From: Brian Candler <B.Candler@...>
Date: 2005-06-05T17:50:57+09:00
Subject: Re: Converting variable-length binary value

> how about:
> 
>   [int.to_s(16)].pack('H*')

That doesn't work if the number of hex digits is odd:

irb(main):006:0> 1234.to_s(16)
=> "4d2"
irb(main):007:0> [1234.to_s(16)].pack("H*")
=> "M "                                         # that's \x4d \x20
irb(main):008:0> ["04d2"].pack("H*")
=> "\004\322"                                   # that's \x04 \xd2
irb(main):009:0> 

> Also, if you aren't tied to this exact format, there's the
> BER-compressed integer option in #pack/#unpack, which handles
> variable-length integers in a nice way

As it happens, I'm unpacking BER. The length field in a BER-encoded element
is encoded as a straightforward N octets. See [**] below:

  def ber_read(io)
    blk = io.read(2)            # minimum: short tag, short length
    tag = blk[0] & 0x1f
    len = blk[1]

    if tag == 0x1f              # long form
      tag = 0
      while true
        ch = io.getc
        blk << ch
        tag = (tag << 7) | (ch & 0x7f)
        break if (ch & 0x80) == 0
      end
      len = io.getc
      blk << len
    end

    if (len & 0x80) != 0        # long form
      len = len & 0x7f
      raise "Indefinite length encoding not supported" if len == 0
      offset = blk.length
      blk << io.read(len)
      # is there a more efficient way of doing this?  [**]
      len = 0
      blk[offset..-1].each_byte { |b| len = (len << 8) | b }
    end

    offset = blk.length
    blk << io.read(len)
    return blk, [blk[0] >> 5, tag], offset
  end

The reason for this is so that I can read a DER element from a stream
(OpenSSL::ASL1::decode requires the data to be in memory first)

You'll notice that the long-form *tag* is encoded in the format you mention;
however I can't use unpack for that since the length isn't known up-front.

Regards,

Brian.