Closed fincham closed 6 years ago
When a UserIDPacket is created using a bare string (e.g. no e-mail or comment) as the UID only the 2nd character from this ends up in the packet. Abbreviated example:
private_key = Crypto.PublicKey.RSA.generate(2048) ggp_private_key = OpenPGP.SecretKeyPacket(( Crypto.Util.number.long_to_bytes(private_key.n), Crypto.Util.number.long_to_bytes(private_key.e), Crypto.Util.number.long_to_bytes(private_key.d), Crypto.Util.number.long_to_bytes(private_key.p), Crypto.Util.number.long_to_bytes(private_key.q), Crypto.Util.number.long_to_bytes(private_key.u) )) pgp_uid = OpenPGP.UserIDPacket('Test')
Once signed etc the resulting UID packet ends up with an 'e' as the UID:
# python test_uid.py | gpg2 --list-packets :secret key packet: version 4, algo 1, created 1486694962, expires 0 skey[0]: [2048 bits] skey[1]: [17 bits] skey[2]: [2047 bits] skey[3]: [1024 bits] skey[4]: [1024 bits] skey[5]: [1024 bits] checksum: 4877 keyid: B99C600A629B4C38 :user ID packet: "e" :signature packet: algo 1, keyid B99C600A629B4C38 version 4, created 1486694963, md5len 0, sigclass 0x13 digest algo 8, begin of digest a1 29 hashed subpkt 2 len 4 (sig created 2017-02-10) hashed subpkt 27 len 1 (key flags: 20) hashed subpkt 16 len 8 (issuer key ID B99C600A629B4C38) data: [2048 bits]
Thanks!
When a UserIDPacket is created using a bare string (e.g. no e-mail or comment) as the UID only the 2nd character from this ends up in the packet. Abbreviated example:
Once signed etc the resulting UID packet ends up with an 'e' as the UID: