Re: UUIDGEN in lisp

[Jesper Harder]

Brad Collins <brad@studiojungle.net> writes:

> But I really do need to use a proper UUID.
>
> I've been trying to see how it's been done by others but it's not
> easy stuff.  I still don't understand about high and low parts of a
> time stamp and how to convert them..

Well, the time-based UUID is annoying to generate -- it requires
system-wide storage of a seed.  I don't see how you can guarantee
that your generator will use the same storage as, say, the Perl
generator and whatever else might generate UUIDs on ms-windows.

This type of UUID also leaks your MAC address.

Random-based UUIDs are much nicer.  Below is an implementation.  It
uses /dev/urandom as a source of high quality random bits on
GNU/Linux, but I don't know if the built-in `random' in Emacs used on
other systems qualifies as a "cryptographic strength random number
generator".

(defun uuid ()
  "Generate a version 4 UUID."
  (let ((bytes (uuid-random)))
    (setf (nth 7 bytes)
          (logior #B01000000 (logand #B01111111 (nth 7 bytes))))
    (setf (nth 8 bytes)
          (logior #B01000000 (logand #B01001111 (nth 8 bytes))))
    (apply 'format
           
"%02x%02x%02x%02x-%02x%02x-%02x%02x-%02x%02x-%02x%02x%02x%02x%02x%02x"
           bytes)))

(defun uuid-random ()
  "Return a list of 16 random bytes."
  (if (file-readable-p "/dev/urandom")
      (let (((coding-system-for-read 'binary)))
        (mapcar 'identity
                (substring
                 (string-as-unibyte
                  (shell-command-to-string
                   "dd count=16 bs=1 < /dev/urandom"))
                 0 16)))
    (random t)
    (mapcar 'random (make-list 16 255))))