# negative numbers and binary formats

Discussion in 'Ruby' started by Paul, Sep 22, 2004.

1. ### PaulGuest

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

Paul, Sep 22, 2004

2. ### Florian GrossGuest

Paul wrote:

Moin!
Does this help?

irb(main):032:0> [-7].pack("c").unpack("H*").first
=> "f9"
Regards,
Florian Gross

Florian Gross, Sep 22, 2004

3. ### Ara.T.HowardGuest

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
--
===============================================================================
| EMAIL :: Ara [dot] T [dot] Howard [at] noaa [dot] gov
| PHONE :: 303.497.6469
| A flower falls, even though we love it;
| and a weed grows, even though we do not love it.
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===============================================================================

Ara.T.Howard, Sep 22, 2004
4. ### MarkusGuest

So, are you wanting it in binary or in hex?

(base 16) instead of binary (base 2) you could write:

(a & 0xff).to_s(16)

for one byte values,

(a & 0xffff).to_s(16)

for two byte values, etc.

If you are wanting it in binary you would instead write:

(a & 0xff).to_s(2)
If you don't know at code-time how large the value will be, but can
determine it at run-time, you could write:

(a & (((1 << (8*n)) - 1)).to_s(16)

where n is the number of bytes in a and the expression involving a makes
a mask if the proper size. Alternatively, you could mess with the

("0"*8 + (a & 0xffffffff).to_s(16))[-n*2..-1]

which pads the result with zeros and then takes the least significant 2n
hexits (i.e, the bottom n bytes).

-- Markus

Markus, Sep 22, 2004
5. ### Mark HubbartGuest

well, if all you need is the hex part, here you go:

class Integer
def internal_hex
# choose a packing strategy
case self
when (-128..127) # char
[self].pack('c').unpack('C').first.to_s 16
when (-32768..32767) # short
[self].pack('s').unpack('S').first.to_s 16
when (-2147483648..2147483647) # long
[self].pack('l').unpack('L').first.to_s 16
else
"too big!" #
end
end
end

-7.internal_hex #=>"f9"

the hex string will be in big-endian (network) byte-order; that's the
way to_s(16) does it.

cheers
Mark

Mark Hubbart, Sep 22, 2004
6. ### Robert KlemmeGuest

Hm...

10:25:20 [source]: irb
irb(main):001:0> printf "%32.32b", -7
00000000000000000000000000001001=> nil
irb(main):002:0> RUBY_VERSION
=> "1.8.1"

robert

Robert Klemme, Sep 23, 2004
7. ### Robert KlemmeGuest

That's nice although it is somewhat limited regarding the size of values:
=> "a8"

After a bit experimenting I came up with this:
=> "f953040"

Florian, what do you think?

Kind regards

robert

Robert Klemme, Sep 23, 2004
8. ### Florian GrossGuest

Very nice, thank you. I wasn't aware of the range limit at first. Only
thing I would change is using .first instead of .shift. (For clarity)
More regards,
Florian Gross

Florian Gross, Sep 23, 2004
9. ### Robert KlemmeGuest

) I deliberately choose #shift in order to make the array a bit
smaller and remove all unnecessary references to the string - kind of GC
paranoid.

Regards

robert

Robert Klemme, Sep 23, 2004
10. ### MarkusGuest

[-7000000].pack("i").unpack("h*").shift.reverse.gsub(/^f+(?=f)/, '')
But:
1. the array can't be referenced after the first/shift returns
(since it was an anonymous link in a message chain), so the
whole array is subject to GC from that point
2. shift can be significantly slower than first
3. how do you know that shift isn't doing something like:

def Array.shift
result = @hypothetical_primitive_array[0]
@hypothetical_primitive_array = @hypothetical_primitive_array[1..-1]
result
end

...in which case you'd still have the (scavengable) copy around
just as if you'd done it yourself?

It only pays to be paranoid if you can trust yourself more than you can
trust "them".

-- Markus

Markus, Sep 23, 2004
11. ### Robert KlemmeGuest

LOL

Lession taken. Thanks! Of course you're right. Now I just need to find
out who is "me" and "them"...

robert

Robert Klemme, Sep 23, 2004