I am trying to write a brute-force dictionary attack on MS-CHAPv2 using python. The two pieces of data that I'm giving to the script are hex representations of the 16-byte Client Challenge Hash and the 24-byte Client Challenge Response. Using a dictionary file, I am generating the Responses using the MS-CHAPv2 algorithm:
Convert the client's password (dictionary entry) to little endian Unicode (16-bit).
Hash the converted password with MD4 --> NT_hash (16 bytes long).
Pad NT_hash with 0's to 21 bytes (add five 0's).
Split NT_hash into three 7-byte chunks ([:14], [14:28], and [28:]).
Add odd-parity values to each byte of each chunk, creating three 8-byte keys.
Encrypt the 16-byte Client Challenge Hash three times with DES, each time with a
different 8-byte key from step 5.
Concatenate the results from step 6 into a response to compare with the original
Client Challenge Response.
The problem I'm having is actually with the first two steps. I have tried several combinations of UTF encoding with various MD4 python implementations with some strange results. What is perplexing me is that in some combinations, I achieve the correct NT hash of some of my passwords, but not others. I'm using the following website to check my hashes:
http://www.arsitech.com/cryptography/windows/password/
Can anyone tell me exactly what hashing function I should be using and/or clarify the inconsistencies I'm discovering? My code (most of it) and example output (using a small dictionary of 11 words) are shown below if that helps. The last entry in the output computes the NT hash for the password "SecREt01", and it matches what this forum post has:
http://www.waraxe.us/ftopic-3900-days0-orderasc-30.html
...but none of the others are right. HELP!
SOURCE CODE
# NOTE: these values came from a user with the password "frog"
user = "user1"
CCH16 = "9c:27:2b:1f:3c:be:53:00"
CCH16 = HexToByte(CCH16)
CCR24 = "e9:cc:b4:9a:08:46:30:8b:4e:44:17:e7:cb:a9:af:4d:85:09:8b:89:4f:8a:18:d6"
CCR24 = CCR24.replace(':', ' ')
CCR24 = CCR24.upper()
print "CCR24 --> ", CCR24 # Client Challenge Response
# Initialize variables for loop
password = ""
password_found = False
# Find password
for guess in dictionary:
print "Guess ---> ", guess
# Create nt_hash for this guess using MD4 hashing algorithm.
# TRYING TO ENCODE guess
uGuess = guess.encode('utf-16le')
print "uGuess --> ", uGuess
byteUGuess = bytes(uGuess)
print "byteUGuess --> ", byteUGuess
hexByteUGuess = ByteToHex(byteUGuess)
print "hexByteUGuess --> ", hexByteUGuess
# VARIOUS MD4 IMPLEMENTATIONS
#nt_hash = MD4.new(hexByteUGuess).hexdigest()
#nt_hash = hashlib.new('md4', hexByteUGuess).hexdigest()
nt_hash = nthash.encrypt(guess)
#nt_hash = hashlib.new('md4', uGuess).hexdigest()
#nt_hash = hmac.new(digest, user.upper().encode('utf-16le')).hexdigest()
print " nt_hash = ", nt_hash
# Split nt_hash into three DES keys.
# Add the parity bits to the DES keys to make them 8-bytes each.
des_key_1 = HexToByte(addParity(nt_hash[0:14]))
des_key_2 = HexToByte(addParity(nt_hash[14:28]))
des_key_3 = HexToByte(addParity(nt_hash[28:] + "0000000000"))
print " des_key_1 = ", ByteToHex(des_key_1)
print " des_key_2 = ", ByteToHex(des_key_2)
print " des_key_3 = ", ByteToHex(des_key_3)
# Create DES encryption objects with keys.
des_1 = DES.new(des_key_1, DES.MODE_ECB)
des_2 = DES.new(des_key_2, DES.MODE_ECB)
des_3 = DES.new(des_key_3, DES.MODE_ECB)
# Calculate 24-byte Client Challenge Response for this guess
# with the DES objects and the 16-byte Client Challenge Hash.
ccr24_part1 = des_1.encrypt(CCH16)
ccr24_part2 = des_2.encrypt(CCH16)
ccr24_part3 = des_3.encrypt(CCH16)
ccr24_guess = ByteToHex(ccr24_part1 + ccr24_part2 + ccr24_part3)
print " ccr24_part1 = ", ByteToHex(ccr24_part1)
print " ccr24_part2 = ", ByteToHex(ccr24_part2)
print " ccr24_part3 = ", ByteToHex(ccr24_part3)
print " ccr24 --> ", ccr24_guess
print "CCR24 -----> ", CCR24, "\n"
# Compare the guess (ccr24_guess) with the original (CCR24).
if ccr24_guess == CCR24:
password_found = True
password = guess
break
# If the password was found, display it.
# Otherwise, display fail message.
if password_found == True:
print "Success - Password found!"
print " - PASSWORD --> ", password
else:
print "Failure - Dictionary exhausted..."
print " - Password not found."
EXAMPLE OUTPUT
CCR24 --> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> skate
uGuess --> s k a t e
byteUGuess --> s k a t e
hexByteUGuess --> 73 00 6B 00 61 00 74 00 65 00 0A 00
nt_hash = e4a8e60472df5891903f148b212e2100
des_key_1 = E5 54 38 C1 46 97 7C B0
des_key_2 = 91 C8 0E E3 49 58 85 5D
des_key_3 = 20 80 01 01 01 01 01 01
ccr24_part1 = AB C0 E5 72 51 8B 9C 23
ccr24_part2 = 23 1E 04 09 51 71 8D 35
ccr24_part3 = 8F 17 C3 5A 08 B8 DC 7B
ccr24 --> AB C0 E5 72 51 8B 9C 23 23 1E 04 09 51 71 8D 35 8F 17 C3 5A 08 B8 DC 7B
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> love
uGuess --> l o v e
byteUGuess --> l o v e
hexByteUGuess --> 6C 00 6F 00 76 00 65 00 0A 00
nt_hash = afe622205005339e583f020ced5596b7
des_key_1 = AE F2 89 45 04 80 15 67
des_key_2 = 9E 2C 0E E0 20 67 B5 AB
des_key_3 = 97 5B C1 01 01 01 01 01
ccr24_part1 = 97 51 2E AB 79 15 FD A6
ccr24_part2 = 5E 05 25 3D 9D 7C 92 D0
ccr24_part3 = 54 0D 1F D8 18 E9 2E E3
ccr24 --> 97 51 2E AB 79 15 FD A6 5E 05 25 3D 9D 7C 92 D0 54 0D 1F D8 18 E9 2E E3
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> dry
uGuess --> d r y
byteUGuess --> d r y
hexByteUGuess --> 64 00 72 00 79 00 0A 00
nt_hash = 45f15741ac05cfa965f3ee726f38c116
des_key_1 = 45 F8 54 E9 1A 61 16 9E
des_key_2 = A8 B3 7C 7C E6 92 BC 70
des_key_3 = C1 8A 80 01 01 01 01 01
ccr24_part1 = E0 E5 5C D3 5B 6B 0C 0B
ccr24_part2 = 16 18 F9 DE 53 59 AA BA
ccr24_part3 = 17 62 17 A1 93 88 2F 38
ccr24 --> E0 E5 5C D3 5B 6B 0C 0B 16 18 F9 DE 53 59 AA BA 17 62 17 A1 93 88 2F 38
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> music
uGuess --> m u s i c
byteUGuess --> m u s i c
hexByteUGuess --> 6D 00 75 00 73 00 69 00 63 00 0A 00
nt_hash = bf8704c94ab6b72ba63f9e8d3354c658
des_key_1 = BF C2 C1 98 94 54 DA 6E
des_key_2 = 2A D3 8F F2 E9 68 CD A8
des_key_3 = C7 2C 01 01 01 01 01 01
ccr24_part1 = 42 E1 2F FF D0 26 2F D9
ccr24_part2 = 3A E1 52 CD 98 D8 65 F2
ccr24_part3 = FC 0A E7 CC 99 E0 B2 26
ccr24 --> 42 E1 2F FF D0 26 2F D9 3A E1 52 CD 98 D8 65 F2 FC 0A E7 CC 99 E0 B2 26
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> german
uGuess --> g e r m a n
byteUGuess --> g e r m a n
hexByteUGuess --> 67 00 65 00 72 00 6D 00 61 00 6E 00 0A 00
nt_hash = 0964cddced56685197ed7271804a903a
des_key_1 = 08 B3 32 BA CE 6B 58 D0
des_key_2 = 51 CB FB AE 26 8C 01 94
des_key_3 = 91 1C 80 01 01 01 01 01
ccr24_part1 = 1A 4C 59 1E 9F 62 F2 EB
ccr24_part2 = BC B5 73 C1 7D A1 F5 3E
ccr24_part3 = 88 09 FB CD 38 EB 27 9F
ccr24 --> 1A 4C 59 1E 9F 62 F2 EB BC B5 73 C1 7D A1 F5 3E 88 09 FB CD 38 EB 27 9F
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> flag
uGuess --> f l a g
byteUGuess --> f l a g
hexByteUGuess --> 66 00 6C 00 61 00 67 00 0A 00
nt_hash = 98da2ab73404f97e65d877b3253e646d
des_key_1 = 98 6D 8A 57 73 A1 13 F2
des_key_2 = 7F 32 76 0E 7A 98 94 7C
des_key_3 = 64 37 40 01 01 01 01 01
ccr24_part1 = D7 F2 66 36 63 30 CB FE
ccr24_part2 = B3 74 C0 C0 D9 96 71 E8
ccr24_part3 = 61 2B 69 DE B5 7B 71 56
ccr24 --> D7 F2 66 36 63 30 CB FE B3 74 C0 C0 D9 96 71 E8 61 2B 69 DE B5 7B 71 56
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> frog
uGuess --> f r o g
byteUGuess --> f r o g
hexByteUGuess --> 66 00 72 00 6F 00 67 00 0A 00
nt_hash = 78d1b2ee58428afdb5fff389c40b0b90
des_key_1 = 79 68 6D 5D E5 C2 0B 15
des_key_2 = FD DA 7F FE 38 4F 10 16
des_key_3 = 0B C8 01 01 01 01 01 01
ccr24_part1 = 76 1D 5C 06 E9 9D 3D 69
ccr24_part2 = C0 44 26 C6 C3 7E F9 ED
ccr24_part3 = 95 9D EB 10 8D 67 8D 6E
ccr24 --> 76 1D 5C 06 E9 9D 3D 69 C0 44 26 C6 C3 7E F9 ED 95 9D EB 10 8D 67 8D 6E
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> horse
uGuess --> h o r s e
byteUGuess --> h o r s e
hexByteUGuess --> 68 00 6F 00 72 00 73 00 65 00 0A 00
nt_hash = ee23c8c25b9443759520c49b9b6a2756
des_key_1 = EF 10 F2 19 25 DC 51 86
des_key_2 = 75 CB 49 19 49 DC 6D D5
des_key_3 = 26 AB 80 01 01 01 01 01
ccr24_part1 = 50 83 2A A8 E0 96 D5 76
ccr24_part2 = 1E 79 0E 0F 46 1C F5 D9
ccr24_part3 = 15 B1 8A EF B5 5E 5B 86
ccr24 --> 50 83 2A A8 E0 96 D5 76 1E 79 0E 0F 46 1C F5 D9 15 B1 8A EF B5 5E 5B 86
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> word
uGuess --> w o r d
byteUGuess --> w o r d
hexByteUGuess --> 77 00 6F 00 72 00 64 00 0A 00
nt_hash = 3cd0adcea3231e4d6f4e313390e25347
des_key_1 = 3D 68 2A B9 EA 19 8C 3D
des_key_2 = 4C B6 D3 C7 13 9D 43 C4
des_key_3 = 52 A2 C1 01 01 01 01 01
ccr24_part1 = DB 21 74 37 87 A2 C8 B6
ccr24_part2 = 49 E0 04 3A 77 D0 2D A5
ccr24_part3 = 22 C9 DB F7 6D E8 5A C7
ccr24 --> DB 21 74 37 87 A2 C8 B6 49 E0 04 3A 77 D0 2D A5 22 C9 DB F7 6D E8 5A C7
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> dense
uGuess --> d e n s e
byteUGuess --> d e n s e
hexByteUGuess --> 64 00 65 00 6E 00 73 00 65 00 0A 00
nt_hash = 2e46b43c23ad709512f8b1898608ef02
des_key_1 = 2F 23 AD 86 C2 1C B5 E0
des_key_2 = 94 89 BF 16 19 4C 19 10
des_key_3 = EF 80 80 01 01 01 01 01
ccr24_part1 = CB C1 1E AB DC 27 58 67
ccr24_part2 = 09 25 34 A4 83 D0 C3 68
ccr24_part3 = 56 22 E7 84 45 B2 F3 6D
ccr24 --> CB C1 1E AB DC 27 58 67 09 25 34 A4 83 D0 C3 68 56 22 E7 84 45 B2 F3 6D
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Guess ---> SecREt01
uGuess --> S e c R E t 0 1
byteUGuess --> S e c R E t 0 1
hexByteUGuess --> 53 00 65 00 63 00 52 00 45 00 74 00 30 00 31 00
nt_hash = cd06ca7c7e10c99b1d33b7485a2ed808
des_key_1 = CD 83 B3 4F C7 F1 43 92
des_key_2 = 9B 8F 4C 76 75 43 68 5D
des_key_3 = D9 04 01 01 01 01 01 01
ccr24_part1 = E7 0A E5 FE 60 61 E9 ED
ccr24_part2 = E5 BB AF 19 78 E0 C0 A6
ccr24_part3 = 12 04 60 76 E6 1F D1 AB
ccr24 --> E7 0A E5 FE 60 61 E9 ED E5 BB AF 19 78 E0 C0 A6 12 04 60 76 E6 1F D1 AB
CCR24 -----> E9 CC B4 9A 08 46 30 8B 4E 44 17 E7 CB A9 AF 4D 85 09 8B 89 4F 8A 18 D6
Failure - Dictionary exhausted...
- Password not found.
Related
I have an array of bytes, they are a raw PNG file that was originally a base64 encoded string.
The encoded string:
stEncodedQR = "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";
The raw bytes:
(src\funcs.cpp:1064) DecodedQRPicture ptr:0x19E51C0 size:0x673
(:0) ---- ----------------------------------------------- ----------------
(:0) 000| 89 50 4E 47 0D 0A 1A 0A 00 00 00 0D 49 48 44 52 .PNG........IHDR
(:0) 010| 00 00 01 40 00 00 01 40 01 00 00 00 00 E5 EB 98 ...#...#.....ek.
(:0) 020| 07 00 00 06 3A 49 44 41 54 78 DA ED 9A 31 8E AB ....:IDATxZm.1.+
(:0) 030| 4A 10 45 0B 11 90 D9 1B 40 EA 6D 90 F5 96 F0 06 J.E...Y.#jm.u.p.
(:0) 040| 30 6C 00 6F 89 8C 6D 20 B1 01 C8 08 10 F5 CF 65 0l.o..m 1.H..uOe
(:0) 050| 22 7B 9E F4 7F 40 BF E0 CB D6 48 A3 B1 0F A3 56 "{.t.#?`KVH#1.#V
(:0) 060| 77 D5 AD 5B D5 36 FF 8F 2F FB 82 5F F0 0B 7E C1 wU-[U6../{._p.~A
(:0) 070| 2F F8 05 FF CF E0 62 F6 F4 79 29 CA C7 30 3D 3C /x..O`bvty)JG0=<
(:0) 080| 1C E6 47 0C FC 59 C7 DD 2C 24 05 57 DF 1F 63 E8 .fG.|YG],$.W_.ch
(:0) 090| DC BB 21 F7 2D 7F 6D 61 1D ED 31 4E CF B1 7C 7A \;!w-.ma.m1NO1|z
(:0) 0A0| 62 70 28 1B 9B 5B 0B EB B6 5B B1 3F 60 B5 F0 F2 bp(..[.k6[1?`5pr
(:0) 0B0| E1 65 5D 24 07 EB 68 F7 31 6F A3 F7 85 DD 8A FD ae]$.khw1o#w.].}
(:0) 0C0| BE 4D 75 51 DE FF 0A D8 14 B9 FB 5E 9B DD DD 7B >MuQ^..X.9{^.]]{
(:0) 0D0| E3 C5 A3 41 FF 20 35 C8 C9 0C 1C 0B 11 61 6C CF cE#A. 5HI....alO
(:0) 0E0| 73 9B 9E 83 82 82 C5 FE 3E C2 6B 41 E2 F1 31 96 s.....E~>BkAbq1.
(:0) 0F0| 7F FC F9 15 B8 17 83 BC 8E 6A 7E 8D 65 E6 F9 51 .|y.8..<.j~.efyQ
(:0) 100| F1 90 3D 87 7C 89 F3 9F D2 F5 62 90 D0 B3 6A EE q.=.|.s.Rub.P3jn
(:0) 110| DC 6A 0B E0 8D 4D 66 F9 3A 2A 33 FA 2A 31 18 E7 \j.`.Mfy:*3z*1.g
(:0) 120| A3 28 9F 23 41 31 1F 55 DE 8D D3 7D B0 A6 DA 33 #(.#A1.U^.S}0&Z3
(:0) 130| AD 3A 2D C8 6F A2 6F A9 90 9C FD 3E 5A 23 9C A0 -:-Ho"o)..}>Z#.
(:0) 140| D8 EB 6A EE 63 5A F0 E0 E4 CF 14 6C 2B 42 72 BF XkjncZp`dO.l+Br?
(:0) 150| 45 AB 15 8C 73 B7 B1 6D 21 29 B8 54 65 36 70 26 E+..s71m!)8Te6p&
(:0) 160| 46 DE DF F4 D0 DE 54 08 1E E9 38 2F 96 16 3C 8A F^_tP^T..i8/..<.
(:0) 170| BC 1B 50 9A D0 16 53 C3 7E 0C AC 2E B4 91 B3 B2 <.P.P.SC~.,.4.32
(:0) 180| 6C 4C 0C 1A 22 47 08 A0 7C BE 14 D3 7D 0B 3E D8 lL.."G. |>.S}.>X
(:0) 190| 73 2B 6F 31 6F CD 13 83 FA E9 46 36 A3 BC 0F 88 s+o1oM..ziF6#<..
(:0) 1A0| EE CF 26 CD AB 97 D9 16 92 82 0B B1 B0 51 E2 CC nO&M+.Y....10QbL
(:0) 1B0| 2A 43 7A D1 00 4E A6 2E F2 B6 40 0C 3C 2D 68 64 *CzQ.N&.r6#.<-hd
(:0) 1C0| 7F 78 6D B3 2B FF A8 F0 96 E9 CD 3D 43 6E 87 90 .xm3+.(p.iM=Cn..
(:0) 1D0| 14 E4 F7 8D E5 0C 94 97 F9 88 14 76 D3 02 8D 37 .dw.e...y..vS..7
(:0) 1E0| BD FB 14 80 8B C1 15 2B E1 EC 90 AF 04 42 25 15 ={...A.+al./.B%.
(:0) 1F0| CC C6 A9 B6 29 DB A6 F7 2C BC 1E F4 01 1F E1 AF LF)6)[&w,<.t..a/
(:0) 200| 21 3F A2 DD 0C ED E1 AC 76 72 02 B9 5D B7 B4 20 !?"].ma,vr.9]74
(:0) 210| C1 F8 F2 D2 0A EF 23 8B 9A 5B 8A 3C 92 2F C1 9B AxrR.o#..[.<./A.
(:0) 220| 78 33 29 B8 C4 09 E7 42 55 6F A8 E4 83 52 DF AA x3)8D.gBUo(d.R_*
(:0) 230| 7C 75 FC 45 38 52 83 C6 EA F2 C3 10 9B E9 56 4D |u|E8R.FjrC..iVM
(:0) 240| C4 42 6F DE A2 B8 3C F4 AE B8 97 83 18 D5 CC 31 DBo^"8<t.8...UL1
(:0) 250| AA 67 0C 56 E5 03 17 89 CB F0 BC 8F 96 6D 9E 14 *g.Ve...Kp<..m..
(:0) 260| 3C 8A 70 6E 86 76 E8 EE B2 70 4F 14 28 52 D8 79 <.pn.vhn2pO.(RXy
(:0) 270| 3F 24 05 DD 65 9A B2 91 92 1E 7A BC 33 69 31 90 ?$.]e.2...z<3i1.
(:0) 280| 88 53 1D CB 77 ED B9 1E 5C 87 9F 35 92 EE 2A EF .S.Kwm9.\..5.n*o
(:0) 290| 2B 4A CF 56 E9 4C C2 12 43 52 10 89 B5 78 2A AE +JOViLB.CR..5x*.
(:0) 2A0| F4 06 0D E0 64 F0 D1 25 19 69 A9 41 FB 49 F4 D0 t..`dpQ%.i)A{ItP
(:0) 2B0| 4A 65 A9 36 BB 92 80 DA CE DE 0C 9E 18 44 E4 64 Je)6;..ZN^...Ddd
(:0) 2C0| 1B 1F F8 88 28 6B F3 1A 26 1C DC AD C0 C0 A6 05 ..x.(ks.&.\-##&.
(:0) 2D0| D7 8D CE 8C FA 06 5B 5A CC 17 19 67 CE 84 D0 50 W.N.z.[ZL..gN.PP
(:0) 2E0| 6F 9A 14 3C 0A AB 2B 95 1A E5 E2 40 9B 22 0D 6E o..<.++..eb#.".n
(:0) 2F0| 0A 79 E7 A5 48 0B BA D3 83 86 D3 41 60 61 E6 36 .yg%H.:S..SA`af6
(:0) 300| 92 FD A1 AF C2 CB 65 DB 93 82 EB 66 34 61 34 67 .}!/BKe[..kf4a4g
(:0) 310| D9 A0 2D 59 C7 FD 3E 90 7F AA F0 0F 4F 0D 22 78 Y -YG}>..*p.O."x
(:0) 320| D2 75 56 67 67 16 D2 8B F7 71 47 F3 96 2A 2D 78 RuVgg.R.wqGs.*-x
(:0) 330| BA 18 AC AB 4E A6 A1 21 DE A8 36 38 29 1A 85 F0 :.,+N&!!^(68)..p
(:0) 340| 61 3E 2E 07 69 4A 5E 58 F5 02 5B 41 16 12 08 F8 a>..iJ^Xu.[A...x
(:0) 350| 38 82 02 1B 95 77 EF D2 7C 39 28 D7 46 E7 AD B2 8....woR|9(WFg-2
(:0) 360| 46 49 9F 5F 3E F3 5C 4D 72 14 A1 7B F7 8F 97 83 FI._>s\Mr.!{w...
(:0) 370| 2E 9F 4E 5C 94 8F B3 1D 6C 6C AF D5 2D ED 19 59 ..N\..3.ll/U-m.Y
(:0) 380| 58 A5 05 0F 4C 84 4D 4F 89 3D D9 80 A1 9B D1 1E X%..L.MO.=Y.!.Q.
(:0) 390| 7A 14 2B 76 8B 69 41 B4 87 D3 E8 25 00 6A 43 51 z.+v.iA4.Sh%.jCQ
(:0) 3A0| 41 3C 1D 11 F1 C2 D2 56 21 31 A8 0E 58 93 AD E2 A<..qBRV!1(.X.-b
(:0) 3B0| 74 EE 84 E7 98 2F 9A FD 4C D9 67 BA 5E 0C EA F3 tn.g./.}LYg:^.js
(:0) 3C0| 4D A7 E1 43 D9 60 DF C6 FC 28 E4 9A 79 BA 7F B7 M'aCY`_F|(d.y:.7
(:0) 3D0| 5C 97 83 72 CD 8E DE E4 CE AE 98 3C BB 56 8D 8D \..rM.^dN..<;V..
(:0) 3E0| D5 50 33 2D B8 A2 AC 83 86 97 6D E4 89 5C E1 2F UP3-8",...md.\a/
(:0) 3F0| E9 DD B1 93 1F 4D C5 E5 E0 A1 FA E6 64 1E 85 85 i]1..MEe`!zfd...
(:0) 400| 8D B9 69 7C 4B 3F 4A 72 F0 81 27 05 D7 91 22 33 .9i|K?Jrp.'.W."3
(:0) 410| 77 23 D2 4E 81 25 F3 24 7B D9 28 A3 B1 7C B8 E6 w#RN.%s${Y(#1|8f
(:0) 420| CB 41 2F 51 1A 04 1E F6 56 E5 3D A9 30 CE 2C F6 KA/Q...vVe=)0N,v
(:0) 430| 3E 96 75 62 D0 A5 79 41 73 4D D7 58 EB 9C 78 95 >.ubP%yAsMWXk.x.
(:0) 440| 52 DC D3 4E 26 06 39 19 F2 DE 6A 4C 04 6B AC 88 R\SN&.9.r^jL.k,.
(:0) 450| 44 4E C9 CE CA 13 92 82 C8 FC 3A 6A BE 25 ED D1 DNINJ...H|:j>%mQ
(:0) 460| 5C 8D EE 44 9E 82 26 69 FD 98 68 5E 0D 3A E1 A0 \.nD..&i}.h^.:a
(:0) 470| D9 03 45 46 16 E6 A8 66 57 AF 80 E2 FA FB BC E7 Y.EF.f(fW/.bz{<g
(:0) 480| 7A F0 50 29 9B A5 F4 63 79 97 97 44 8D C2 61 41 zpP).%tcy..D.BaA
(:0) 490| B3 AE 2A 2D E8 2A 29 B2 C9 28 3D 6E AE 93 A7 43 3.*-h*)2I(=n..'C
(:0) 4A0| 12 C8 48 0C 5D 5A 70 A1 DB DE 66 3E E9 39 8A CD .HH.]Zp![^f>i9.M
(:0) 4B0| EE 72 52 B2 6F AB C4 3E A4 05 23 91 48 6B 22 17 nrR2o+D>$.#.Hk".
(:0) 4C0| 49 A1 D3 1C D7 D4 25 60 A3 DA 22 2D 88 6B 5E AA I!S.WT%`#Z"-.k^*
(:0) 4D0| B3 2F 89 14 9C BC 1B 66 C2 B3 31 55 D7 F7 89 E6 3/...<.fB31UWw.f
(:0) 4E0| F5 E0 A2 FC 43 D7 95 0D 84 A1 8A DB 59 6A D4 30 u`"|CW...!.[YjT0
(:0) 4F0| 7D 4C 3E 2E 07 89 47 DF 9B 6A 6A A2 2E 6A 5E AE }L>...G_.jj".j^.
(:0) 500| FB A8 FB 38 53 6D 2C A6 05 71 CD D4 93 D6 66 17 {({8Sm,&.qMT.Vf.
(:0) 510| 82 73 9F CF 0E 89 0E D5 DB C4 E0 21 B9 3D 6F 20 .s.O...U[D`!9=o
(:0) 520| C7 DD 2A 45 25 E6 11 23 F9 18 7F 35 E2 57 83 4B G]*E%f.#y..5bW.K
(:0) 530| C4 B4 6A 66 5F C7 FC 1C 6B B1 64 82 62 6A 64 AC D4jf_G|.k1d.bjd,
(:0) 540| 3C 29 88 7F EC 0B DD 0E 51 57 EF 83 8C 55 2F C9 <)..l.].QWo..U/I
(:0) 550| A7 EC E0 6A 13 83 BA FD 23 E7 D0 78 D6 58 AA 29 'l`j..:}#gPxVX*)
(:0) 560| 94 E2 6A F6 F3 31 60 4B 00 E6 6D 35 F7 72 AF 14 .bjvs1`K.fm5wr/.
(:0) 570| F3 F3 86 6A D3 0D E4 13 57 B5 A5 05 35 BE 8D AA ss.jS.d.W5%.5>.*
(:0) 580| 2D 78 A8 26 DA 39 D8 23 46 E6 25 CE 1F 23 92 CB -x(&Z9X#Ff%N.#.K
(:0) 590| 41 BC 43 87 E2 6E 08 8F 86 6A BA 9D 38 A7 D7 9A A<C.bn...j:.8'W.
(:0) 5A0| 7C BC DB 99 CB 41 8D 51 9D ED D1 0D 24 6D E8 8B |<[.KA.Q.mQ.$mh.
(:0) 5B0| 84 C0 48 46 B6 0A D9 F3 A4 E0 8F F2 FD C4 45 A3 .#HF6.Ys$`.r}DE#
(:0) 5C0| AF 33 D0 99 FD 74 FF BF 26 9A 57 83 BA 6D F6 89 /3P.}t.?&.W.:mv.
(:0) 5D0| A6 1F 0F AB 9B CF 8D 8D 91 A7 63 AB 6A F3 A4 20 &..+.O...'c+js$
(:0) 5E0| EF 3D 37 3B BF 52 A1 AF 51 DC CE 39 6E 37 68 1C o=7;?R!/Q\N9n7h.
(:0) 5F0| F2 E9 9A 2F 07 95 76 34 46 F2 8F 7D 85 B3 90 79 ri./..v4Fr.}.3.y
(:0) 600| 7C 8E A8 82 7A 88 E4 E0 39 6D D2 2D 8D 5A 13 96 |.(.z.d`9mR-.Z..
(:0) 610| B6 DF 4C F3 9E 6E FB 0B A0 AE BB 5B DB 6B 09 7F 6_Ls.n{. .;[[k..
(:0) 620| EE E7 AC EB 16 D3 83 6A FD 71 2E FB 7D 0B BA 83 ng,k.S.j}q.{}.:.
(:0) 630| DD D4 89 A2 3A 1C CE 2B 31 78 7A 0A 8A 8C DD CC ]T.":.N+1xz...]L
(:0) 640| CF CE 7B 7E 0D 54 DA BC FB 75 35 74 39 F8 FD 76 ON{~.TZ<{u5t9x}v
(:0) 650| D8 17 FC 82 5F F0 0B 7E C1 2F F8 AF AF 7F 00 2C X.|._p.~A/x//..,
(:0) 660| B0 6E EF BE 03 93 F3 00 00 00 00 49 45 4E 44 AE 0no>..s....IEND.
(:0) 670| 42 60 82 B`.
But when I try to write a file.png with the raw bytes, it doesn't write anything and the created png is blank. Any ideas how can I work around this?
size_t szDecodedLength = 0;
unsigned char * stDecodedQR = base64_decode(encodedPictureBase64,inLenEncodedPicture, &szDecodedLength);
ofstream outPicture("./flash/file.png", std::ios::out | std::ios::binary | std::ios::trunc);
string outString(reinterpret_cast<char*>(stDecodedQR), szDecodedLength);
if( outPicture && !outPicture.is_open() ){
return;
}
outPicture.write( reinterpret_cast<char*> (stDecodedQR), szDecodedLength);
// outPicture << outString; I've also tried creating a string and doing this but result is the same.
outPicture.flush();
outPicture.close();
This code solves the issue, it looks like it might be permission on the device I'm working on. Tested below code in Visual Studio with C++ and it generates the wanted file(picture) correctly, is a QR code that I can scan with my smartphone.
size_t szDecodedLength = 0;
unsigned char* stDecodedQR = base64_decode(encodedPictureBase64, inLenEncodedPicture, &szDecodedLength);
const size_t szArr = szDecodedLength;
std::vector<unsigned char> vDecodedQR;
for (auto i = 0; i < szDecodedLength; i++) {
vDecodedQR.push_back(stDecodedQR[i]);
}
std::ofstream outPicture("mypng.png", std::ios::out | std::ios::binary | std::ios::trunc);
if (outPicture && !outPicture.is_open()) {
std::cerr << " Error creating PNG file" << std::endl;
return;
}
outPicture.write(reinterpret_cast<char*> (stDecodedQR), szDecodedLength);
outPicture.close();
since the 2000th years we were using a compression-lib called addCRYPT.dll from a company called littleBIGware for our Windows-apps. Lib and company doesn't exists anymore, but users of our apps created lots of files up to now which are compressed with this lib.
Now for a new version I don't want to use this old lib, but must still be able to decompress the existing data.
I tried Zip, GZip and some others, but with no success.
Does anybody now, how to decompress these data? Here is a hex-dump of the first 64 bytes of some compressed blocks. There is a kind of a pattern visible, but what?
015104: 40 9c 00 00 42 5a 68 39 31 41 59 26 53 59 63 98 00 a8 00 30 ba 70 00 7f ff fa 10 8f fb e0 51 7c 77 db ed 7d bb 92 35 77 63 91 75 89 5d 60 65 91 26 b0 bb 32 ed aa 14 b5 a5 35 8d 34 ae dd c6 fa ...
017640: 40 9c 00 00 42 5a 68 39 31 41 59 26 53 59 2d fc 84 c1 00 49 2e 70 00 7f f8 7a 10 ef fb e0 5a ee 3e 9f 1e da ed a5 68 16 d5 ad ad 6b 3d cf 5e a4 f7 6d d6 d3 6b 47 bb ab 76 39 34 d8 ad bd ec ea ...
015760: 40 9c 00 00 42 5a 68 39 31 41 59 26 53 59 a5 a4 8a 00 00 4a 8b f0 00 7f ff fa 14 cf fb e0 51 dd b6 d5 ad 5e d6 fa 9b b9 bb 1a 77 77 6e dd 54 80 52 aa e5 98 cd 6d ad 92 18 bb 7a cb af 53 b1 da ...
015336: 40 9c 00 00 42 5a 68 39 31 41 59 26 53 59 4f 3c 3e 6f 00 2d cd f0 00 7f ff fe 10 8f fb e0 53 9f 6f 66 cb ec f6 6f 64 a2 da da 61 8d 2a 85 6b 4d 04 80 03 2b 1b 20 16 d9 de cf 6a b3 d5 a0 32 52 ...
016240: 40 9c 00 00 42 5a 68 39 31 41 59 26 53 59 af b2 df 80 00 27 bd f0 00 7f fd fa 10 8f fb e0 57 bd 5e 9f 19 f7 67 02 35 b0 dd ba 5d 67 b3 72 ce cd db 7a eb ba ce bd de f6 af 0c d6 ba d7 57 dc 1c ...
I am looking for an example of LUC algorithm, but I can't find anything. I know that it is in Crypto++, but I don't know C++ too well to use it.
I look for an example of algorithm of LUC...
It kind of depends on what you want to do. You might want to browse luc.h to see some of the things Crypto++ offers for LUC. There's a LUCES, a LUCSS and a LUC_IES. The *ES is encryption scheme, the *SS is a signature scheme, and the *IES is an integrated encryption scheme (which includes a key agreement algorithm and mask function).
Generally speaking, LUC is a public key encryption system. Using it is like using any other public key encryption system offered by Crypto++. That's because all the public key encryption systems inherit from the same classes (more correctly, base interfaces). You can see the design in the comments for file pubkey.h.
$ grep -R LUCES *
...
typedef LUCES<OAEP<SHA> >::Decryptor LUCES_OAEP_SHA_Decryptor;
typedef LUCES<OAEP<SHA> >::Encryptor LUCES_OAEP_SHA_Encryptor;
And that's pretty much all you need, though you may not know it.
Here's the easier problem to solve. How do you perform RSA encryption in Crypto++?
$ grep -R RSAES *
...
typedef RSAES<PKCS1v15>::Decryptor RSAES_PKCS1v15_Decryptor;
typedef RSAES<PKCS1v15>::Encryptor RSAES_PKCS1v15_Encryptor;
typedef RSAES<OAEP<SHA> >::Decryptor RSAES_OAEP_SHA_Decryptor;
typedef RSAES<OAEP<SHA> >::Encryptor RSAES_OAEP_SHA_Encryptor;
If you find an RSAES_PKCS1v15_Decryptor or RSAES_OAEP_SHA_Decryptor example, you just copy/replace with LUCES_OAEP_SHA_Decryptor and it will work just fine. And if you find an RSAES_PKCS1v15_Encryptor or RSAES_OAEP_SHA_Encryptor example, you just copy/replace with LUCES_OAEP_SHA_Encryptor and it will work just fine.
You can find the examples of using RSAES_OAEP_SHA_Encryptor and RSAES_OAEP_SHA_Decryptor on the Crypto++ wiki page for RSA Encryption Schemes. Or you can use the ECIES examples at Elliptic Curve Integrated Encryption Scheme (remember, all the public key systems inherit from the same base interfaces, so they all have the same methods and you use them the same way).
This should get you started. It creates a private key, saves it, then creates a public key, and saves it.
try
{
AutoSeededRandomPool prng;
FileSink fs1("lucs-private.der", true);
FileSink fs2("lucs-public.der", true);
InvertibleLUCFunction params;
params.GenerateRandomWithKeySize(prng, 2048);
LUC::PrivateKey privateKey(params);
privateKey.DEREncode(fs1);
LUCES_OAEP_SHA_Decryptor decryptor(privateKey);
// ...
LUC::PublicKey publicKey(params);
publicKey.DEREncode(fs2);
LUCES_OAEP_SHA_Encryptor encryptor(publicKey);
// ...
}
catch(CryptoPP::Exception& ex)
{
cerr << ex.what() << endl;
}
If you don't want to use InvertibleLUCFunction, the do something like this to generate the key. Note: RSA has an InvertibleRSAFunction.
LUC::PrivateKey privateKey;
privateKey.Initialize(prng, 2048);
...
LUC::PublicKey publicKey(privateKey);
...
An here's yet another way to do it:
FileSink fs1("lucs-private.der", true);
FileSink fs2("lucs-public.der", true);
LUCES_OAEP_SHA_Decryptor decryptor;
decryptor.AccessKey().Initialize(prng, 2048);
decryptor.AccessKey().DEREncode(fs1);
...
LUCES_OAEP_SHA_Encryptor encryptor(decryptor);
encryptor.AccessKey().DEREncode(fs2);
...
And here's a dump of the private key created by the test program:
$ dumpasn1 lucs-private.der
0 662: SEQUENCE {
4 1: INTEGER 0
7 257: INTEGER
: 00 B8 7A CA 6A 61 D9 CF 2F D8 89 5C A4 7D 74 7B
: AC F5 10 4C 3D 95 BF DD 2E F5 4E E5 F4 20 CF CD
: 44 7F C7 27 41 48 6B 83 E0 7C D9 66 16 8D 54 36
: 97 B9 CE 2D 80 A6 F6 E5 25 87 83 6E B9 41 45 DC
: 2A EB EC 4E EC D9 C0 17 B4 E0 04 F0 58 61 60 F8
: 87 18 27 16 58 BA 56 4E DD 9B C8 CD 18 46 28 38
: A2 6A A6 14 36 D0 A6 FF 9C B8 A8 B5 0F 3A 11 B5
: 00 08 44 B3 31 58 AF 01 F8 57 17 E8 FC 68 B2 5F
: [ Another 129 bytes skipped ]
268 1: INTEGER 17
271 129: INTEGER
: 00 C8 DF 47 D0 B2 6F C2 1A E4 B7 E8 3D 12 BB FF
: 04 F7 34 40 A0 0E ED DC F7 24 7B D9 46 EE 10 C4
: D5 E2 9C 93 05 CF 13 53 40 F4 50 EC 1F 6D D7 33
: FF FF 46 42 88 8D FC F4 EE 7F 0C 8B 71 71 51 D2
: 3C 32 E3 9A 11 B7 D8 CF EA 10 B2 07 49 3F 93 CD
: A0 3F 71 A9 23 27 35 1F 6A C9 1D FE CE 24 75 33
: 8F 53 71 B9 0B DE BC 05 93 98 A3 EA 94 8E 04 B1
: 29 A1 4F 4C 82 34 7A 08 3A 0E 07 98 8B 00 30 D7
: 5B
403 129: INTEGER
: 00 EB 1B D0 EF 5C 0F FC FC B7 56 A7 70 8C AA B7
: A6 90 C8 1F AA AD A0 0B 66 E5 33 75 F2 BE 68 35
: 29 2E 57 AC E0 E0 C8 04 A7 C4 13 1D 10 30 8B 50
: 20 17 0C 83 A7 14 4A 7D 25 31 77 50 66 08 36 13
: BE 9D C0 4E F4 44 74 7A BB D2 92 D0 F7 AE 7C EB
: 8E 84 5C 27 61 2C C9 7A D1 D0 C5 A0 13 98 96 E3
: 76 CD B0 E7 E8 7E 4E 0A 2D 00 86 07 57 DB 8A 51
: 1E 59 76 EA 88 44 4D DA F3 D6 AB 75 CB A6 45 F3
: F3
535 128: INTEGER
: 2E 6A AA BA B4 E8 DD 11 2D 31 A4 D5 F7 08 AB E3
: 1A 9A 15 58 AE C8 59 BE C4 75 85 90 6D 5D A4 18
: 39 27 8F FF 1C 9A FD 0F 0C 29 05 98 9C 16 FE 84
: A4 5C 85 15 F7 98 E6 D5 5B 23 CA 2F A2 27 8A 00
: 6E B1 BB 02 6E 93 53 85 30 30 61 F5 1C 49 5D 19
: EF DF CD 6F 11 7C 6D DC AE F6 A2 06 53 BB 7E 03
: C3 E5 4E E9 59 E0 D8 5F C3 28 0E E0 17 5C 63 6E
: 8E A6 18 FC AD A5 9B 08 D1 8B 7B 28 9D E2 CF E2
: }
0 warnings, 0 errors.
I have to write a small authentication server that, when a client connects, sends a RSA public exponent and modulus to it, and the client returns the username and password encrypted with them. The part where I get the exponent and modulus works fine, the client receives them, and sends the encoded block back; that's not a problem. The problem is in decoding it: the RSA Decryptor always returns 0 as MaxPlaintextLength, and no matter how many different ways I try to move the data around, I can't get it to work.
I'm providing hex dumps of the different parts after the code.
// These are static in the namespace to keep the key loaded.
static CryptoPP::AutoSeededRandomPool rsaRng;
static CryptoPP::InvertibleRSAFunction rsaParameters;
static CryptoPP::RSA::PrivateKey rsaPrivateKey(rsaParameters);
// This is done when the client connects.
{
rsaPrivateKey.GenerateRandomWithKeySize(rsaRng, 1024);
// This is where GetPublicExponent() and GetModulus() are
// encoded and sent to the client. They are sent correctly
// and I receive the encrypted data in a char buffer.
}
// This runs when I receive the encrypted data, which starts
// at &dataBuffer[7] and has a length of rsaPayloadLen.
{
int rsaPlainSize;
byte *rsaBuffer;
rsaBuffer = new byte[rsaPayloadLen];
memcpy(&rsaBuffer[0], &dataBuffer[7], rsaPayloadLen);
CryptoPP::SecByteBlock rsaCypher(rsaBuffer, rsaPayloadLen);
CryptoPP::RSAES_OAEP_SHA_Decryptor rsaDecryptor(rsaPrivateKey);
// At this point I inspected rsaCypher and it does contain the
// expected binary data, and rsaCypher.size() returns 256.
rsaPlainSize = rsaDecryptor.MaxPlaintextLength(rsaCypher.size());
// rsaPlainSize is 0. I have no idea why!
}
Hex dumps of everything at the time of calling MaxPlaintextLength:
rsaPrivateKey.GetPublicExponent():
000000: 11 .
rsaPrivateKey.GetPrivateExponent():
000000: 10 7a fd fd 9e a9 72 8c c3 5d 5b 80 e8 f4 6f bc .z....r..][...o.
000010: bc 6a 7a 51 4f 9f af d3 e4 76 a5 4a 9d fe 17 37 .jzQO....v.J...7
000020: 03 cf 82 24 33 e2 a0 d8 97 26 0a 6b ac 9d b1 de ...$3....&.k....
000030: 39 d5 3a 93 aa 65 66 be 17 43 3c 00 20 77 68 0a 9.:..ef..C<. wh.
000040: ac 2f 77 1e b8 c4 7f 64 52 54 7c 17 54 b6 e6 a4 ./w....dRT|.T...
000050: 95 49 60 7b 7b 16 6a 41 72 54 03 a2 2d 3a 80 8b .I`{{.jArT..-:..
000060: aa 74 fa 77 22 5d 0a d9 81 b2 b2 48 01 db 43 e8 .t.w"].....H..C.
000070: 16 1c c4 c3 a6 bf 45 7e 90 d3 6a 37 10 40 9f 71 ......E~..j7.#.q
rsaPrivateKey.GetModulus():
000000: d2 20 26 61 a6 f0 74 82 ba e6 4e ab 9a 2c 90 a6 . &a..t...N..,..
000010: 62 4d 97 8c b7 34 01 cd a0 e8 bb 77 5e 67 a7 fd bM...4.....w^g..
000020: 70 95 bb 4d 95 89 82 c9 87 25 04 dc d8 da 9b d1 p..M.....%......
000030: 61 5e aa da bc 8c dd f7 a8 99 3d 01 9d f2 6e 89 a^........=...n.
000040: e4 75 ec 91 31 e9 86 f4 da 43 4a ca a4 66 6b 04 .u..1....CJ..fk.
000050: c2 c9 a1 18 1d fa 81 b0 6e ef a5 13 04 44 88 89 ........n....D..
000060: 42 41 be 9c 7c 77 75 96 50 07 70 ad eb 60 e5 05 BA..|wu.P.p..`..
000070: aa a8 d8 27 03 28 cf bb c7 f5 cb 0d b3 b3 96 7f ...'.(..........
rsaPrivateKey.GetPrime1():
000000: d7 9e af ac e4 04 42 e4 58 9c 39 19 0e 56 7c ef ......B.X.9..V|.
000010: b3 bf b6 26 73 25 d8 ab d7 5e d1 e0 56 49 ae 66 ...&s%...^..VI.f
000020: c4 d8 81 bc d0 be c2 ef f4 6a 09 72 ef 72 35 7e .........j.r.r5~
000030: 15 f4 f9 3b f8 be f9 3a a1 0d 3e d0 eb c8 34 11 ...;...:..>...4.
rsaPrivateKey.GetPrime2():
000000: f9 7a 0e 1c 9a 1b eb d1 67 f1 e3 88 1d f3 f1 62 .z......g......b
000010: 9f a2 5c cb 49 76 de 42 25 e1 a4 de ed 50 f3 2d ..\.Iv.B%....P.-
000020: c0 15 c3 70 b5 96 68 51 25 f7 06 24 e4 43 0d b8 ...p..hQ%..$.C..
000030: 7a c5 12 2c 7c 63 20 73 70 61 01 fe b8 b3 71 8f z..,|c spa....q.
Plain text buffer that was encrypted:
000000: 73 74 72 69 6e 67 62 75 66 66 65 72 00 00 00 00 stringbuffer....
000010: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000020: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000030: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000040: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ................
000050: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 ..............
rsaCypher:
000000: 0e 9e bd 34 10 16 98 a5 b8 e4 0c 9b 4f 23 71 6d ...4........O#qm
000010: af d6 e8 c1 4d 97 b9 32 cb 25 eb 01 fe 4f 5c 79 ....M..2.%...O\y
000020: 2d d8 32 c4 4f fa e9 2e 58 dd fd 37 7f 08 97 d8 -.2.O...X..7....
000030: 95 bb 6f 04 46 fa 83 77 05 01 43 75 ca be b4 4a ..o.F..w..Cu...J
000040: 60 f9 e7 4a 91 3d bc ac fb e9 41 f3 9d b7 df d3 `..J.=....A.....
000050: a7 03 80 3a 7f 35 98 46 ca 06 b1 f3 d1 7b 56 83 ...:.5.F.....{V.
000060: 1b 00 7d 97 59 39 be 46 d5 cf 6d 2c b3 a7 8e 30 ..}.Y9.F..m,...0
000070: 39 ca ca d5 59 a2 71 43 e7 7e 75 b3 3c d6 a3 a5 9...Y.qC.~u.<...
000080: aa 89 e3 e9 32 e1 a9 c1 a5 a8 f5 66 be 7f c9 ba ....2......f....
000090: 65 35 0f 61 a0 d4 fa c7 ac 8e 28 7c 39 26 3f 01 e5.a......(|9&?.
0000a0: 34 ad 82 69 5e c4 ab 92 48 47 42 04 02 48 79 c4 4..i^...HGB..Hy.
0000b0: 39 6e f2 2c 7c 19 71 34 36 38 1c eb c1 f2 33 f0 9n.,|.q468....3.
0000c0: 49 b9 7e bb c3 16 ed d7 f7 3e 10 a7 cc 2b 8c 31 I.~......>...+.1
0000d0: f1 17 c7 a5 49 ce dd a3 c6 e2 9c 3c 2f 37 e4 97 ....I......</7..
0000e0: ac b7 24 17 b3 f8 75 6f 2a 85 cb 23 7a e1 77 72 ..$...uo*..#z.wr
0000f0: 02 0b 90 28 9b 9b ff 5d 6f 9b 11 11 d3 8b dd 4b ...(...]o......K
rsaCypher.size(): 256
rsaDecryptor.MaxPlaintextLength(rsaCypher.size()): 0
I'm really completely at a loss and the only reason I've spent the whole weekend fighting this is because I'm going to need CryptoPP later for Blowfish and Adler32, otherwise I would have just used another RSA library.
RSAES_OAEP_SHA_Decryptor MaxPlaintextLength returning zero...
Your calculation of rsaPayloadLen is probably not correct. I would suggest you check the size of the ciphertext before you transmit it and after you recover it. I think its size is rsaPayloadLen in you code above.
From the comments in cryptlib.h:
//! maximum length of plaintext for a given ciphertext length
/*! \note This function returns 0 if ciphertextLength is not valid (too long or too short). */
virtual size_t MaxPlaintextLength(size_t ciphertextLength) const =0;
I cannot duplicate when I take plain text, encrypt it, and then pass ciphertext.size() into MaxPlaintextLength. I can duplicate if I pass ciphertext.size() - 1 into MaxPlaintextLength.
As a work around to get you a little further along, you might try using FixedMaxPlaintextLength(). It takes no arguments and only depends upon the public key length. You can use it to find the largest buffer based on the key. But you could fail on decryption.
You should probably show us your encryption code, too. You might not be sizing a buffer correctly.
The whole problem was the key size. The doc said I should use a 1024-bit key, but the other program was expecting a 1152-byte key; apparently that change was never documented. Everything works fine after changing a single line:
rsaPrivateKey.GenerateRandomWithKeySize(rsaRng, 1152);
Leaving this answer here in case someone else ever has this problem. Check the key sizes, and don't trust the documentation!
I'm working on understanding and drawing my own DLL for PDF417 (2d barcodes). Anyhow, the actual drawing of the file is perfect, and in correct boundaries of 32 bits (as monochrome result). At the time of writing the data, the following is a memory dump as copied from C++ Visual Studio memory dump of the pointer to the bmp buffer. Each row is properly allocated to 36 wide before the next row.
Sorry about the wordwrap in the post, but my output was intended to be the same 36 bytes wide as the memory dump so you could better see the distortion.
The current drawing is 273 pixels wide by 12 pixels high, monochrome...
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 7e d0 29 e8 14 f4 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0a 8e 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c db ce 6d 5f be 02 eb 00 00
Here is the code to WRITE the file out -- verbatim immediately at the time of the memory dump from above
FILE *stream;
if( fopen_s( &stream, cSaveToFile, "w+" ) == 0 )
{
fwrite( &bmfh, 1, (UINT)sizeof(BITMAPFILEHEADER), stream );
fwrite( &bmi, 1, (UINT)sizeof(BITMAPINFO), stream );
fwrite( &RGBWhite, 1, (UINT)sizeof(RGBQUAD), stream );
fwrite( ppvBits, 1, (UINT)bmi.bmiHeader.biSizeImage, stream );
fclose( stream );
}
Here's what ACTUALLY Gets written to the file.
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab a8 61 d7 18 ed 18 f7 a3 89 1c dd 70 86 f5 f7 1a 20 91 3b c9 27 e7 67 12 1c 68 ae 3c b7 3e 02 eb 00 00
00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb 00
00 00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02 eb
00 00 00 ab 81 4b ca 07 6b 9c 11 40 9a e6 0c 76 0d 0a fc a3 33 70 bb 30 55 87 e9 c4 10 58 d9 ea 0d 48 3e 02
eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c 0d 0a
8e 02 eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79 bc 5c
0d 0a 8e 02 eb 00 00 00 ab 85 7e d0 29 e8 14 f4 0d 0a 7a 05 3c 37 ba 86 87 04 db b6 09 dc a0 62 fc d1 31 79
bc 5c 0d 0a 8e 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00 00 ab 85 43 c5 30 e2 26 70 4a 1a f3 e4 4d ce 2a 3f 79 cd bc e6 de 73 6f 39 b7 9c
db ce 6d 5f be 02 eb 00 00
Notice the start of the distortion with the "0d" in the result from reading the file back in the 4th line, about the 15th byte over... Then, there are a few more staggered around which in total, skew the image off by 9 bytes worth...
Obviously, the drawing portion is working ok as everything remains properly aligned in memory for the 12 lines.
Shouldn't you open the file in a compound mode i.e. writable & binary as in wb+?
Notice the start of the distortion with the "0d"
That's ASCII code for Carriage Return (CR) -- added on some OSes with newline (where a newline is actually a sequence of CR/LF). This should go away once you start writing the output in binary mode.
Your code looks neat otherwise. Cheers!
Your 0x0A (\n) gets converted to DOS format 0x0D0A (\r\n), becouse you're write the file in text mode. Switch to binary mode.
I actually just did a similar thing in java (printing bmp data to a thermal receipt printer). There are a couple of things i want to share with you:
bmp image data != an image format from microsoft. the MS bitmap has about 54 bytes of header information before any image data. (i spent a day or two working on this before I realized the difference)
bmp image data reads left to right, top to bottom, with the most significant bit on the left.
make sure the barcode image has a bitdepth of 1. this means 1 bit = 1 pixel. hexidecimal "ab" is 10101011 in binary, those 8 pixels will be filled in accordingly.
if you have a barcode 36 bytes wide, the barcode resolution is 288 x 12, not 273 x 12. (36 * 8 = 288).
the image data should be 432 bytes in size (12 rows of 36 bytes).
i dont know what this means:
Anyhow, the actual drawing of the file is perfect, and in correct boundaries of 32 bits (as monochrome result).
monochrome means its either 1 color or another. the pixel (think bit) is either filled in or it isnt.
Hope this helps