PyCryptodome AES CBC encryption does not give desired output
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I'm trying to encrpyt & decrypt a simple text in AES with CBC mode in Python (2.7.14) with Pycryptodome (3.7.0)
Here's my code to attempt encryption:
from Crypto.Cipher import AES
from Crypto.Util import Padding
import base64
encryption_key = "1111111111111111111111111111111111111111111111111111111111111111".decode("hex")
text = "Test text"
text_padded = Padding.pad(text, AES.block_size)
iv = "0000000000000000"
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
encrypted = iv + cipher_enc
print encrypted
print base64.b64encode(encrypted)
print encrypted.encode("hex")
print base64.b64encode(encrypted).encode("hex")
And the outputs are
0000000000000000X???]????H?
MDAwMDAwMDAwMDAwMDAwMFje9RzRXc3LHt8GBBLTSPQ=
3030303030303030303030303030303058def51cd15dcdcb1edf060412d348f4
4d4441774d4441774d4441774d4441774d4441774d466a6539527a525863334c4874384742424c545350513d
But when I enter the same key, text and initial vector values to http://aes.online-domain-tools.com/, I got different results.
Output is : 6a56bc5c0b05892ae4e63d0ca6b3169b
Here's the screenshot:

What am I doing wrong? How can I get the output value at the online encryption website by pycrypto?
python encryption pycrypto pycryptodome
add a comment |
I'm trying to encrpyt & decrypt a simple text in AES with CBC mode in Python (2.7.14) with Pycryptodome (3.7.0)
Here's my code to attempt encryption:
from Crypto.Cipher import AES
from Crypto.Util import Padding
import base64
encryption_key = "1111111111111111111111111111111111111111111111111111111111111111".decode("hex")
text = "Test text"
text_padded = Padding.pad(text, AES.block_size)
iv = "0000000000000000"
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
encrypted = iv + cipher_enc
print encrypted
print base64.b64encode(encrypted)
print encrypted.encode("hex")
print base64.b64encode(encrypted).encode("hex")
And the outputs are
0000000000000000X???]????H?
MDAwMDAwMDAwMDAwMDAwMFje9RzRXc3LHt8GBBLTSPQ=
3030303030303030303030303030303058def51cd15dcdcb1edf060412d348f4
4d4441774d4441774d4441774d4441774d4441774d466a6539527a525863334c4874384742424c545350513d
But when I enter the same key, text and initial vector values to http://aes.online-domain-tools.com/, I got different results.
Output is : 6a56bc5c0b05892ae4e63d0ca6b3169b
Here's the screenshot:

What am I doing wrong? How can I get the output value at the online encryption website by pycrypto?
python encryption pycrypto pycryptodome
add a comment |
I'm trying to encrpyt & decrypt a simple text in AES with CBC mode in Python (2.7.14) with Pycryptodome (3.7.0)
Here's my code to attempt encryption:
from Crypto.Cipher import AES
from Crypto.Util import Padding
import base64
encryption_key = "1111111111111111111111111111111111111111111111111111111111111111".decode("hex")
text = "Test text"
text_padded = Padding.pad(text, AES.block_size)
iv = "0000000000000000"
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
encrypted = iv + cipher_enc
print encrypted
print base64.b64encode(encrypted)
print encrypted.encode("hex")
print base64.b64encode(encrypted).encode("hex")
And the outputs are
0000000000000000X???]????H?
MDAwMDAwMDAwMDAwMDAwMFje9RzRXc3LHt8GBBLTSPQ=
3030303030303030303030303030303058def51cd15dcdcb1edf060412d348f4
4d4441774d4441774d4441774d4441774d4441774d466a6539527a525863334c4874384742424c545350513d
But when I enter the same key, text and initial vector values to http://aes.online-domain-tools.com/, I got different results.
Output is : 6a56bc5c0b05892ae4e63d0ca6b3169b
Here's the screenshot:

What am I doing wrong? How can I get the output value at the online encryption website by pycrypto?
python encryption pycrypto pycryptodome
I'm trying to encrpyt & decrypt a simple text in AES with CBC mode in Python (2.7.14) with Pycryptodome (3.7.0)
Here's my code to attempt encryption:
from Crypto.Cipher import AES
from Crypto.Util import Padding
import base64
encryption_key = "1111111111111111111111111111111111111111111111111111111111111111".decode("hex")
text = "Test text"
text_padded = Padding.pad(text, AES.block_size)
iv = "0000000000000000"
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
encrypted = iv + cipher_enc
print encrypted
print base64.b64encode(encrypted)
print encrypted.encode("hex")
print base64.b64encode(encrypted).encode("hex")
And the outputs are
0000000000000000X???]????H?
MDAwMDAwMDAwMDAwMDAwMFje9RzRXc3LHt8GBBLTSPQ=
3030303030303030303030303030303058def51cd15dcdcb1edf060412d348f4
4d4441774d4441774d4441774d4441774d4441774d466a6539527a525863334c4874384742424c545350513d
But when I enter the same key, text and initial vector values to http://aes.online-domain-tools.com/, I got different results.
Output is : 6a56bc5c0b05892ae4e63d0ca6b3169b
Here's the screenshot:

What am I doing wrong? How can I get the output value at the online encryption website by pycrypto?
python encryption pycrypto pycryptodome
python encryption pycrypto pycryptodome
edited Nov 15 '18 at 13:23
erdimeola
asked Nov 15 '18 at 13:05
erdimeolaerdimeola
769516
769516
add a comment |
add a comment |
1 Answer
1
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oldest
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first in python 3: python 3 is a lot stricter about bytes vs strings.
this reproduces the given example:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.hex())
print(iv.hex())
print(cipher_enc.hex())
# 1111111111111111111111111111111111111111111111111111111111111111
# 00000000000000000000000000000000
# 6a56bc5c0b05892ae4e63d0ca6b3169b
note that there is no need for encrypted = iv + cipher_enc; you are running AES in CBC mode already.
got it to run on python 2 as well:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.encode('hex'))
print(iv.encode('hex'))
print(cipher_enc.encode('hex'))
Thank you very much, My problem was not using the b'' notation and adding initial vector value again I suppose.
– erdimeola
Nov 15 '18 at 13:43
1
you are welcome! good luck with pycrypto!
– hiro protagonist
Nov 15 '18 at 13:44
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
first in python 3: python 3 is a lot stricter about bytes vs strings.
this reproduces the given example:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.hex())
print(iv.hex())
print(cipher_enc.hex())
# 1111111111111111111111111111111111111111111111111111111111111111
# 00000000000000000000000000000000
# 6a56bc5c0b05892ae4e63d0ca6b3169b
note that there is no need for encrypted = iv + cipher_enc; you are running AES in CBC mode already.
got it to run on python 2 as well:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.encode('hex'))
print(iv.encode('hex'))
print(cipher_enc.encode('hex'))
Thank you very much, My problem was not using the b'' notation and adding initial vector value again I suppose.
– erdimeola
Nov 15 '18 at 13:43
1
you are welcome! good luck with pycrypto!
– hiro protagonist
Nov 15 '18 at 13:44
add a comment |
first in python 3: python 3 is a lot stricter about bytes vs strings.
this reproduces the given example:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.hex())
print(iv.hex())
print(cipher_enc.hex())
# 1111111111111111111111111111111111111111111111111111111111111111
# 00000000000000000000000000000000
# 6a56bc5c0b05892ae4e63d0ca6b3169b
note that there is no need for encrypted = iv + cipher_enc; you are running AES in CBC mode already.
got it to run on python 2 as well:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.encode('hex'))
print(iv.encode('hex'))
print(cipher_enc.encode('hex'))
Thank you very much, My problem was not using the b'' notation and adding initial vector value again I suppose.
– erdimeola
Nov 15 '18 at 13:43
1
you are welcome! good luck with pycrypto!
– hiro protagonist
Nov 15 '18 at 13:44
add a comment |
first in python 3: python 3 is a lot stricter about bytes vs strings.
this reproduces the given example:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.hex())
print(iv.hex())
print(cipher_enc.hex())
# 1111111111111111111111111111111111111111111111111111111111111111
# 00000000000000000000000000000000
# 6a56bc5c0b05892ae4e63d0ca6b3169b
note that there is no need for encrypted = iv + cipher_enc; you are running AES in CBC mode already.
got it to run on python 2 as well:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.encode('hex'))
print(iv.encode('hex'))
print(cipher_enc.encode('hex'))
first in python 3: python 3 is a lot stricter about bytes vs strings.
this reproduces the given example:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.hex())
print(iv.hex())
print(cipher_enc.hex())
# 1111111111111111111111111111111111111111111111111111111111111111
# 00000000000000000000000000000000
# 6a56bc5c0b05892ae4e63d0ca6b3169b
note that there is no need for encrypted = iv + cipher_enc; you are running AES in CBC mode already.
got it to run on python 2 as well:
from Crypto.Cipher import AES
encryption_key = 32 * b'x11'
text = "Test text".encode()
text_padded = text + (AES.block_size - (len(text) % AES.block_size)) * b'x00'
iv = 16 * b'x00'
cipher = AES.new(encryption_key, AES.MODE_CBC, iv)
cipher_enc = cipher.encrypt(text_padded)
print(encryption_key.encode('hex'))
print(iv.encode('hex'))
print(cipher_enc.encode('hex'))
edited Nov 15 '18 at 13:39
answered Nov 15 '18 at 13:33
hiro protagonisthiro protagonist
20.7k74264
20.7k74264
Thank you very much, My problem was not using the b'' notation and adding initial vector value again I suppose.
– erdimeola
Nov 15 '18 at 13:43
1
you are welcome! good luck with pycrypto!
– hiro protagonist
Nov 15 '18 at 13:44
add a comment |
Thank you very much, My problem was not using the b'' notation and adding initial vector value again I suppose.
– erdimeola
Nov 15 '18 at 13:43
1
you are welcome! good luck with pycrypto!
– hiro protagonist
Nov 15 '18 at 13:44
Thank you very much, My problem was not using the b'' notation and adding initial vector value again I suppose.
– erdimeola
Nov 15 '18 at 13:43
Thank you very much, My problem was not using the b'' notation and adding initial vector value again I suppose.
– erdimeola
Nov 15 '18 at 13:43
1
1
you are welcome! good luck with pycrypto!
– hiro protagonist
Nov 15 '18 at 13:44
you are welcome! good luck with pycrypto!
– hiro protagonist
Nov 15 '18 at 13:44
add a comment |
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