From: Ara.T.Howard@...
Date: 2004-09-23T03:44:34+09:00
Subject: Re: negative numbers and binary formats

On Wed, 22 Sep 2004, Paul wrote:

> Im trying to take a negative integer value, convert it to its binary
> equivalent, and save its hex value
>
> -7 -> 1111 1001 -> F9
>
>
> I was hoping to use sprintf, but:
>
> irb(main):017:0> b = sprintf("%4b" , a)
> => "..1001"
> irb(main):018:0>
>
> The .. that appear break any further processing. So my questions are:
>
> Why the dots, and what are they?
> How do I do what Im trying to do - given my -7 in the example may be a
> 1 byte, 2 byte or 4 byte value. ( Im sure its some magic with
> pack....)
>
> Thanks
>
> Paul

you have a couple of options:

the [] method of Fixnum returns the bit:

   harp:~ > cat a.rb
   class Fixnum
     def to_bin
       packed = [self].pack 'N'
       n_bytes = packed.size
       n_bits = n_bytes * 8
       s = ''
       n_bits.times{|bit| s << self[n_bits - bit].to_s}
       s
     end
   end

   p 42.to_bin
   p -7.to_bin

   harp:~ > ruby a.rb
   "00000000000000000000000000010101"
   "11111111111111111111111111111100"

but perhaps only printf will suffice?

   irb(main):003:0> printf "%32.32b", -7
   11111111111111111111111111111001=> nil

   irb(main):009:0> printf "%4.4o", -7
   7771=> nil

but i'm not exactly sure where you are headed with this.

the '...' above is just showing you the extension of the sign bit.

-a
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