Padding Oracle
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CBC - Cipher Block Chaining
In CBC mode the previous encrypted block is used as IV to XOR with the next block:
![](https://book.hacktricks.xyz/~gitbook/image?url=https%3A%2F%2Fdefuse.ca%2Fimages%2Fcbc_encryption.png&width=768&dpr=4&quality=100&sign=b64f8495&sv=1)
To decrypt CBC the opposite operations are done:
![](https://book.hacktricks.xyz/~gitbook/image?url=https%3A%2F%2Fdefuse.ca%2Fimages%2Fcbc_decryption.png&width=768&dpr=4&quality=100&sign=c360e6cd&sv=1)
Notice how it's needed to use an encryption key and an IV.
Message Padding
As the encryption is performed in fixed size blocks, padding is usually needed in the last block to complete its length.
Usually PKCS7 is used, which generates a padding repeating the number of bytes needed to complete the block. For example, if the last block is missing 3 bytes, the padding will be \x03\x03\x03
.
Let's look at more examples with a 2 blocks of length 8bytes:
byte #0 | byte #1 | byte #2 | byte #3 | byte #4 | byte #5 | byte #6 | byte #7 | byte #0 | byte #1 | byte #2 | byte #3 | byte #4 | byte #5 | byte #6 | byte #7 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
P | A | S | S | W | O | R | D | 1 | 2 | 3 | 4 | 5 | 6 | 0x02 | 0x02 |
P | A | S | S | W | O | R | D | 1 | 2 | 3 | 4 | 5 | 0x03 | 0x03 | 0x03 |
P | A | S | S | W | O | R | D | 1 | 2 | 3 | 0x05 | 0x05 | 0x05 | 0x05 | 0x05 |
P | A | S | S | W | O | R | D | 0x08 | 0x08 | 0x08 | 0x08 | 0x08 | 0x08 | 0x08 | 0x08 |
Note how in the last example the last block was full so another one was generated only with padding.
Padding Oracle
When an application decrypts encrypted data, it will first decrypt the data; then it will remove the padding. During the cleanup of the padding, if an invalid padding triggers a detectable behaviour, you have a padding oracle vulnerability. The detectable behaviour can be an error, a lack of results, or a slower response.
If you detect this behaviour, you can decrypt the encrypted data and even encrypt any cleartext.
How to exploit
You could use https://github.com/AonCyberLabs/PadBuster to exploit this kind of vulnerability or just do
Ili kujaribu kama cookie ya tovuti ina udhaifu unaweza kujaribu:
Encoding 0 inamaanisha kwamba base64 inatumika (lakini zingine zinapatikana, angalia menyu ya msaada).
Unaweza pia kutumia udhaifu huu kuandika data mpya. Kwa mfano, fikiria kwamba maudhui ya cookie ni "user=MyUsername", kisha unaweza kubadilisha kuwa "_user=administrator_" na kuongeza mamlaka ndani ya programu. Unaweza pia kufanya hivyo ukitumia paduster
ukitaja -plaintext parameter:
Ikiwa tovuti ina udhaifu, padbuster
itajaribu moja kwa moja kubaini wakati kosa la padding linapotokea, lakini unaweza pia kuonyesha ujumbe wa kosa hilo ukitumia -error parameter.
Nadharia
Kwa muhtasari, unaweza kuanza kufungua data iliyosimbwa kwa kubashiri thamani sahihi ambazo zinaweza kutumika kuunda paddings tofauti. Kisha, shambulio la padding oracle litaanza kufungua byte kutoka mwisho hadi mwanzo kwa kubashiri ni ipi itakuwa thamani sahihi inayounda padding ya 1, 2, 3, n.k..
![](https://book.hacktricks.xyz/~gitbook/image?url=https%3A%2F%2F3018635170-files.gitbook.io%2F%7E%2Ffiles%2Fv0%2Fb%2Fgitbook-x-prod.appspot.com%2Fo%2Fspaces%252FCtGnfY5n0YCGcWtBkgwe%252Fuploads%252Fgit-blob-d80b715c6d845dde617ba3ab1d8d0387dd7c73b1%252Fimage%2520%28561%29.png%3Falt%3Dmedia&width=768&dpr=4&quality=100&sign=6dfe365b&sv=1)
Fikiria una maandiko yaliyosimbwa yanayochukua blocks 2 yaliyoundwa na byte kutoka E0 hadi E15. Ili kufungua block ya mwisho (E8 hadi E15), block nzima inapita kupitia "block cipher decryption" ikizalisha byte za kati I0 hadi I15. Hatimaye, kila byte ya kati inafanywa XOR na byte zilizopita zilizofichwa (E0 hadi E7). Hivyo:
C15 = D(E15) ^ E7 = I15 ^ E7
C14 = I14 ^ E6
C13 = I13 ^ E5
C12 = I12 ^ E4
...
Sasa, inawezekana kubadilisha E7
hadi C15
iwe 0x01
, ambayo pia itakuwa padding sahihi. Hivyo, katika kesi hii: \x01 = I15 ^ E'7
Hivyo, kupata E'7, inawezekana kuyakadiria I15: I15 = 0x01 ^ E'7
Ambayo inaturuhusu kuyakadiria C15: C15 = E7 ^ I15 = E7 ^ \x01 ^ E'7
Kujua C15, sasa inawezekana kuyakadiria C14, lakini wakati huu kwa kubashiri padding \x02\x02
.
Hii BF ni ngumu kama ile ya awali kwani inawezekana kukadiria E''15
ambayo thamani yake ni 0x02: E''7 = \x02 ^ I15
hivyo inahitajika tu kupata E'14
inayozalisha C14
inayolingana na 0x02
.
Kisha, fanya hatua hizo hizo kufungua C14: C14 = E6 ^ I14 = E6 ^ \x02 ^ E''6
Fuata mnyororo huu hadi ufungue maandiko yote yaliyosimbwa.
Ugunduzi wa udhaifu
Jisajili na ujiandikishe na akaunti hii. Ikiwa unafanya kuingia mara nyingi na kila wakati unapata keki ile ile, kuna uwezekano kuna kitu sijakamilika katika programu. Keki inayotumwa nyuma inapaswa kuwa ya kipekee kila wakati unapoingia. Ikiwa keki ni daima ile ile, kuna uwezekano itakuwa daima halali na hakuna njia ya kuifuta.
Sasa, ikiwa unajaribu kubadilisha keki, unaweza kuona unapata kosa kutoka kwa programu. Lakini ikiwa unafanya BF padding (ukitumia padbuster kwa mfano) unafanikiwa kupata keki nyingine halali kwa mtumiaji tofauti. Hali hii ina uwezekano mkubwa wa kuwa na udhaifu kwa padbuster.
Marejeleo
Learn & practice AWS Hacking:HackTricks Training AWS Red Team Expert (ARTE)
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HackTricks Training GCP Red Team Expert (GRTE)
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