Edgar Pettijohn wrote:
> thought I read somewhere that gpg creates version 4 packets.
True. But the version 4 public-key packet specification only tells you what
information will be contained in the packet, not the format used for the packet
header.
- fuzzy
On Apr 12, 2018 3:39 AM, Werner Koch wrote:
>
> On Thu, 12 Apr 2018 05:29, ed...@pettijohn-web.com said:
>
> > did a hexdump of the file and the first word is `99' which in binary
> > would be `10011001'. I was expecting to encounter `11000110'. I'm
>
> OpenPGP (RFC-4880) has
On Apr 12, 2018 2:30 AM, FuzzyDrawrings via Gnupg-users
wrote:
>
> Edgar Pettijohn wrote:
>
> > the first word is `99' which in binary would be
> > `10011001'. I was expecting to encounter `11000110'.
>
> You were expecting the packet header to be written in the "new"
Edgar Pettijohn wrote:
> the first word is `99' which in binary would be
> `10011001'. I was expecting to encounter `11000110'.
You were expecting the packet header to be written in the "new" format, but it
is actually written in the "old" format (indicated by it beginning with "10" vs
"11").
On Thu, 12 Apr 2018 05:29, ed...@pettijohn-web.com said:
> did a hexdump of the file and the first word is `99' which in binary
> would be `10011001'. I was expecting to encounter `11000110'. I'm
OpenPGP (RFC-4880) has several ways to encode a packet header. This
first byte is called the CTB
I'm trying to learn the pgp packet syntax. I created a new key with gpg2
--gen-key and then gpg2 --export > pubkey.key and then gpg2 --dearmor
pubkey.key. Which left me with a pubkey.key.gpg file. I then did a
hexdump of the file and the first word is `99' which in binary would be
`10011