RSA with PyCrypto

[Tommy Lindgren]

"A.M. Kuchling" <amk at amk.ca> writes:

> On 11 Dec 2002 23:47:47 +0100, 
> 	Tommy Lindgren <tomyltomyl at linuxlinux.nu> wrote:

> You could do that, or A could sign it with their private key.
> Something like:
> 
> msg = 'blah blah'
> signed_msg = privkeyA.sign(msg)
> encrypted_msg = pubkeyB.encrypt(msg, '')

... and then B can verify the source with 

  pubkeyA.verify(decrypted_msg, signed_msg) 

?

My revised example:

from Crypto.PublicKey import RSA
from Crypto.Util.randpool import RandomPool
rpool = RandomPool()

privkeyA = RSA.generate(368, rpool.get_bytes)
privkeyB = RSA.generate(368, rpool.get_bytes)
pubkeyA = privkeyA.publickey()
pubkeyB = privkeyB.publickey()

msg = 'blah blah'
signed_msg = privkeyA.sign(msg, '') # Assumed the second parameter
encrypted_msg = pubkeyB.encrypt(msg, '')
decrypted_msg = privkeyB.decrypt(encrypted_msg)

print pubkeyA.verify(decrypted_msg, signed_msg) # Should return true
print pubkeyB.verify(decrypted_msg, signed_msg) # Should return false


The first verify always succeeds and returns '1' (as excepted). The
second sometimes returns '0' (as excepted) and sometimes throws the
exception 'Crypto.PublicKey.RSA.error: Plaintext too large'. Is this
a bug or should I just enclose the call in a try block? Or are the
parameters bad?

-- 
Tommy Lindgren  |  o y @ i u . u
41A942131CAA5C  | t m l l n x n
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