Using java to encrypt integers

I'm trying to encrypt some integers in java using java.security and javax.crypto.

The problem seems to be that the Cipher class only encrypts byte arrays. I can't directly convert an integer to a byte string (or can I?). What is the best way to do this?

Should I convert the integer to a string and the string to byte? This seems too inefficient.

Does anyone know a quick/easy or efficient way to do it?

Please let me know.

Thanks in advance.

jbu

7 Answers
7

You can turn ints into a byte using a DataOutputStream, like this:

ByteArrayOutputStream baos = new ByteArrayOutputStream ();
DataOutputStream dos = new DataOutputStream (baos);
dos.writeInt (i);
byte data = baos.toByteArray();
// do encryption


Then to decrypt it later:

byte decrypted = decrypt (data);
ByteArrayInputStream bais = new ByteArrayInputStream (data);
DataInputStream dis = new DataInputStream (bais);
int j = dis.readInt();


You can also use BigInteger for conversion:

BigInteger.valueOf(integer).toByteArray();


Just use NIO. It's designed for this specific purpose. ByteBuffer and IntBuffer will do what you need quickly, efficiently, and elegantly. It'll handle big/little endian conversion, "direct" buffers for high performance IO, and you can even mix data types into the byte buffer.

Convert integers into bytes:

ByteBuffer bbuffer = ByteBuffer.allocate(4*theIntArray.length);
IntBuffer ibuffer = bbuffer.asIntBuffer(); //wrapper--doesn't allocate more memory
ibuffer.put(theIntArray); //add your int's here; can use
//array if you want
byte rawBytes = bbuffer.array(); //returns array backed by bbuffer--
//i.e. *doesn't* allocate more memory


Convert bytes into integers:

ByteBuffer bbuffer = ByteBuffer.wrap(rawBytes);
IntBuffer ibuffer = bbuffer.asIntBuffer();
while(ibuffer.hasRemaining())
System.out.println(ibuffer.get()); //also has bulk operators


I have found the following code that may help you, since Integer in Java is always 4 bytes long.

public static byte intToFourBytes(int i, boolean bigEndian) {
if (bigEndian) {
byte data = new byte[4];
data[3] = (byte) (i & 0xFF);
data[2] = (byte) ((i >> 8) & 0xFF);
data[1] = (byte) ((i >> 16) & 0xFF);
data[0] = (byte) ((i >> 24) & 0xFF);
return data;

} else {
byte data = new byte[4];
data[0] = (byte) (i & 0xFF);
data[1] = (byte) ((i >> 8) & 0xFF);
data[2] = (byte) ((i >> 16) & 0xFF);
data[3] = (byte) ((i >> 24) & 0xFF);
return data;
}
}


You can find more information about the bigEndian parameter here:
http://en.wikipedia.org/wiki/Endianness

create a 4-byte array and copy the int to the array in 4 steps, with bitwise ANDs and bitshifting, like Paulo said.

But remember that block algorithms such as AES and DES work with 8 or 16 byte blocks so you will need to pad the array to what the algorithm needs. Maybe leave the first 4 bytes of an 8-byte array as 0's, and the other 4 bytes contain the integer.

Just use:

Integer.toString(int).getBytes();


Make sure you use your original int and getBytes() will return a byte array. No need to do anything else complicated.

To convert back:

Integer.parseInt(encryptedString);


My Simple Solution is that Encrypt Integer to the String by shifting ASCII Value of the Integer by the secret key you Provide.

Here is the Solution:

public String encodeDiscussionId(int Id) {

String tempEn = Id + "";
String encryptNum ="";
for(int i=0;i<tempEn.length();i++) {
int a = (int)tempEn.charAt(i);
a+=148113;
encryptNum +=(char)a;
}
return encryptNum;
}

public Integer decodeDiscussionId(String encryptText) {

String decodeText = "";
for(int i=0;i<encryptText.length();i++) {
int a= (int)encryptText.charAt(i);
a -= 148113;
decodeText +=(char)a;
}
int decodeId = Integer.parseInt(decodeText);
return decodeId;
}


Steps to Encode:

String temp = givenInt + ""

Integer

String encryptNum

Steps to Decode:

decodeText

As previous encode output is always String '???' and vary according to number of digits of input Id.

String '???'

Id

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