kGen | Experimenting with code to generate and optimise K/APL code
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Experimenting with code to generate and optimise K/APL code.
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QUESTION
I have kind of old code where we run sonarqube. I am not expert on Java and Exception descendants etc. so I hope someone will be able to help me to fix this issue, as Sonar says it is a blocker.
This is the code:
...ANSWER
Answered 2020-Aug-19 at 12:05Thanks to all comments:
This is solution (removal of generic exception and adding explicit exceptions):
QUESTION
I have a function written in C#. Basically the function is used to generate a token on the basis of parameters like text and key.
...ANSWER
Answered 2020-Jul-02 at 09:12The AES algorithm used in the C# code is AES 128-bit in ECB mode.
We can perform the same encryption in Node.js (and decrypt as well if we wish), using the following code:
Node.js Code
QUESTION
I'm want to encrypt and decrypt files from SD-card using AES. In order to do it we always need a seed (usually a string which is inserted by user as a password):
...ANSWER
Answered 2019-Oct-14 at 13:32Updated 2019-10-14
TL;DRNo, you can't access the fingerprint. You can only get a "thumbs up" or "thumbs down" from the Biometric API. This is intentional by design.
You can, however, leverage the Android Keystore for hardware-backed cryptographic operations, and require user re-authentication to release the key. This pretty much does what you want.
The long answerGenerating a password-like seed from a fingerprint is impossible. As James K Polk commented, fingerprints vary when scanned, and they are never legibly stored directly on the device.
When a fingerprint is being enrolled, its image is temporarily stored on secure device memory, where it is processed to generate validation data and a fingerprint template (these are all inaccessible to the Android OS). The raw image is then discarded. When a finger is scanned, the image is compared to the validation data generated before, and if it matches to a certain degree of certainty, a user is deemed as authenticated.
Biometric operations are conducted inside of Android's Trusted Execution Environment (TEE). This is a completely isolated OS running either on a protected part of the CPU on a separate coprocessor on modern devices (SE).
It's a virtually untouchable environment with a restricted interface and hardware barriers put in place to protect against tampering with the chip and forced extraction of biometric validation data and cryptographic keys.
SolutionGoing back to your original question, no, you can't get any unique finger identification. This would be inherently insecure, as any application could read the secret!
What you can do, is leverage Android's hardware-backed Keystore
and require device-level authentication to release hardware-backed cryptographic keys (setUserAuthenticationRequired(true)
). This means generating a random secret which is securely saved to the Keystore, requiring a finger swipe to release the key to userspace. I can't stress the word hardware-backed enough.
You have no control over which finger is can be used and whether vendor-specific implementations allow bypassing of biometrics with the device unlock pattern, for example.
Android KeystoreThe Keystore's purpose is to protect cryptographic keys. Keys can only be retrieved by the application that owns them once sufficient requirements have been met, such as recent or immediate biometric authentication.
Keys can be protected against malicious extraction, and on modern devices, hardware bound, meaning they never leave the secure hardware (TEE/SE), and therefore are never exposed to your Android application. Any cryptographic operations, such as AES encryption/decryption, are securely executed outside of userspace (on secure hardware), and enrolling new fingerprints/changing the lock pattern will permanently invalidate the key. In this mode of operation, the Keystore entry merely serves as an "interface" to conduct crypto operations inside of the secure hardware, the true secret is never exposed to your application.
In summaryThere is a Fingerprint/Biometric API, which is there purely for convenience, allowing you to quickly confirm an action by requiring the user to authenticate. It boils down to a "yes"/"no" answer from the TEE/SE, and vary greatly depending on the phone manufacturer!
The Keystore is a hardware-backed vault for cryptographic keys. Devices running API-level 28+ also have access to Strongbox Keymaster, if the device hardware supports it, which restricts cryptographic operations to a dedicated security CPU with more secure storage.
These features are device/vendor specific! And could be compromised/insecure! Warn users before enabling fingerprint authentication if you aren't sure about the device. The only truly secure encryption method is prompting the user every time for the decrypt key (in this case, the mind is the hardware-backed store). Having it stored anywhere, even in live memory, is always a calculated risk.
Doing cryptography right is extremely difficult. I highly advise that you research and try to understand the basics, and what additional security Android has to offer, before attempting to use this in production.
QUESTION
Here's a link to [the source code for the keras Sequence class][1].
I have created what I believe to be the exact same object by doing the following:
- copying and pasting the same
__init__
,__getitem__
, and__len__
methods from theInfiniteGenerator
class to theKGen
class. - copying the iter method from the source code of the
Sequence
class the theInfiniteGenerator
class.
My hypothesis is that these two generators should both create a finite list, but the Sequence
object will continue iterating forever. Why?
ANSWER
Answered 2019-Jan-08 at 15:19I ran into this as well using tensorflow.keras version 1.10. You can see in the source code they defined __iter__()
to return an infinite generator. I added the following function to all of my Sequence classes to create a one-shot iterator for the situations where I need one.
QUESTION
I made a class which has a method to encrypt data using SHA1PRNG and AES algorithm.
...ANSWER
Answered 2018-Nov-26 at 08:18You're printing the Encryption
object instead of the result of the function call.
You can do this instead:
QUESTION
I'm trying to encrypt and decrypt a message in Java I get the same error all the time:
java.security.InvalidKeyException: Key is too long for unwrapping
I'm trying to simulate sending encripted messages from a Server to a Client.
1st: The client generate a private and public key.
2nd The client sends the public key to the server.
3rd: The server generates the symmetric AES key.
4t: The server encrypts the text that he wants to send to the client with his AES key.
5e: The server encrypts the AES key with the public key received from the client.
6e: The server sends the encrypted text and the encrypted AES key to the client.
7e: The client decrypt the AES key with his private key.
8e: The client decrypts the text received from the server with the decrypted AES key.
...ANSWER
Answered 2018-Apr-07 at 23:17The problem is that you cannot just assume that RSA or AES encryption results in bytes that represent a string. When you generate the strings using new String
it will silently remove all bytes that do not represent characters. You can use CharsetDecoder
if you want to have more control and throw an exception (which should have been the default).
The problem is that when you reconvert to bytes that some bytes may be missing. That means that the number may be significantly smaller, which will throw an exception one way or another as the input of unwrap is now too small.
Here are some changes that show how to get it working. I just commented out your code so you can compare.
QUESTION
I'm trying to generate AES key of size 256 bit and I downloaded the JCE security policy of version 8 and I placed the US_export_policy.jar and the local_policy.jar in the jre/lib/security folder. Is there anything I've left out in the code and what could be the cause for the exception?
This is what I've coded and it gives me an illegal key size exception.
Here's the code :
...ANSWER
Answered 2018-Feb-21 at 07:36Use jdk1.8.0_91
package, it will work after changing JCE policy files.
In jdk1.8.0_144
, it won't work even changing JCE files. Seems it has some issues in that version. Ask you query in Java8 forum regarding this.
QUESTION
In case the question wasn't that clear, I was wondering in the process of writing data into a file, what types of input or operations can make them turn from text files to binary files?
Reason for the question:
I was just starting my way into (Java) cryptography and I made a simple program that reads the text from an existing file and, after encrypting it, it writes the encrypted message back in that file.
Problem:
Sometimes the file stays "text", showing some ramdom symbols and letters from the encryption, but other times it turns into a binary file and despite running some tests, I can't seem to understand what could be the cause.
Example:
-with document containing the word "test",
expected output:�Ҋ5�)>;
result in File: �Ҋ5�)>;
-with document containing the word "Message",
expected output:�̮�~��<-�L��%
result in File: File turned into binary
Note: I'm pretty much positive it does not depend on the word, and I had a guess it could be related to what encryption symbols it generates, but I could be wrong.
In case it's useful: I'm currently using Ubunto and creating the text file with "gedit".
Code: (I appologize for the try/catches)
...ANSWER
Answered 2017-Oct-09 at 02:17All files are binary files. There isn't really such a thing as a "text file" in the sense that you seem to understand it.
There is such a thing as a file that, by chance, happens to contain a sequence of encoded bytes that perfectly represent a UTF8 string of characters. But it's still a binary file.
The same can be said for any other encoding ever.
QUESTION
I'm barely learning flow, but according to the documentation, the following code should be giving me an error since I'm destructuring the object and assigning a default value to 'overrideKeyProp' that is not a boolean.
This is Ok, no errors:
...ANSWER
Answered 2017-Aug-11 at 22:10The key thing is that when you haven't given a type to Flow for a given variable declaration, it will infer the type.
QUESTION
I am creating a program to generate a key pair but i have this error:
...ANSWER
Answered 2017-Mar-15 at 22:15This line is wrong:
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