Converting BitSet to Byte[] - java

I have a BitSet, which needs to be converted to a Byte[]. However, by using BitSet.toByteArray(), I don't get the correct output. I have tried converting the Byte[] to its binary form in order to check whether the Bitset and the binary form of the Byte[] are similiar.
public static void generate() {
BitSet temp1 = new BitSet(64);
for (int i = 0; i < 64; i++) {
if(i % 8 != 0 && i < 23) {
temp1.set(i, true);
}
}
StringBuilder s = new StringBuilder();
for (int i = 0; i < 64; i++) {
s.append(temp1.get(i) == true ? 1 : 0);
}
System.out.println(s);
byte[] tempByteKey1 = temp1.toByteArray();
for (byte b : tempByteKey1) {
System.out.print(Integer.toBinaryString(b & 255 | 256).substring(1));
}
}
Output:
Bitset: 0111111101111111011111100000000000000000000000000000000000000000
Converted Byte: 1111111011111110011111100000000000000000000000000000000000000000
They are both 64 bits, but the first 0 in the BitSet is placed somewhere else after the conversion. Why is this happening, and how can I fix it?

From BitSet#toByteArray() javadoc:
Returns a new byte array containing all the bits in this bit set.
More precisely, if..
byte[] bytes = s.toByteArray();
then
bytes.length == (s.length()+7)/8
and
s.get(n) == ((bytes[n/8] & (1<<(n%8))) != 0)
for all n < 8 * bytes.length.
#return a byte array containing a little-endian representation of all the bits in this bit set
#since 1.7
Attention: toByteArray() doesn't even claim to know size(), it is only "reliable" regarding length()!
..So I would propose as implementation (alternative for your toBinaryString()) a method like:
static String toBinaryString(byte[] barr, int size) {
StringBuilder sb = new StringBuilder();
int i = 0;
for (; i < 8 * barr.length; i++) {
sb.append(((barr[i / 8] & (1 << (i % 8))) != 0) ? '1' : '0');
}
for (; i < size; i++) {
sb.append('0');
}
return sb.toString();
}
..to call it like:
System.out.println(toBinaryString(bitSet.toByteArray(), 64);
run:
0111111101111111011111100000000000000000000000000000000000000000
0111111101111111011111100000000000000000000000000000000000000000
BUILD SUCCESSFUL (total time: 0 seconds)

Related

How can I create a stream of bits from a byte array? [duplicate]

How can i iterate bits in a byte array?
You'd have to write your own implementation of Iterable<Boolean> which took an array of bytes, and then created Iterator<Boolean> values which remembered the current index into the byte array and the current index within the current byte. Then a utility method like this would come in handy:
private static Boolean isBitSet(byte b, int bit)
{
return (b & (1 << bit)) != 0;
}
(where bit ranges from 0 to 7). Each time next() was called you'd have to increment your bit index within the current byte, and increment the byte index within byte array if you reached "the 9th bit".
It's not really hard - but a bit of a pain. Let me know if you'd like a sample implementation...
public class ByteArrayBitIterable implements Iterable<Boolean> {
private final byte[] array;
public ByteArrayBitIterable(byte[] array) {
this.array = array;
}
public Iterator<Boolean> iterator() {
return new Iterator<Boolean>() {
private int bitIndex = 0;
private int arrayIndex = 0;
public boolean hasNext() {
return (arrayIndex < array.length) && (bitIndex < 8);
}
public Boolean next() {
Boolean val = (array[arrayIndex] >> (7 - bitIndex) & 1) == 1;
bitIndex++;
if (bitIndex == 8) {
bitIndex = 0;
arrayIndex++;
}
return val;
}
public void remove() {
throw new UnsupportedOperationException();
}
};
}
public static void main(String[] a) {
ByteArrayBitIterable test = new ByteArrayBitIterable(
new byte[]{(byte)0xAA, (byte)0xAA});
for (boolean b : test)
System.out.println(b);
}
}
Original:
for (int i = 0; i < byteArray.Length; i++)
{
byte b = byteArray[i];
byte mask = 0x01;
for (int j = 0; j < 8; j++)
{
bool value = b & mask;
mask << 1;
}
}
Or using Java idioms
for (byte b : byteArray ) {
for ( int mask = 0x01; mask != 0x100; mask <<= 1 ) {
boolean value = ( b & mask ) != 0;
}
}
An alternative would be to use a BitInputStream like the one you can find here and write code like this:
BitInputStream bin = new BitInputStream(new ByteArrayInputStream(bytes));
while(true){
int bit = bin.readBit();
// do something
}
bin.close();
(Note: Code doesn't contain EOFException or IOException handling for brevity.)
But I'd go with Jon Skeets variant and do it on my own.
I needed some bit streaming in my application. Here you can find my BitArray implementation. It is not a real iterator pattern but you can ask for 1-32 bits from the array in a streaming way. There is also an alternate implementation called BitReader later in the file.
I know, probably not the "coolest" way to do it, but you can extract each bit with the following code.
int n = 156;
String bin = Integer.toBinaryString(n);
System.out.println(bin);
char arr[] = bin.toCharArray();
for(int i = 0; i < arr.length; ++i) {
System.out.println("Bit number " + (i + 1) + " = " + arr[i]);
}
10011100
Bit number 1 = 1
Bit number 2 = 0
Bit number 3 = 0
Bit number 4 = 1
Bit number 5 = 1
Bit number 6 = 1
Bit number 7 = 0
Bit number 8 = 0
You can iterate through the byte array, and for each byte use the bitwise operators to iterate though its bits.
Alternatively, you can use BitSet for this:
byte[] bytes=...;
BitSet bitSet=BitSet.valueOf(bytes);
for(int i=0;i<bitSet.length();i++){
boolean bit=bitSet.get(i);
//use your bit
}

Unable to generate Image from modified Byte array in java

In my code I'm reading image convert it to byte array and modifying that byte array with some logic and trying to generate image from that modified byte array, but i'm unable to generate image from that code
my code sample:
//1. Convert Image to byte code
ByteArrayOutputStream baos=new ByteArrayOutputStream();
BufferedImage img=ImageIO.read(new File(dirName,"MyImg.png"));
ImageIO.write(img, "png", baos);
baos.flush();
byte[] bytes = baos.toByteArray();
byte[] modified = baos.toByteArray();
String temp_string = new String();
for (int i = 0; i < bytes.length; i++)
{
// conversion of byte to unsign byte
int b = bytes[i] & 0xFF;
/*
* convert byte array to an 8 bit string
*/
int temp,count = 1;
byte b1 = (byte)b;
String uv = String.format("%8s", Integer.toBinaryString(b1 & 0xFF)).replace(' ', '0');
String tempStr = "";
for(int zx = 0 ; zx < uv.length() ; zx++ )
{
temp = Character.getNumericValue(uv.charAt(zx));
if(temp == 1)
{
temp += count;
count = temp;
if(temp % 2 == 0)
temp = 0;
else
temp = 1;
tempStr += temp;
}
else if(temp == 0)
{
tempStr += 0;
}
}
temp_string += tempStr;
if(i < bytes.length)
{
temp_string +=",";
}
}
String[] string_ByteArray = temp_string.split(",");
for(int a =0 ; a < string_ByteArray.length ; a++)
{
int aaa = Integer.parseInt(string_ByteArray[a],2);
modified[a] = (byte) aaa;
}
//3. Convert byte code to Image
ByteArrayInputStream bis = new ByteArrayInputStream(modified);
BufferedImage bImage2 = ImageIO.read(bis);
ImageIO.write(bImage2, "png", new File("output.png") );
in this code i'm getting error in 3rd step:
Exception in thread "main" java.lang.IllegalArgumentException: image == null!
at javax.imageio.ImageTypeSpecifier.createFromRenderedImage(Unknown Source)
at javax.imageio.ImageIO.getWriter(Unknown Source)
at javax.imageio.ImageIO.write(Unknown Source)
at mypack.Img_conversion.main(Img_conversion.java:96)
That is an impressively roundabout way of doing bit manipulation. There is no valid reason to use Strings. I suggest you either use bitwise operators, or use a BitSet.
Iterating through bits mathematically:
int b = bytes[i] & 0xFF;
for (int j = 7; j >= 0; j--) {
int bit = (b >> j) & 1;
temp = /* ... */;
if (temp != 0) {
b |= (1 << j); // set bit j
} else {
b &= ~(1 << j); // clear bit j
}
}
modified[i] = (byte) b;
Iterating through bits with a BitSet:
byte b = bytes[i];
BitSet bits = BitSet.valueOf(new byte[] { b });
for (int j = 7; j >= 0; j--) {
int bit = bits.get(j) ? 1 : 0;
temp = /* ... */;
bits.set(j, temp != 0);
}
modified[i] = bits.toByteArray()[0];
You might notice that since BitSet.valueOf takes an array of bytes, it’s wasteful to keep creating new BitSets. Instead, you could just do BitSet.valueOf(bytes) once, and run through all the bits in that single BitSet:
BitSet bits = BitSet.valueOf(bytes);
for (int i = bits.cardinality() - 1; i >= 0; i--) {
int bit = bits.get(i) ? 1 : 0;
temp = /* ... */;
bits.set(i, temp != 0);
}
byte[] modified = bits.toByteArray();
However…
A PNG image is (usually) compressed. This means the bits do not directly correspond to pixels. Modifying those bits creates an invalid compressed data block, which is why your attempt to read it with ImageIO.read fails and returns null.
If you want bytes you can directly manipulate, get them from the raw BufferedImage, not from a PNG representation:
int[] pixels = img.getData().getPixels(
0, 0, img.getWidth(), img.getHeight(),
new int[0]);
byte[] bytes = pixels.length * 4;
ByteBuffer.wrap(bytes).asIntBuffer().put(pixels);
It would be much easier for others to help you, if you took the time to give your variables meaningful names. temp and uv and zx are cryptic and meaningless. Better names would be:
temp_string → allByteValues
uv → bitsOfByte
tempStr → newBits
zx → bitIndex (or just a typical secondary indexing variable, like j)
temp → bit
When you’re done modifying the bytes, you still have raw image data, not a PNG representation, so you cannot make a ByteArrayInputStream from those bytes and pass them to ImageIO.read. Attempting to pass off those bytes as a PNG representation will always fail.
Instead, overwrite your image with the pixel data:
int[] pixels = new int[bytes.length / 4];
ByteBuffer.wrap(bytes).asIntBuffer().get(pixels);
img.getRaster().setPixels(0, 0, img.getWidth(), img.getHeight(), pixels);
ImageIO.write(img, "png", new File("output.png"));
As stated in the documentation if any of the parameter of the write method is null it will throw IllegalArgumentException.
You call like this:
ImageIO.write(bImage2, "png", new File("output.png") );
The only parameter which can be null is the bImage2.
Please check it if it's really null.

Convert a array having bits into one byte

I m trying to convert 8 bits into one byte. The way the bits are represented are by using a byte object that only contains a 1 or a 0.
If i have a 8 length byte array with these bits, how can i convert them into one byte.
public byte bitsToByte(byte[] bits) {
//Something in here. Each byte inside bits is either a 1 or a 0.
}
Can anyone help?
Thanks
public static byte bitsToByte(byte[] bits){
byte b = 0;
for (int i = 0; i < 8; i++)
b |= (bits[i]&1) << i;
return b;
}
//as an added bonus, the reverse.
public static byte[] bitsToByte(byte bits){
byte[] b = new byte[8];
for (int i = 0; i < 8; i++)
b[i] = (byte) ((bits&(1 << i)) >>> i);
return b;
}
left shift by 1 first for each bit in the array and then add the bit to the byte.
The implementation is based on the assumption that first bit in the array is sign bit and following bits are the magnitude in higher to lower positions of the byte.
public byte bitsToByte(byte[] bits) {
byte value = 0;
for (byte b : bits) {
value <<=1;
value += b;
}
return value;
}
Test the method:
public static void main(String[] args) {
BitsToByte bitsToByte = new BitsToByte();
byte bits[] = new byte[]{0,0,1,0,1,1,0,1}; // 1 + 0 + 4 + 8 + 0 + 32 + 0 + 0
byte value = bitsToByte.bitsToByte(bits);
System.out.println(value);
}
output:
45
Covert the byte array into a byte value (in the same order):
public static byte bitsToByte1(byte[] bits){
byte result = 0;
for (byte i = 0; i < bits.length; i++) {
byte tmp = bits[i];
tmp <<= i; // Perform the left shift by "i" times. "i" position of the bit
result |= tmp; // perform the bit-wise OR
}
return result;
}
input: (same array in reverse)
byte bits1[] = new byte[]{1,0,1,1,0,1,0,0};
value = bitsToByte1(bits1);
System.out.println(value);
output:
45

Java: convert byte[] to base36 String

I'm a bit lost. For a project, I need to convert the output of a hash-function (SHA256) - which is a byte array - to a String using base 36.
So In the end, I want to convert the (Hex-String representation of the) Hash, which is
43A718774C572BD8A25ADBEB1BFCD5C0256AE11CECF9F9C3F925D0E52BEAF89
to base36, so the example String from above would be:
3SKVHQTXPXTEINB0AT1P0G45M4KI8U0HR8PGB96DVXSTDJKI1
For the actual conversion to base36, I found some piece of code here on StackOverflow:
public static String toBase36(byte[] bytes) {
//can provide a (byte[], offset, length) method too
StringBuffer sb = new StringBuffer();
int bitsUsed = 0; //will point how many bits from the int are to be encoded
int temp = 0;
int tempBits = 0;
long swap;
int position = 0;
while((position < bytes.length) || (bitsUsed != 0)) {
swap = 0;
if(tempBits > 0) {
//there are bits left over from previous iteration
swap = temp;
bitsUsed = tempBits;
tempBits = 0;
}
//fill some bytes
while((position < bytes.length) && (bitsUsed < 36)) {
swap <<= 8;
swap |= bytes[position++];
bitsUsed += 8;
}
if(bitsUsed > 36) {
tempBits = bitsUsed - 36; //this is always 4
temp = (int)(swap & ((1 << tempBits) - 1)); //get low bits
swap >>= tempBits; //remove low bits
bitsUsed = 36;
}
sb.append(Long.toString(swap, 36));
bitsUsed = 0;
}
return sb.toString();
}
Now I'm doing this:
// this creates my hash, being a 256-bit byte array
byte[] hash = PBKDF2.deriveKey(key.getBytes(), salt.getBytes(), 2, 256);
System.out.println(hash.length); // outputs "256"
System.out.println(toBase36(hash)); // outputs total crap
the "total crap" is something like
-7-14-8-1q-5se81u0e-3-2v-24obre-73664-7-5-5cor1o9s-6h-4k6hr-5-4-rt2z0-30-8-2u-8-onz-4a2j-6-8-18-8trzza3-3-2x-6-4153to-4e3l01me-6-azz-2-k-4ckq-nav-gu-irqpxx-el-1j-6-rmf8hs-1bb5ax-3z25u-2-2r-t5-22-6-6w1v-1p
so it's not even close to what I want. I tried to find a solution now, but it seems I'm a bit lost here. How do I get the base36-encoded String representation of the Hash that I need?
Try using BigInteger:
String hash = "43A718774C572BD8A25ADBEB1BFCD5C0256AE11CECF9F9C3F925D0E52BEAF89";
//use a radix of 16, default would be 10
String base36 = new BigInteger( hash, 16 ).toString( 36 ).toUpperCase();
This might work:
BigInteger big = new BigInteger(your_byte_array_to_hex_string, 16);
big.toString(36);

Store binary sequence in byte array?

I need to store a couple binary sequences that are 16 bits in length into a byte array (of length 2). The one or two binary numbers don't change, so a function that does conversion might be overkill. Say for example the 16 bit binary sequence is 1111000011110001. How do I store that in a byte array of length two?
String val = "1111000011110001";
byte[] bval = new BigInteger(val, 2).toByteArray();
There are other options, but I found it best to use BigInteger class, that has conversion to byte array, for this kind of problems. I prefer if, because I can instantiate class from String, that can represent various bases like 8, 16, etc. and also output it as such.
Edit: Mondays ... :P
public static byte[] getRoger(String val) throws NumberFormatException,
NullPointerException {
byte[] result = new byte[2];
byte[] holder = new BigInteger(val, 2).toByteArray();
if (holder.length == 1) result[0] = holder[0];
else if (holder.length > 1) {
result[1] = holder[holder.length - 2];
result[0] = holder[holder.length - 1];
}
return result;
}
Example:
int bitarray = 12321;
String val = Integer.toString(bitarray, 2);
System.out.println(new StringBuilder().append(bitarray).append(':').append(val)
.append(':').append(Arrays.toString(getRoger(val))).append('\n'));
I have been disappointed with all of the solutions I have found to converting strings of bits to byte arrays and vice versa -- all have been buggy (even the BigInteger solution above), and very few are as efficient as they should be.
I realize the OP was only concerned with a bit string to an array of two bytes, which the BitInteger approach seems to work fine for. However, since this post is currently the first search result when searching "bit string to byte array java" in Google, I am going to post my general solution here for people dealing with huge strings and/or huge byte arrays.
Note that my solution below is the only solution I have ran that passes all of my test cases -- many online solutions to this relatively simple problem simply do not work.
Code
/**
* Zips (compresses) bit strings to byte arrays and unzips (decompresses)
* byte arrays to bit strings.
*
* #author ryan
*
*/
public class BitZip {
private static final byte[] BIT_MASKS = new byte[] {1, 2, 4, 8, 16, 32, 64, -128};
private static final int BITS_PER_BYTE = 8;
private static final int MAX_BIT_INDEX_IN_BYTE = BITS_PER_BYTE - 1;
/**
* Decompress the specified byte array to a string.
* <p>
* This function does not pad with zeros for any bit-string result
* with a length indivisible by 8.
*
* #param bytes The bytes to convert into a string of bits, with byte[0]
* consisting of the least significant bits in the byte array.
* #return The string of bits representing the byte array.
*/
public static final String unzip(final byte[] bytes) {
int byteCount = bytes.length;
int bitCount = byteCount * BITS_PER_BYTE;
char[] bits = new char[bitCount];
{
int bytesIndex = 0;
int iLeft = Math.max(bitCount - BITS_PER_BYTE, 0);
while (bytesIndex < byteCount) {
byte value = bytes[bytesIndex];
for (int b = MAX_BIT_INDEX_IN_BYTE; b >= 0; --b) {
bits[iLeft + b] = ((value % 2) == 0 ? '0' : '1');
value >>= 1;
}
iLeft = Math.max(iLeft - BITS_PER_BYTE, 0);
++bytesIndex;
}
}
return new String(bits).replaceFirst("^0+(?!$)", "");
}
/**
* Compresses the specified bit string to a byte array, ignoring trailing
* zeros past the most significant set bit.
*
* #param bits The string of bits (composed strictly of '0' and '1' characters)
* to convert into an array of bytes.
* #return The bits, as a byte array with byte[0] containing the least
* significant bits.
*/
public static final byte[] zip(final String bits) {
if ((bits == null) || bits.isEmpty()) {
// No observations -- return nothing.
return new byte[0];
}
char[] bitChars = bits.toCharArray();
int bitCount = bitChars.length;
int left;
for (left = 0; left < bitCount; ++left) {
// Ignore leading zeros.
if (bitChars[left] == '1') {
break;
}
}
if (bitCount == left) {
// Only '0's in the string.
return new byte[] {0};
}
int cBits = bitCount - left;
byte[] bytes = new byte[((cBits) / BITS_PER_BYTE) + (((cBits % BITS_PER_BYTE) > 0) ? 1 : 0)];
{
int iRight = bitCount - 1;
int iLeft = Math.max(bitCount - BITS_PER_BYTE, left);
int bytesIndex = 0;
byte _byte = 0;
while (bytesIndex < bytes.length) {
while (iLeft <= iRight) {
if (bitChars[iLeft] == '1') {
_byte |= BIT_MASKS[iRight - iLeft];
}
++iLeft;
}
bytes[bytesIndex++] = _byte;
iRight = Math.max(iRight - BITS_PER_BYTE, left);
iLeft = Math.max((1 + iRight) - BITS_PER_BYTE, left);
_byte = 0;
}
}
return bytes;
}
}
Performance
I was bored at work so I did some performance testing comparing against the accepted answer here for when N is large. (Pretending to ignore the fact that the BigInteger approach posted above doesn't even work properly as a general approach.)
This is running with a random bit string of size 5M and a random byte array of size 1M:
String -> byte[] -- BigInteger result: 39098ms
String -> byte[] -- BitZip result: 29ms
byte[] -> String -- Integer result: 138ms
byte[] -> String -- BitZip result: 71ms
And the code:
public static void main(String[] argv) {
int testByteLength = 1000000;
int testStringLength = 5000000;
// Independently random.
final byte[] randomBytes = new byte[testByteLength];
final String randomBitString;
{
StringBuilder sb = new StringBuilder();
Random rand = new Random();
for (int i = 0; i < testStringLength; ++i) {
int value = rand.nextInt(1 + i);
sb.append((value % 2) == 0 ? '0' : '1');
randomBytes[i % testByteLength] = (byte) value;
}
randomBitString = sb.toString();
}
byte[] resultCompress;
String resultDecompress;
{
Stopwatch s = new Stopwatch();
TimeUnit ms = TimeUnit.MILLISECONDS;
{
s.start();
{
resultCompress = compressFromBigIntegerToByteArray(randomBitString);
}
s.stop();
{
System.out.println("String -> byte[] -- BigInteger result: " + s.elapsed(ms) + "ms");
}
s.reset();
}
{
s.start();
{
resultCompress = zip(randomBitString);
}
s.stop();
{
System.out.println("String -> byte[] -- BitZip result: " + s.elapsed(ms) + "ms");
}
s.reset();
}
{
s.start();
{
resultDecompress = decompressFromIntegerParseInt(randomBytes);
}
s.stop();
{
System.out.println("byte[] -> String -- Integer result: " + s.elapsed(ms) + "ms");
}
s.reset();
}
{
s.start();
{
resultDecompress = unzip(randomBytes);
}
s.stop();
{
System.out.println("byte[] -> String -- BitZip result: " + s.elapsed(ms) + "ms");
}
s.reset();
}
}
}

Categories

Resources