On Mon, Sep 29, 2003, Dr Stephen Henson wrote:

> There isn't a command to do this but the standard OpenSSL S/MIME code does the
> search. If you look at around line 401 in pk7_doit.c you'll see a look which
> checks each RecipientInfo structure against each certificate and breaks out
> when it finds the right one or errors out if it can't. You should be able to
> adapt that to your needs.

Yes !
Thanks for your accurate answer Steve, I could get it to work !

Aside from issuer name and serial number, I also have other questions:

On Mon, Sep 29, 2003, I wrote:

> (I first wanted to encrypt the cleartext file with each of those 20 
> certificates files, and then to compare the MD5 checksum against the MD5 sum 
> of the "mysterious" encrypted file. I guess this doesnt work because it 2 
> successive encryptions of a given file with a given key seem to yield 
> different encrypted files)


This is what I actually did :

# openssl smime -encrypt -in cleartext -out test1 certificate.crt
# openssl smime -encrypt -in cleartext -out test2 certificate.crt
# openssl smime -encrypt -in cleartext -out test3 certificate.crt
# openssl smime -encrypt -in cleartext -out test4 certificate.crt
# openssl smime -encrypt -in cleartext -out test5 certificate.crt
# cp test1 copy1
# md5sum *


719ad63dbda4a607480ab8fa00c99a3b  copy1
<snip>
719ad63dbda4a607480ab8fa00c99a3b  test1
5e368e517c75d1307a23fc85076dc3b0  test2
b45ac4071a1b133a3505c906838ea3bd  test3
c539a908bb79792a8ff98e912efef7d9  test4
178dac41201d36379275dd2ee06b4498  test5


QUESTION 1:
So I wanted to ask why it is beneficial/necessary that "2 encryptions of a 
given file with a given key yield different encrypted files". And also how
is it achieved ?


# openssl smime -encrypt -in cleartext -out file certificate.crt

>From the man page, this command uses the RC2 40bit algorithm.
I have had a quick look at the RC2 description but it didn't mention any 
pseudo-random operation.
(http://www.ipa.go.jp/security/rfc/RFC2268EN.html)



QUESTION 2:
On a side note, the certificate was generated from a 2048-bit private key.
The encryption command didn't report any warning/error, so how does it handle
that key? ... 40bit RC2 with a 2048bit key ... I'm lost here.


Dave

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