J
Size: a a a
2020 May 15
PK
yeah
you have padding_length in your packets
PK
if they are decrypted, just read it and cut it tho
a
Ну вот это скрипт распарсил
a
молодец
a
теперь он знает
J
An encryption algorithm and a key will be negotiated during the key
exchange. When encryption is in effect, the packet length, padding
length, payload, and padding fields of each packet MUST be encrypted
with the given algorithm.
exchange. When encryption is in effect, the packet length, padding
length, payload, and padding fields of each packet MUST be encrypted
with the given algorithm.
yeah kex length can be divided by 8
a
рип полоумные рукожёпки
PK
yeah kex length can be divided by 8
whaaaat?
MO
Розумем, ама не знам да го говорам
Я такоже але ов е заjмливы експеримент језыкознавства
J
you have padding_length in your packets
but,can't send packets with random padding.
PK
Each packet is in the following format:
uint32 packet_length
byte padding_length
byte[n1] payload; n1 = packet_length - padding_length - 1
byte[n2] random padding; n2 = padding_length
byte[m] mac (Message Authentication Code - MAC); m = mac_length
uint32 packet_length
byte padding_length
byte[n1] payload; n1 = packet_length - padding_length - 1
byte[n2] random padding; n2 = padding_length
byte[m] mac (Message Authentication Code - MAC); m = mac_length
AA
Я такоже але ов е заjмливы експеримент језыкознавства
I understand this completely, with a bit of effort
PK
but,can't send packets with random padding.
yap, while using OpenSSH
AA
If I speak Macedonian, големи се шансите дека ќе ме разбереш
J
whaaaat?
I mean packet length
V
Я такоже але ов е заjмливы експеримент језыкознавства
Так, вот это уже непонятно
J
Each packet is in the following format:
uint32 packet_length
byte padding_length
byte[n1] payload; n1 = packet_length - padding_length - 1
byte[n2] random padding; n2 = padding_length
byte[m] mac (Message Authentication Code - MAC); m = mac_length
uint32 packet_length
byte padding_length
byte[n1] payload; n1 = packet_length - padding_length - 1
byte[n2] random padding; n2 = padding_length
byte[m] mac (Message Authentication Code - MAC); m = mac_length
yeah , padding is the problem.
a
I have a serious question now that you mentioned crypto. There's a CTF challenge asking me to provide a file and HMAC, and says the HMAC should be 16-byte (32 hex characters). I tried MD4 and MD5 HMAC, but it says "HMAC mismatch".