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Mix byte array android

I would like to mix audio byte array, but I didn't succeed to sum the array.(note i already added some silent bytes of 0 as padding before).
I have an ArrayList of byte[] which contains:
the first byte[] is header (44 bytes).
Following byte[] are raw data byte array to be mixed
Here is my code:
ArrayList<byte[]> ListAudio = new ArrayList<byte[]>();
byte[] header= WriteHeader(); //wav header 44 bytes
ListAudio.add(header);
for (byte[] b : audioTreatment.ListDataByte) {
ListAudio.add(b);
}
//calculate total length of audio
int length = 0;
for (byte[] array : ListAudio) {
length += array.length;
}
final int len = length;
final byte[] mixBytes = new byte[len];
for (byte[] array : ListAudio) {
for (int i = 44; i < len; ++i) {
mixBytes[i] += array[i];
// mixBytes[i]=(byte) ((bytes1[i]+bytes2[i]) / 2);
}
}
I found somewhere that the method to mix digital byte array is :
mixBytes[i]=(byte) ((bytes1[i]+bytes2[i]) / 2);
I don't arrive to include the calcul above, to sum the byte array.
How can i sum the bytes array from my ArrayList ?

you have to declare your sources to merge them
byte[] source1 = ListAudio.get(0); //first from list
byte[] source2 = ListAudio.get(1); //second from list
int length = Math.min(source1.length, source2.length);//length of new array
length = length - 44; //skipping 44 byte
byte[] dest = new byte[length];
for(int index = 0; index < length; index ++){
byte b1 = source1[index+44];
byte b2 = source2[index+44];
dest[index] = (byte) ((b1+b2) / 2);
}
That would merge the first two byte[] from your list.
If you want to merge other sources you can change them by selecting other byte[] from your List.
HINT
The length of the destination is declared as Math.min(a,b) but you can fill missing bytes with zeros if you want...
if you want to merge all arrays, you have to adjust your merge operation
mixing two bytes: mixBytes[i]=(byte) ((bytes1[i]+bytes2[i]) / 2);
mixing three bytes: mixBytes[i]=(byte) ((bytes1[i]+bytes2[i]+bytes3[i]) / 3);
mixing N bytes: mixBytes[i]=(byte) ((bytes1[i]+bytes2[i]+bytes3[i]+...+bytesN[i]) / N);
ok, for your code snipped it would be:
int length = ...;//length of result, either min or max as mentioned above, see HINT
byte[] mixBytes = new byte[length];
int amountAudio = ListAudio.size(); //amount of tracks in your list aka 'N'
int sum;
for(int index = 0; index < length; index++){
sum = 0;
for(byte[] source: ListAudio){
//adding all byte into one big integer
sum = sum + source[index]; //NOTE: watch for indexOutOfBoundsException
}
//afterward divide the big int through amount of Audio tracks in your list
mixBytes[index] = (byte)(sum / amountAudio);
}

What will this constructor look like?

public byte[] d
In this byte array, each byte represents a digits, where d[0] is the least significant digit, and a[d.length-1] is the most significant digit. For example, 543210 is stored as {0,1,2,3,4,5}. The most significant digit can't be a zero;
In the real implementation, this array should be private.
Constructor Detail: AdditionOnlyInt
public AdditionOnlyInt(java.lang.String a)
This is a constructor that construct an AdditionOnlyInt with value presented by the input string. For example, if input a = "00012340", this constructor will set the byte array to have value {0,4,3,2,1}. Note that the leading zeros in the input string should not be stored.
Parameters:a - is a string such as "00012340"
I do not know how to do this constructor does anyone?
I know its very wrong but I tried this
public AdditionOnlyInt(String number) {
int counter = number.length();
number.replace("0","");
data = new byte[number.length()];
int i = 0;
while(i<number.length()) {
data[i] = (byte)number.charAt(counter);
i++;
counter--;
}
}
and I do know converting to byte gives you different values.

You're in luck: only a handful of modifications need to be made to your program. You weren't entirely wrong. :)
First, these three statements are giving you fits:
int counter = number.length();
number.replace("0","");
data = new byte[number.length()];
You get a counter, which is the unbiased String (that is, with zeroes), which will undoubtedly be smaller than the String, without zeroes. You'd also be creating the array with the size of the unbiased list.
Well...it would be if your second statement did something. String is immutable, so anything that's done to modify it would only generate a new String, leaving the old one unmodified. That's fixable by this:
number = number.replace("0", "");
...but in reality, given your input set, it should be fixed by this:
number = String.valueOf(Integer.parseInt(number));
This way, you keep internal zeroes.
Now assuming that your byte[] is actually called data and not d, there's one little issue we have to fix: numbers in terms of a byte are quite large (that is, the character for '7' is 0x37, which is 55).
So we need to bias it. Whatever our byte number is, we need to subtract '0' from it; that is, we need to subtract 48 from it, to give us our correct value. I'll show you what that looks like in a moment.
Now, for your loop:
int i = 0;
while(i<number.length()) {
data[i] = (byte)number.charAt(counter);
i++;
counter--;
}
I'm not sold on the necessity of counter, so let's get rid of it. Now we'll use i from now on. Essentially, what this means is that we have to move charAt from the end of the String to the front of the String, placing the values into the array as such. What that (mostly) looks like is this:
data[i] = (byte) (number.charAt(number.length() - 1 - i);
Pay close attention here - we have to subtract 1 from the length right off the bat, since we don't have a place on the String that's exactly equal to its maximum length. We then subtract i from that, so we get the effect of moving backwards on the String.
That is, for a string of length 10 without zeroes, if we start at i = 0, we get the character at 10 - 1 - 0 = 9, or the last character.
Remember what I said about biasing the result of that, though? After you've got the data, be sure to subtract '0' from it.
data[i] = (byte) (number.charAt(number.length() - 1 - i) - '0';
And really, that's all there is to it. You mostly got it, except the iteration and sanitization was a bit wonky.

public AdditionOnlyInt(String input)
{
//remove trailing 0s
int relevStart = 0;
while(input.charAt(relevStart) == '0')
{
relevStart++;
}
String relevantTerms = input.substring(relevStart);
//flip the remaining chars
int length = relevantTerms.length();
data = new byte[length];
for(int iter = 0; iter < length; iter++)
{
data[iter] = (byte) (relevantTerms.charAt(length - iter - 1) - '0');
}
}

Hope this helps.
Step 1: Check whether its a number by using Integer value = Integer.valueOf(args);
Step 2: convert the Integer into a byte array by calling byte array[] = value.toString().getBytes();
Step 3: the byte array contains the value in forward fashion, So swapping all the values in the array will make the digit reverse as user requested.
for (int i = 0, j = array.length - 1; i < array.length / 2; i++, j--) {
byte temp = array[i];
array[i] = array[j];
array[j] = temp;
}
find the complete program below:
public class AdditionOnlyInt {
public static void main(String[] args) {
// TODO Auto-generated method stub
new AdditionOnlyInt("01010120");
}
public AdditionOnlyInt(String args) {
try {
Integer value = Integer.valueOf(args);
byte array[] = value.toString().getBytes();
for (int i = 0, j = array.length - 1; i < array.length / 2; i++, j--) {
byte temp = array[i];
array[i] = array[j];
array[j] = temp;
}
for (int i = 0; i < array.length; i++) {
System.out.print((char) array[i]);
}
} catch (Exception e) {
}
}
}

Size of 2d array in Java

I defined a 2d array in Java. As I read about it(i.e. 2d array), the first dimension of this 2d array is a pointer (I do not know that is it right or not, please tell me about it). So If I consider it as pointer, in a 64-bit system, what will be the size of below code after execution?
short [][] array1 = new short [10][];
short[][] array2 = new short[10][];
for (int i = 0; i < 10; i++)
array1[i] = new short[1];
for (int i = 0; i < 10; i++)
array2[i] = array1[i];
Please tell me about the size of above code.

For every one dimensional array you have a 24 byte overhead in addition to the space for the data in the array.
So your first two lines of code each create an array of 10 pointers - you are right about that - which take 8 bytes each on 64-bit system. This means you are allocating 2 * (24 + 10 * 8) = 208 bytes.
In the first for loop you are creating 10 arrays which are 24 + 2 * 10 = 44 bytes each. These are padded to at least 8 byte boundaries and thus take up 48 bytes or 480 bytes in total.
In the second loop, you are not allocating any new memory.
In total you are using 208 + 480 = 688 bytes.
Note that the actual usage depends on the JVM. For example:
Some JVMs compress pointers.
Some JVMs only use a 12 byte header for arrays.
In order to see the difference between outer and inner arrays it might be helpful to rewrite your code like this:
short[][] outerArray = new short[10][]; // Array of references to short arrays
short[] innerArray; // Array of shorts
for (int i = 0; i < 10; i++) {
innerArray = new short[1];
outerArray[i] = innerArray;
}

How to convert image to Boolean Array in Java (Android)?

I'm trying to store a 32 x 32 Boolean array in a 32 x 32 black and white image (either bitmap or PNG), to then be mapped to a Boolean[32][32] array with black pixels being true and white being false.
This is to store frames of animation to display on a virtual 32 x 32 display. Here's what I have so far below.
Bitmap bmp = BitmapFactory.decodeResource(context.getResources(), R.raw.f1);
bmp.compress(Bitmap.CompressFormat.PNG, 100, o_stream);
byte[] byteArray = o_stream.toByteArray();
What do I do with byteArray to make it a Boolean[32][32] array or am I going about this all wrong in the first place?

While I never did anything with images (so I don't know if this is anything close to what one should do to get the black-and-white of a pic), I suppose you need a rule to decide whether a pixel is closer to black or closer to white. But I'm curious, how can a byte represent a color? Even if it's RGB, you need at least three bytes, don't you?

if(src!= null){
ByteArrayOutputStream os=new ByteArrayOutputStream();
src.compress(android.graphics.Bitmap.CompressFormat.PNG, 100,(OutputStream) os);
byte[] byteArray = os.toByteArray();
//Log.d("byte=",""+byteArray.length);//returns length.
//str = Base64.encodeToString(byteArray,Base64.DEFAULT);//returns string
}
where src is the bitmap....

If you just want to encode an array of Booleans into a bitmap to save storage space, why use an image? That's a lot of extra overhead. Why not just create a bitmap yourself, like this:
Boolean[][] booleanArray = ... // this is your Boolean array
int[] bits = new int[32]; // One int holds 32 bits
for (int i = 0; i < 32; i++) {
for (int j = 0; j < 32; j++) {
if (booleanArray[i][j]) {
// Set bit at the corresponding position in the bits array
bits[i] |= 1 << j;
}
}
}
// Now you have the data in a int array which you can write to a file
// using DataOutputStream. The file would contain 128 bytes.
// To recreate the Boolean[32][32] from the int array, do this:
Boolean[][] booleanArray = new Boolean[32][32];
int[] bits = ... // This is the data you read from the file using DataInputStream
for (int i = 0; i < 32; i++) {
for (int j = 0; j < 32; j++) {
if ((bits[i] & (1 << j)) != 0) {
// Set corresponding Boolean
booleanArray[i][j] = true;
}
}
}

Fast reading of little endian integers from file

I need to read a binary file consisting of 4 byte integers (little endian) into a 2D array for my Android application. My current solution is the following:
DataInputStream inp = null;
try {
inp = new DataInputStream(new BufferedInputStream(new FileInputStream(procData), 32768));
}
catch (FileNotFoundException e) {
Log.e(TAG, "File not found");
}
int[][] test_data = new int[SIZE_X][SIZE_Y];
byte[] buffer = new byte[4];
ByteBuffer byteBuffer = ByteBuffer.allocate(4);
for (int i=0; i < SIZE_Y; i++) {
for (int j=0; j < SIZE_X; j++) {
inp.read(buffer);
byteBuffer = ByteBuffer.wrap(buffer);
test_data[j][SIZE_Y - i - 1] = byteBuffer.order(ByteOrder.LITTLE_ENDIAN).getInt();
}
}
This is pretty slow for a 2k*2k array, it takes about 25 seconds. I can see in the DDMS that the garbage collector is working overtime, so that is probably one reason for the slowness.
There has to be a more efficient way of using the ByteBuffer to read that file into the array, but I'm not seeing it at the moment. Any idea on how to speed this up?

Why not read into a 4-byte buffer and then rearrange the bytes manually? It will look like this:
for (int i=0; i < SIZE_Y; i++) {
for (int j=0; j < SIZE_X; j++) {
inp.read(buffer);
int nextInt = (buffer[0] & 0xFF) | (buffer[1] & 0xFF) << 8 | (buffer[2] & 0xFF) << 16 | (buffer[3] & 0xFF) << 24;
test_data[j][SIZE_Y - i - 1] = nextInt;
}
}
Of course, it is assumed that read reads all four bytes, but you should check for the situation when it's not. This way you won't create any objects during reading (so no strain on the garbage collector), you don't call anything, you just use bitwise operations.

If you are on a platform that supports memory-mapped files, consider the MappedByteBuffer and friends from java.nio
FileChannel channel = new RandomAccessFile(procData, "r").getChannel();
MappedByteBuffer map = channel.map(FileChannel.MapMode.READ_ONLY, 0, 4 * SIZE_X * SIZE_Y);
map.order(ByteOrder.LITTLE_ENDIAN);
IntBuffer buffer = map.asIntBuffer();
int[][] test_data = new int[SIZE_X][SIZE_Y];
for (int i=0; i < SIZE_Y; i++) {
for (int j=0; j < SIZE_X; j++) {
test_data[j][SIZE_Y - i - 1] = buffer.get();
}
}
If you need cross-platform support or your platform lacks memory-mapped buffers, you may still want to avoid performing the conversions yourself using an IntBuffer. Consider dropping the BufferedInputStream, allocating a larger ByteBuffer yourself and obtaining a little-endian IntBuffer view on the data. Then in a loop reset the buffer positions to 0, use DataInputStream.readFully to read the large regions at once into the ByteBuffer, and pull int values out of the IntBuffer.

First of all, your 'inp.read(buffer)' is unsafe, as read contract does not guarantee that it will read all 4 bytes.
That aside, for quick transformation use the algorithm from DataInputStream.readInt
I've adapted for you case of byte array of 4 bytes:
int little2big(byte[ ] b) {
return (b[3]&0xff)<<24)+((b[2]&0xff)<<16)+((b[1]&0xff)<<8)+(b[0]&0xff);
}

I don't think it is necessary to reinvent the wheel and perform the byte reordering for endianness again. This is error prone and there is a reason a class like ByteBuffer exists.
Your code can be optimized in the sense that it wastes objects. When a byte[] is wrapped by a ByteBuffer the buffer adds a view, but the original array remains the same. It does not matter wheather the original array is modified/read from directly or the ByteBuffer instance is used.
Therefore, you only need to initialize one instance of ByteBuffer and also have to set the ByteOrder once.
To start again, just use rewind() to set the counter again to the beginning of the buffer.
I have taken your code and modified it as desribed. Be aware that it does not check for errors if there are not enough bytes in the input left. I would suggest to use inp.readFully, as this will throw EOFException if not enough bytes to fill the buffer are found.
int[][] test_data = new int[SIZE_X][SIZE_Y];
ByteBuffer byteBuffer = ByteBuffer.wrap(new byte[4]).order(ByteOrder.LITTLE_ENDIAN);
for (int i=0; i < SIZE_Y; i++) {
for (int j=0; j < SIZE_X; j++) {
inp.read(byteBuffer.array());
byteBuffer.rewind();
test_data[j][SIZE_Y - i - 1] = byteBuffer.getInt();
}
}