aesCbc | aes-cbc加密解密 | Encryption library
kandi X-RAY | aesCbc Summary
kandi X-RAY | aesCbc Summary
aes-cbc加密解密
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- mcryptSetKey sets the encryption key to use .
- mcryptEncrypt encrypts data using rinst .
- mcryptDecrypt decrypts data using rinst .
- mcryptRijndaelGentables generates the LM Rijael tables
- NewAesCipher128 creates a new AesCipher128 .
- mcryptGenericInit is used to initialize a new cipher .
- mcrypt encrypts plaintext using the previous ciphertext .
- mdecrypt decrypts the given ciphertext .
- Converts a byte to another byte
- invMixCol returns the inverse of x .
aesCbc Key Features
aesCbc Examples and Code Snippets
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Trending Discussions on aesCbc
QUESTION
NodeJs:
I am trying decrypt text using AES CBC PKCS7 in NodeJs and PKCS5 in java. I am getting error: Given final block not properly padded. Such issues can arise if a bad key is used during decryption.
UPDATED
...ANSWER
Answered 2022-Jan-08 at 09:46There are a few issues in the CryptoJS part, apply the following fixes:
QUESTION
ANSWER
Answered 2021-Nov-03 at 07:16The signature certificate of the HarmonyOS application is different from the signature certificate of the Android application.
The signature certificate of the HarmonyOS application is a .p12 file. Like following:
You could check the signature certificate fingerprint as follows:
QUESTION
I have encrypted a file using below commands
openssl rand 32 > test.key
openssl enc -aes-256-cbc -iter 10000 -pbkdf2 -salt -in test.txt -out test.txt.enc -pass file:test.key
Now i am trying to decrypt it using java. tring since last few days but no success.
Can anyone help ?
my code
...ANSWER
Answered 2020-May-06 at 16:48You have several problems. The most obvious is that you are trying to read the IV from the file, but openssl enc
in its default password-based mode derives both key and IV from password and salt -- even when using PBKDF2. However, both the standard (Sun/Oracle/OpenJDK) and BouncyCastle providers in Java implement PBKDF2 to derive only a key -- the way it is used in PBES2.
Even without that, your method of generating the 'password' as random bytes wouldn't work either. The PKCS5 standard actually defines PBKDF2 to take the password as
an octet string of arbitrary length whose interpretation as a text string is unspecified. In the interest of interoperability, however, it is recommended that applications follow some common text encoding rules. ASCII and UTF-8 [RFC3629] are two possibilities. (ASCII is a subset of UTF-8.)
Many systems take interoperable encoding more seriously, and Java in particular (which was designed from its inception to be worldwide) defines PBEKeySpec
to contain characters -- char[]
in Java is UTF-16 -- which are encoded as UTF-8 when doing PBKDF2. In contrast openssl
is a C program dating from before the turn of the century when C started admitting the existence of countries other than the USA, so it only knows about bytes -- bytes which might be ASCII, or some other single-byte code like EBCDIC, but maybe not characters at all and certainly not any of those weird foreign characters that don't fit in a byte. The probability of a sequence of 32 random bytes being valid UTF-8 is very low; it's too much work for me to figure analytically, but I ran a test of 100 million random values and got only one that would work with your scheme. (I was going to test a billion but got tired of waiting.)
Plus, since a password is supposed to be text, openssl
reads -pass file:
as a text file and treats it as a string. That means if any of the random bytes is a null byte or a byte corresponding to the newline character, the remainder of the data in the file is discarded and ignored for the key-and-IV derivation. This will occur on average about 1 in 4 times for random 32-byte values, and about 1 in 20 times it will occur early enough in the file to make the result cryptographically weak and breakable.
Which raises the point: why are you using password-based encryption at all? If your 'key' is 32 bytes from a decent secure RNG -- which openssl rand
is -- you don't need to strengthen it, it's already valid as a key. You can use openssl enc
to do key-based encryption, not password-based, and it's more efficient, more secure, AND much easier in Java -- a massive win. IF you use a new, random key for each encryption you don't even have to use a real IV, you can just use a zero IV as I did below. But if you are going to reuse the/any key, you need to use a unique and unpredictable -- normally random -- IV for each encryption, and convey it with the data, perhaps by just putting it at the front.
So anyway, here's a fairly simple Java program which can handle either case: the openssl form of pdbkf2 with a 'password' that isn't actually a password and isn't UTF-8, or the more sensible key-based form (but for this demo with zero IV):
QUESTION
I have next code that was copied from the right answer here:
...ANSWER
Answered 2020-Apr-14 at 08:05The encrypt
-method must return the data in OpenSSL format, which consists of the ASCII encoding of Salted__, followed by the 8 bytes randomly generated salt and the actual ciphertext, whereby the data are Base64 encoded after their concatenation.
Note, however, that the key derivation function used for the OpenSSL format is insecure and is not a standard, here. A possible extension of the encrypt
method could be:
QUESTION
I am learning TLS protocol 1.0/1.1/1.2 recently. I notice AESCBC128/256 can be used for TLS 1.0 although it is not mentioned in the initial TLS1.0 RFC. I am wonderring how the AESCBC IV/salt is exchanged between client and server? Is it the same as TLS 1.2 that always exchange IV within the application data (the first 16 bytes of the data)? If any official materials describe this would be grateful.
...ANSWER
Answered 2020-Jan-30 at 14:56TLS_RSA_WITH_AES_128_CBC_SHA is mandatory to implement in TLS 1.2 (see RFC 5246).
Section §6.2.3.2 explains how CBC works:
For block ciphers (such as 3DES or AES), the encryption and MAC functions convert TLSCompressed.fragment structures to and from block TLSCiphertext.fragment structures.
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