Learning AES: the KeyBytes

Posted by Tom Brito on Stack Overflow See other posts from Stack Overflow or by Tom Brito
Published on 2010-05-25T17:31:49Z Indexed on 2010/05/25 17:41 UTC
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I got the following example from here:

import java.security.Security;

import javax.crypto.Cipher;
import javax.crypto.spec.SecretKeySpec;

public class MainClass {
  public static void main(String[] args) throws Exception {
    Security.addProvider(new org.bouncycastle.jce.provider.BouncyCastleProvider());    
    byte[] input = "www.java2s.com".getBytes();
    byte[] keyBytes = new byte[] { 0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x09,
        0x0a, 0x0b, 0x0c, 0x0d, 0x0e, 0x0f, 0x10, 0x11, 0x12, 0x13, 0x14, 0x15, 0x16, 0x17 };

    SecretKeySpec key = new SecretKeySpec(keyBytes, "AES");

    Cipher cipher = Cipher.getInstance("AES/ECB/PKCS7Padding", "BC");

    System.out.println(new String(input));

    // encryption pass
    cipher.init(Cipher.ENCRYPT_MODE, key);

    byte[] cipherText = new byte[cipher.getOutputSize(input.length)];
    int ctLength = cipher.update(input, 0, input.length, cipherText, 0);
    ctLength += cipher.doFinal(cipherText, ctLength);
    System.out.println(new String(cipherText));
    System.out.println(ctLength);

    // decryption pass
    cipher.init(Cipher.DECRYPT_MODE, key);
    byte[] plainText = new byte[cipher.getOutputSize(ctLength)];
    int ptLength = cipher.update(cipherText, 0, ctLength, plainText, 0);
    ptLength += cipher.doFinal(plainText, ptLength);
    System.out.println(new String(plainText));
    System.out.println(ptLength);
  }
}

I imagine that the byte[] keyBytes should be random generated, so I gone to test the max size before do it.

When adding one more byte 0x18 to the array, the exception raised: InvalidKeyException: Key length not 128/192/256 bits. But the original 18 bytes (from 0 to 17) are not multiple of nither 128, 192 or 256.

I would like to understand the math here.. can anyone explain me? Thanks!

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