Skip to Main Content

Java Security

Announcement

For appeals, questions and feedback about Oracle Forums, please email oracle-forums-moderators_us@oracle.com. Technical questions should be asked in the appropriate category. Thank you!

cryptix-openpgp - (public/private) key format

843811Jun 17 2010 — edited Jun 18 2010
hi,
first of all - this is my 'first time' with cryptography. i must encrypt/decrypt messages using PGP. i've chosen cryptix-openpgp.
To this library, some examples are added (for encryption/decryption). As i did not have keys, i run GenerateAndWriteKeys class (which is added in examples) to generate keys. Then when i used Encrypt/Decrypt clasess (from examples) i could easily encrypt/decrypt messages.

The problem is the key format. i've noticed that public key is constructed like this:

1) header "-----BEGIN PGP PUBLIC KEY BLOCK-----\r\n
Version: Cryptix OpenPGP 0.20050418\r\n\r\n"

2) after that, 20 blocks of 64 signs. each block is ended with "\r\n" (to actually each line has length = 66)

3) 21st line has 60signs, ended with "\r\n"

4) Last line has 5 signs, it begins with '=' (CRC?) and after that it has -----END PGP PUBLIC KEY
BLOCK-----


So, if i use key: "ujyhjyt965tyhgfrt9656ujhgt967....."(so on) it won't work! cause it is not digited to blocks of 64 ended with "\r\n".

Similar situation is with private key.

My questions are:

1. Can i use cryptix-openpgp to encrypt/decrypt data using a key in a format of a signs row which are not digited? Or do i need to format it to the way i've described?

2. Are this last 5 signs (which begins with '=') CRC? Does every key ends with CRC?

3. What should be the proper key format to use cryptix-openpgp?

Please, don't answer me 'use BouncyCastle', as this question is about cryptix-openpgp.


regards and thx for all suggestions
Comments
Locked Post
New comments cannot be posted to this locked post.
Post Details
Locked on Jul 16 2010
Added on Jun 17 2010
8 comments
705 views