From: Warren Brown <WBrown@...>
Date: 2004-09-23T04:01:55+09:00
Subject: Re: negative numbers and binary formats

Paul,

> 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?

    I'm not sure about the "why", but the "what" is fairly easy.  Think
of the ".." as "continue the first digit out to the left as far as you
need to".  For example, "%b" % -1 yields "..1", which means 1111,
11111111, 1111111111111111, or however many bits you need in your
representation of -1.  "%x" % -1 yields "..f", which means ff, ffff,
ffffffff, or however many hex digits you need to represent -1.  Does
this make sense?

> 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.

    The most straight-forward way of handling this would be to replicate
the digit following the ".." as many times as you need and remove the
"..".  The general form of this function would be:

def myformat(val,fmtstr,len)
    fill = "0"
    str = fmtstr % val
    if str =~ /^\.\./
        str = str[2..-1]
        fill = str[0..0]
    end
    str.rjust(len,fill)[-len..-1]
end

    Then myformat(-7,"%x",2) would yield "f9" and myformat(-7,"%x",4)
would yield "fff9".

    I hope this helps.

    - Warren Brown