Python AES-CBC implementation using AES-ECB
0
$begingroup$
I am looking for ways to make this code more "pythonic" and any issues with my implementation. This was designed to solve Crytopal's Challenge 10. This challenge involves recreating the AES-CBC cipher using a library-provided AES-ECB function.
from Cryptodome.Cipher import AES
def decrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.decrypt(data)
def encrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.encrypt(data)
def pkcs7(val, block_size=16):
remaining = block_size - len(val) % block_size
if remaining == block_size:
remaining = 16
ret = val + chr(remaining).encode() * remaining
return ret
def unpkcs7(val, block_size=16):
pad_amount = val[-1]
if pad_amount == 0:
raise Exception
for i in range(len(val) - 1, len(val) - (pad_amount + 1), -1):
if val[i] != pad_amount:
raise Exception
return val[:-pad_amount]
def decrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
prev_chunk = iv
decrypted =
for i in range(0, len(data), 16):
chunk = data[i : i + 16]
decrypted += xor(decrypt_aes_ecb(chunk, key), prev_chunk)
prev_chunk = chunk
if pad:
return unpkcs7(bytes(decrypted))
return bytes(decrypted)
def encrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
if pad:
padded = pkcs7(data)
else:
padded = data
prev_chunk = iv
encrypted =
for i in range(0, len(padded), 16):
chunk = padded[i : i + 16]
encrypted_block = encrypt_aes_ecb(xor(chunk, prev_chunk), key)
encrypted += encrypted_block
prev_chunk = encrypted_block
return bytes(encrypted)
python python-3.x programming-challenge cryptography
asked 14 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
$begingroup$
I am looking for ways to make this code more "pythonic" and any issues with my implementation. This was designed to solve Crytopal's Challenge 10. This challenge involves recreating the AES-CBC cipher using a library-provided AES-ECB function.
from Cryptodome.Cipher import AES
def decrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.decrypt(data)
def encrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.encrypt(data)
def pkcs7(val, block_size=16):
remaining = block_size - len(val) % block_size
if remaining == block_size:
remaining = 16
ret = val + chr(remaining).encode() * remaining
return ret
def unpkcs7(val, block_size=16):
pad_amount = val[-1]
if pad_amount == 0:
raise Exception
for i in range(len(val) - 1, len(val) - (pad_amount + 1), -1):
if val[i] != pad_amount:
raise Exception
return val[:-pad_amount]
def decrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
prev_chunk = iv
decrypted =
for i in range(0, len(data), 16):
chunk = data[i : i + 16]
decrypted += xor(decrypt_aes_ecb(chunk, key), prev_chunk)
prev_chunk = chunk
if pad:
return unpkcs7(bytes(decrypted))
return bytes(decrypted)
def encrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
if pad:
padded = pkcs7(data)
else:
padded = data
prev_chunk = iv
encrypted =
for i in range(0, len(padded), 16):
chunk = padded[i : i + 16]
encrypted_block = encrypt_aes_ecb(xor(chunk, prev_chunk), key)
encrypted += encrypted_block
prev_chunk = encrypted_block
return bytes(encrypted)
python python-3.x programming-challenge cryptography
asked 14 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
0
0
$begingroup$
I am looking for ways to make this code more "pythonic" and any issues with my implementation. This was designed to solve Crytopal's Challenge 10. This challenge involves recreating the AES-CBC cipher using a library-provided AES-ECB function.
from Cryptodome.Cipher import AES
def decrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.decrypt(data)
def encrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.encrypt(data)
def pkcs7(val, block_size=16):
remaining = block_size - len(val) % block_size
if remaining == block_size:
remaining = 16
ret = val + chr(remaining).encode() * remaining
return ret
def unpkcs7(val, block_size=16):
pad_amount = val[-1]
if pad_amount == 0:
raise Exception
for i in range(len(val) - 1, len(val) - (pad_amount + 1), -1):
if val[i] != pad_amount:
raise Exception
return val[:-pad_amount]
def decrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
prev_chunk = iv
decrypted =
for i in range(0, len(data), 16):
chunk = data[i : i + 16]
decrypted += xor(decrypt_aes_ecb(chunk, key), prev_chunk)
prev_chunk = chunk
if pad:
return unpkcs7(bytes(decrypted))
return bytes(decrypted)
def encrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
if pad:
padded = pkcs7(data)
else:
padded = data
prev_chunk = iv
encrypted =
for i in range(0, len(padded), 16):
chunk = padded[i : i + 16]
encrypted_block = encrypt_aes_ecb(xor(chunk, prev_chunk), key)
encrypted += encrypted_block
prev_chunk = encrypted_block
return bytes(encrypted)
python python-3.x programming-challenge cryptography
asked 14 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am looking for ways to make this code more "pythonic" and any issues with my implementation. This was designed to solve Crytopal's Challenge 10. This challenge involves recreating the AES-CBC cipher using a library-provided AES-ECB function.
from Cryptodome.Cipher import AES
def decrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.decrypt(data)
def encrypt_aes_ecb(data, key):
cipher = AES.new(key, AES.MODE_ECB)
return cipher.encrypt(data)
def pkcs7(val, block_size=16):
remaining = block_size - len(val) % block_size
if remaining == block_size:
remaining = 16
ret = val + chr(remaining).encode() * remaining
return ret
def unpkcs7(val, block_size=16):
pad_amount = val[-1]
if pad_amount == 0:
raise Exception
for i in range(len(val) - 1, len(val) - (pad_amount + 1), -1):
if val[i] != pad_amount:
raise Exception
return val[:-pad_amount]
def decrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
prev_chunk = iv
decrypted =
for i in range(0, len(data), 16):
chunk = data[i : i + 16]
decrypted += xor(decrypt_aes_ecb(chunk, key), prev_chunk)
prev_chunk = chunk
if pad:
return unpkcs7(bytes(decrypted))
return bytes(decrypted)
def encrypt_aes_cbc(data, key, iv = b'x00' * 16, pad=True):
if pad:
padded = pkcs7(data)
else:
padded = data
prev_chunk = iv
encrypted =
for i in range(0, len(padded), 16):
chunk = padded[i : i + 16]
encrypted_block = encrypt_aes_ecb(xor(chunk, prev_chunk), key)
encrypted += encrypted_block
prev_chunk = encrypted_block
return bytes(encrypted)
python python-3.x programming-challenge cryptography
python python-3.x programming-challenge cryptography
asked 14 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 mins ago
InfuzionInfuzion
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 mins ago
InfuzionInfuzion
101
asked 14 mins ago
InfuzionInfuzion
101
101
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Infuzion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "196"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Infuzion is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f215904%2fpython-aes-cbc-implementation-using-aes-ecb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Infuzion is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Infuzion is a new contributor. Be nice, and check out our Code of Conduct.
Infuzion is a new contributor. Be nice, and check out our Code of Conduct.
Infuzion is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Code Review Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f215904%2fpython-aes-cbc-implementation-using-aes-ecb%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
