Creating 8 byte IV in Java

Question

Im new to cryptography in Java and I am trying to write a program to encrypt and decrypt a phrase using DES symmetric cipher, based on CBC mode of operation.

Can anyone tell me how to go about creating an 8-byte initialization vector and how to cast the new IV into AlgorithmParameterSpec class?

Also, which packages should I import?

Edit: Right now I have these lines:

SecureRandom sr = new SecureRandom(); //create new secure random
byte [] iv = new byte[8]; //create an array of 8 bytes 
sr.nextBytes(iv); //create random bytes to be used for the IV (?) Not too sure.
IvParameterSpec IV = new IvParameterSpec(iv); //creating the IV 

Is my above approach correct?

Thanks.

Answer 1

Yes. Till Now You are right.
the Class IvParameterSpec is used to pass Initial Vector to Class Cipher
After this create a Cipher as below

Cipher cipherInstance = Cipher.getInstance("DES/CBC/NoPadding");

here DES/CBC/NoPadding is used because you are useing DES encryption in CBC mode.

Next is initializing it.

cipherInstance.init(Cipher.DECRYPT_MODE, Key, ivParameterSpec);

Parameters are:
1st is mode of encryption either Cipher.DECRYPT_MODE or Cipher.ENCRYPT_MODE
2nd is Secret Key You have generated using Class SecretKey
3rd is generated ivParameterSpec

and last line is

outData = cipherInstance.doFinal(input);

if mode is Decrypt it will return decrypted data and if mode is Encrypt it will return encrypted data.

Last but important catch all the Exceptions properly.
And you are done