The output of command fortune | gpg -vaer <receipent-id> has a line that begins with:

gpg: RSA/AES256.CFB encrypted for ...

What is AES256.CFB doing here when the key is a RSA key? Isn't the data supposed to be encrypted with RSA given this is public key cryptography?


1 Answer 1


For Elgamal and RSA, it's only generally possible to encrypt a small amount of data. For ECDH and other DH variants, no data is encrypted at all; only a set of shared keys can be generated.

As a result, when we encrypt data with a public key cryptosystem, we usually use a hybrid method (called “hybrid encryption”), where the bulk of the data is encrypted with a symmetric key algorithm, and we either derive the symmetric key from a DH or ECDH approach or we derive the secret key randomly and encrypt it with the RSA or Elgamal public key.

In this case, what GnuPG is telling you is that this a hybrid of RSA for the public key and AES256 with CFB (and, presumably, the modification detection code). You may also see “OCB” instead of “CFB” if you're using a newer version of GnuPG. The data here is then encrypted with AES256 in the specified mode, and the symmetric key is encrypted with RSA.

  • $\begingroup$ It is probably worth mentioning this paradigm is called "Hybrid Encryption". Your answer includes these words separately, but it's a useful term to know to search on for things like this. $\endgroup$
    – Mark Schultz-Wu
    Commented Mar 8 at 23:48
  • $\begingroup$ Thanks for your answer. In newer gpg version uses "OCB"? In version 2.4.5 I still see "CFB". $\endgroup$
    – aackmann
    Commented Mar 9 at 11:17
  • 1
    $\begingroup$ If the key has a preference for supporting AEAD, then it will say “OCB” because that will be used. Otherwise, it will probably default to CFB. $\endgroup$
    – bk2204
    Commented Mar 9 at 16:15

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