I am attempting to duplicate wireshark's packet capture decryption for a TLS HTTP session, where I control the private key of the server.

The cipher suite number is 0x00002f, TLS_RSA_AES_128_CBC_SHA

I wrote a utility that takes apart the packet captures using libpcap, and so far can match what the logs from wireshark say they have in terms of MAC keys, Write keys, and IV's; printing out the binary data as hex characters.

Visually, I can confirm that the keyblock and ciphertext are identical to what wireshark's log display.

However, after I perform the following decryption steps:

... decrypt pre_master_secret, generate keyblock
... initialize cipher context

if (!EVP_CipherInit_ex(&myevp, cipher, NULL, wKey, IV, 0))
    LogError("evp_cipherinit_ex:", ERR_error_string(ERR_get_error(), NUL));

dec = EVP_DecryptUpdate(&myevp, outb, outlen, data, datalen);
totalcount += outlen;

if (!dec) 
    LogError("evp_decryptupdate:", ERR_error_string(ERR_get_error(), NUL));

if (dec && blocklen != 1) {
    dec = EVP_DecryptFinal_ex(&myevp, outb + totalcount, outlen);
    totalcount += outlen;    

...get padding from last bit of outb, and remove padding from outb (plaintext)
...remove the MAC from the plaintext

The resultant plaintext I get is identical to what wireshark spits out in their logs, with the execption of the first block of decrypted data.

I know that in general, this is usually caused by not having a correct initialization vector when you begin the decryption. However, all my computed keyblock values seem correct (that is to say, they jive with what wireshark is spitting out).

here's the first few blocks of what wireshark shows...

47 45 54 20 2f 68 65 6c 6c 6f 2e 68 74 6d 6c 20
48 54 54 50 2f 31 2e 31 0d 0a 48 6f 73 74 3a 20
31 37 32 2e 31 36 2e 32 34 34 2e 31 39 36 0d 0a

and mine

e1 a2 f4 b3 2a 93 d1 c4 1b f4 60 d8 c9 92 03 25
48 54 54 50 2f 31 2e 31 0d 0a 48 6f 73 74 3a 20
31 37 32 2e 31 36 2e 32 34 34 2e 31 39 36 0d 0a

It's just that first block. I can verify before the Decryption step that the evp context has the correct original IV value and the current are there.

LogData("evp oiv", &myevp->evp.oiv, &myevp->evp.cipher->iv_len);
LogData("evp iv", &myevp->evp.iv, &myevp->evp.cipher->iv_len);

here's the one from wireshark

Client Write IV[16]:
8b 55 c0 c5 3c 77 dc 91 09 22 18 8b 0c 53 16 66

here's mine

evp iv[16]
8b 55 c0 c5 3c 77 dc 91 09 22 18 8b 0c 53 16 66

Likewise, the keys are the same. I don't think I need to illustrate those values, as the remaining blocks do decrypt correctly.

I've also checked to make sure that first block isn't gzipped, or whatever. I checked the wireshark code, and they are doing the same equivalent stuff I'm doing, albeit in libgpgcrypt.

Anybody ever seen anything like this?


1 Answer 1


For TLS the IV for the first packet is generated from the shared secrets; quoting the RFC 2246:

To generate the key material, compute

  key_block = PRF(SecurityParameters.master_secret,
                     "key expansion",
                     SecurityParameters.server_random +

until enough output has been generated. Then the key_block is partitioned as follows:


I suppose that you have generated these, and checked them? (You need 6 ·16 bytes in your cipher suite, and the last 32 are the IV's, one for each direction.)

Also, the IV for the next TLS-record is the last cipher text block of the previous record in the same direction. So you need to keep track of these as well. So the first IV is never sent, and you need to keep updating the IV as we get new packets in, or send new ones.

  • 2
    $\begingroup$ The last paragraph would be my guess, too. In TLS, the IV generated from the key block will be used for the Finished message at the end of the handshake, and the record for the "GET /..." will use as IV the last 16 bytes of that previous (encrypted) record. From TLS-1.1 onward, each record has its own transmitted IV, but for SSL-3.0 and TLS-1.0 you must use the last block from the previous record. $\endgroup$ Sep 17, 2011 at 14:13
  • $\begingroup$ The issue was, as Thomas pointed out, the last paragraph. I wasn't updating the IV from the Finished message. I knew this conceptually (update after last block), but somehow overlooked the idea that it spans across TLS messages themselves as well. Seems obvious in hindsight. Thanks! $\endgroup$
    – madczar
    Sep 19, 2011 at 12:00

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