
Code:
bnot 2#101 band 2#111

bnot has higher precedence than band, so the example works without parentheses.
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sexmith
Posted: Tue Jan 27, 2009 1:52 am Reply with quote
Joined: 03 Jan 2008 Posts: 5 Location: Ohio USA
What I hear your say is:

Erlang is a 8bit/byte machine. All bitwise operators (band bxor bor bnot bsl bsr) operate on 8bits.

Erlang's definition of a bit string is: an area of untyped memory which Erlang uses to store bit strings.
[my guess: in this memory area the pointer/offset is
at the bit level].

Ok in my example I capture 3 bits via Y:3.

Y:3 becomes Y for computing. So when Y:3 becomes Y it becomes a padded 8bits. The bitwise operator bxor worked for me because 2#111 becomes padded 8bits 00000111 coupled with the nature of xor where the Zeros do not affect the result.
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