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How to convert 16-bit audio created with Android's AudioRecord to 12-bit audio through bit shifting?

I am attempting to convert 16 bit audio into 12 bit audio. However, I am quite inexperienced with such conversions and believe my approach is possibly incorrect or flawed.
The use case, as context for the code snippets below, is an Android app which the user can speak into and that audio is transmitted to an IoT device for immediate playback. The IoT device expects audio in mono 12 bit, 8k sample rate, little endian, unsigned, with the data stored in the first twelve bits (0-11) and final four bits (12-15) are zeroes. Audio data needs to be received in packets of 1000 bytes.
The audio is being created in the Android app through the use of AudioRecord. The instantiation of which is as follows:
int bufferSize = 1000;
this.audioRecord = new AudioRecord(
MediaRecorder.AudioSource.MIC,
8000,
AudioFormat.CHANNEL_IN_MONO,
AudioFormat.ENCODING_PCM_16BIT,
bufferSize
);
In a while loop, the AudioRecord is being read from by 1000 byte packets and modified to the specifications in the use case. Not sure this part is relevant, but for completeness:
byte[] buffer = new byte[1000];
audioRecord.read(buffer, 0, buffer.length);
byte[] modifiedBytes = convert16BitTo12Bit(buffer);
Then the modifiedBytes are sent off to the device.
Here are the methods which modify the bytes. Basically, to conform to the specifications, I am shifting the bits in each 16 bit set (tossing the least significant 4) and adding zeroes to the final four spots. I do this through BitSet.
/**
* Takes a byte array presented as 16 bit audio and converts it to 12 bit audio through bit
* manipulation. Packets must be of 1000 bytes or no manipulation will occur and the input
* will be immediately returned.
*/
private byte[] convert16BitTo12Bit(byte[] input) {
if (input.length == 1000) {
for (int i = 0; i < input.length; i += 2) {
Log.d(TAG, "convert16BitTo12Bit: pass #" + (i / 2));
byte[] chunk = new byte[2];
System.arraycopy(input, i, chunk, 0, 2);
if (!isEmptyByteArray(chunk)) {
byte[] modifiedBytes = convertChunk(chunk);
System.arraycopy(
modifiedBytes,
0,
input,
i,
modifiedBytes.length
);
}
}
return input;
}
Log.d(TAG, "convert16BitTo12Bit: Failed - input is not 1000 in length; it is " + input.length);
return input;
}
/**
* Converts 2 bytes 16 bit audio into 12 bit audio. If the input is not 2 bytes, the input
* will be returned without manipulation.
*/
private byte[] convertChunk(byte[] chunk) {
if (chunk.length == 2) {
BitSet bitSet = BitSet.valueOf(chunk);
Log.d(TAG, "convertChunk: bitSet starts as " + bitSet.toString());
modifyBitSet(bitSet);
Log.d(TAG, "convertChunk: bitSet ends as " + bitSet.toString());
return bitSet.toByteArray();
}
Log.d(TAG, "convertChunk: Failed = chunk is not 2 in length; it is " + chunk.length);
return chunk;
}
/**
* Removes the first four bits and shifts the rest to leave the final four bits as 0.
*/
private void modifyBitSet(BitSet bitSet) {
for (int i = 4; i < bitSet.length(); i++) {
bitSet.set(i - 4, bitSet.get(i));
}
if (bitSet.length() > 8) {
bitSet.clear(12, 16);
} else {
bitSet.clear(4, 8);
}
}
/**
* Returns true if the byte array input contains all zero bits.
*/
private boolean isEmptyByteArray(byte[] input) {
BitSet bitSet = BitSet.valueOf(input);
return bitSet.isEmpty();
}
Unfortunately, this approach produces subpar results. The audio is quite noisy and it is difficult to make out what someone is saying (but you can hear that words are being spoken).
I also have been playing around with just saving the bytes to a file and playing it back on Android through AudioTrack. I noticed that if I just remove the first four bits and do not shift anything, the audio actually sounds good, as such:
private void modifyBitSet(BitSet bitSet) {
bitSet.clear(0, 4);
}
However, when played through the device, it sounds even worse, and I don't even think I can make out any words.
Clearly, my approach is not working here. Central question is how would one convert a 16 bit chunk into 12 bit audio and maintain audio quality given the requirement that the final four bits must be zero? Additionally, given my larger approach of using AudioRecord to obtain the audio, would such a solution for the prior question fit this use case?
Please let me know if there is anything more I can provide to clarify these questions and my intent.

Given that the audio is 16 bits but must be changed to 12 with four zeros at the end, four bits somewhere do have to be tossed.
Yes, of course and there is no other way, is there?
This is something quick that I can comeout with right now. Certainly not fully tested though. Only tested with input of 2 and 4 bytes. I'll leave it to you to test it.
//Reminder :: Convert as many as possible.
//Reminder :: To calculate the required size for store:
//if((bytes.length & 1) == 0) Math.round((bytes.length * 6) / 8F) : Math.round(((bytes.length - 1) * 6) / 8F).
//Return :: Amount of converted bytes.
public static final int convert16BitTo12Bit(final byte[] bytes, final byte[] store)
{
final int size = bytes.length;
int storeIndex = 0;
//Copy the first 2 bytes into store.
store[storeIndex++] = bytes[0]; store[storeIndex] = bytes[1];
if(size < 4) {
store[storeIndex] = (byte)(store[storeIndex] & 0xF0);
return 2;
}
final int result;
final byte tmp;
// 11111111 11110000 00000000 00000000
//+ 11111111 11110000 (<< 12)
//= 11111111 11111111 11111111 00000000 (1)
//-----------------------------------------
// 11111111 00000000 00000000 00000000 (1)
//+ 11111111 11110000 (<< 16)
//= 11111111 11111111 11110000 00000000 (2)
//-----------------------------------------
// 11110000 00000000 00000000 00000000 (2)
//+ 1111 11111111 0000 (<< 20)
//= 11111111 11111111 00000000 00000000 (3)
//-----------------------------------------
// 00000000 00000000 00000000 00000000 (3)
//+ 11111111 11110000 (<< 24)
//= 11111111 11110000 00000000 00000000
for(int i=2, shiftBits = 12; i < size; i += 2) {
if(shiftBits == 24) {
//Copy 2 bytes from bytes[] into store[] and move on.
store[storeIndex] = bytes[i];
//Never store byte 0 (Garbage).
tmp = (byte)(bytes[i + 1] & 0xF0); //Bit order: 11110000.
if(tmp != 0) store[++storeIndex] = tmp;
shiftBits = 12; //Reset
} else if(shiftBits == 20) {
result = ((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
| (((bytes[i] & 0xFF) << 20) | ((bytes[i + 1] & 0xFF) << 12));
store[storeIndex] = (byte)((result >> 24) & 0xFF);
tmp = (byte)((result >> 16) & 0xFF);
//Never store byte 0 (Garbage).
if(tmp != 0) store[++storeIndex] = tmp;
shiftBits = 24;
} else if(shiftBits == 16) {
result = ((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
| (((bytes[i] & 0xFF) << 16) | ((bytes[i + 1] & 0xFF) << 8));
store[storeIndex] = (byte)((result >> 16) & 0xFF);
tmp = (byte)((result >> 8) & 0xF0);
//Never store byte 0 (Garbage).
if(tmp != 0) store[++storeIndex] = tmp;
shiftBits = 20;
} else {
result = ((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
| (((bytes[i] & 0xFF) << 12) | ((bytes[i + 1] & 0xFF) << 4));
store[storeIndex] = (byte)((result >> 16) & 0xFF);
tmp = (byte)((result >> 8) & 0xFF);
//Never store byte 0 (Garbage).
if(tmp != 0) store[++storeIndex] = tmp;
shiftBits = 16;
}
}
return ++storeIndex;
}
Explanations
result = ((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
| (((bytes[i] & 0xFF) << 20) | ((bytes[i + 1] & 0xFF) << 12));
What this does is basically merge two integers into one.
((store[storeIndex - 1] << 24) | ((store[storeIndex] & 0xFF) << 16))
The first one is make an integer with same constant bit position.
(((bytes[i] & 0xFF) << 20) | ((bytes[i + 1] & 0xFF) << 12));
The latter is for 2 current bytes with different bit positions.
(...) | (...)
Pipe or vertical bar at the middle is to merge these two integers we've just created into one.
Usage
To use this method is pretty straight forward.
byte[] buffer = new byte[1000];
byte[] store;
if((buffer.length & 1) == 0) { //Even.
store = new byte[Math.round((bytes.length * 6) / 8F)];
} else { //Odd.
store = new byte[Math.round(((bytes.length - 1) * 6) / 8F)];
}
audioRecord.read(buffer, 0, buffer.length);
int convertedByteSize = convert16BitTo12Bit(buffer, store);
System.out.println("size: " + convertedByteSize);

I have discovered a solution that produces clear audio. First, it is important to recount the requirements for the use case, which is 12 bit unsigned mono audio which will be read in little endian by the device in packets of 1000 bytes.
The initialization and configuration of the AudioRecord as described in the question is fine.
Once the 1000 bytes of audio is read from AudioRecord, it can be put into a ByteBuffer and defined as little endian for modification, and then put into a ShortBuffer to do manipulation on the 16 bit level:
// Audio specifications of device is in little endian.
ByteBuffer byteBuffer = ByteBuffer.wrap(input).order(ByteOrder.LITTLE_ENDIAN);
// Turn into a ShortBuffer so bitwise manipulation can occur on the 16 bit level.
ShortBuffer shortBuffer = byteBuffer.asShortBuffer();
Next, in a loop, take each short and modify it to 12 bit unsigned:
for (int i = 0; i < shortBuffer.capacity(); i++) {
short currentShort = shortBuffer.get(i);
shortBuffer.put(i, convertShortTo12Bit(currentShort));
}
This can be accomplished by shifting the 16 bits four spaces to the right to turn it into 12 bit signed. Then, to convert to unsigned, add 2048. For our purposes as a safety step, we also mask the least significant four bits as required by device, but given the shifting and adding, it shouldn't be the case that any bits actually remain there:
private static short convertShortTo12Bit(short input) {
int inputAsInt = input;
inputAsInt >>>= 4;
inputAsInt += 2048;
input = (short) (inputAsInt & 0B0000111111111111);
return input;
}
If one wishes to return 12 bits to 16 bits, do the reverse for each short (subtract 2048 and shift four spaces to the left).

How to edit binary file in Android?

I want to create an app that can open binary file, read and edit it's content in decimal values, and then save it back to binary file.
This file that I want to open have multiple parameters written in 2 bytes. e.g. 00 06 00 50 01 AB = 6, 80, 427
So far I managed to open bin file, put all data from file into byte[].
Here is the code to open and read file:
String path = getFilesDir().getAbsolutePath();
File file = new File(path + "/file.bin");
int size = (int) file.length();
byte[] bytes = new byte[size];
FileInputStream fis = new FileInputStream(file);
try {
fis.read(bytes);
...
} finally {
fis.close();
}
Here is first attempt to convert byte[] into deciaml values:
for (int i = 0; i < size / 2; i = i + 2) {
intArray[i] = ((bytes[i] & 0xff) << 8) | (bytes[i + 1] & 0xff);
}
But in this code only every second value is correct. It's shows 6,0,80,0,1 and should be 6,80,1.
The second idea was to convert byte[] into String[] and then into e.g. int[] but It doesn't work as intended. In this fragment I didn't parse whole array, but only hexArray[4] which was 50 in hex and 80 in dec, but the result was 50.
for (int i = 0; i < size; i = i + 2) {
num1 = String.format("%02X", bytes[i] & 0xFF);
num2 = String.format("%02X", bytes[i + 1] & 0xFF);
String twoByte = num1 + num2;
hexArrayWithNull[i] = twoByte;
}
String[] hexArray = Arrays.stream(hexArrayWithNull).filter(Objects::nonNull).toArray(String[]::new);
num = String.format("%d", Integer.parseInt(hexArray[4]));
I'm quite stuck here and need some guidance. I must say also that I'm new in programming.
Thanks

You increment i in each iteration by 2 and you store in intArray[i]. So, you store in index [0] then index [2] then index [4] and so on. This explains the zeros at odd indices [1], [3], ...
Also, you must continue the loop until size, not size/2.
Here is the correct loop:
for (int i = 0; i < size; i = i + 2) {
intArray[i/2] = ((bytes[i] & 0xff) << 8) | (bytes[i + 1] & 0xff);
}
Here is the another possibility:
for (int i = 0; i < size/2; i = i + 1) {
intArray[i] = ((bytes[2*i] & 0xff) << 8) | (bytes[2*i + 1] & 0xff);
}
[EDIT]
The reverse operation would be:
for (int i = 0; i < size/2; i = i + 1) {
bytes[2*i] = (intArray[i] & 0xff00) / 256; //or >>8 instead of /256
bytes[2*i+1] = intArray[i] & 0xff;
}

Byte array to int getting weird int result

I know there are many similar questions in here, but I have some weird case. What I want to do is to convert a byte[4] to an int.
Here is the conversion from int to byte:
int data24bit = 51;
for (int i = 0; i < 3; i++) {
data8bit = (byte)(data24bit & 0x0000FF);
data24bit = data24bit >> 8;
byte_file.put(data8bit);
}
So far is clear enough. After that I want to read this 4 bytes to get the 51 back. I tried to do it by different ways:
Reading 4 bytes:
byte[] bytes = new byte[4];
for (int i = 0; i < 3; i++) {
byte b = dis.readByte();
bytes[i] = b;
}
// bytes[3] = (byte)(0x000000);
Convert bytes to int:
int value = 0;
value = ((0xFF & bytes[0]) << 24) | ((0xFF & bytes[1]) << 16) |
((0xFF & bytes[2]) << 8) | (0xFF & bytes[3]);
or
value = ByteBuffer.wrap(bytes).getInt();
or
value = new BigInteger(bytes).intValue();
I always get 855638016 as result where 51 is expected.
When I debug the code and look into the byte array I can see the following content: [51, 0, 0, 0].
What am I doing wrong?

The problem is that you're writing the bytes in little-endian (least significant byte first), but read it back assuming big-endian.
After writing it out, your byte array looks like this:
[51, 0, 0, 0]
Then you're trying to convert that back into an integer, like in this example from your post:
value = ((0xFF & bytes[0]) << 24)
| ((0xFF & bytes[1]) << 16)
| ((0xFF & bytes[2]) << 8)
| (0xFF & bytes[3]);
If you fill in the actual values, that calculation is basically this:
value = 51 * 256 * 256 * 256
+ 0 * 256 * 256
+ 0 * 256
+ 0
= 855638016
While what you actually want is this:
value = 0 * 256 * 256 * 256
+ 0 * 256 * 256
+ 0 * 256
+ 51
= 51
The fixed calculation would thus be this:
value = ((0xFF & bytes[3]) << 24)
| ((0xFF & bytes[2]) << 16)
| ((0xFF & bytes[1]) << 8)
| (0xFF & bytes[0]);

Ok stupid enough but I just didn't preserve the byte order.
[51, 0, 0, 0] -> is 855638016
[0, 0, 0, 51] -> is 51

why is the base64 encode java code doing this

So I'm trying to understand base64 encoding better and I came across this implementation on wikipedia
private static String base64Encode(byte[] in) {
StringBuffer out = new StringBuffer((in.length * 4) / 3);
int b;
for (int i = 0; i < in.length; i += 3) {
b = (in[i] & 0xFC) >> 2;
out.append(codes.charAt(b));
b = (in[i] & 0x03) << 4;
if (i + 1 < in.length) {
b |= (in[i + 1] & 0xF0) >> 4;
out.append(codes.charAt(b));
b = (in[i + 1] & 0x0F) << 2;
if (i + 2 < in.length) {
b |= (in[i + 2] & 0xC0) >> 6;
out.append(codes.charAt(b));
b = in[i + 2] & 0x3F;
out.append(codes.charAt(b));
} else {
out.append(codes.charAt(b));
out.append('=');
}
} else {
out.append(codes.charAt(b));
out.append("==");
}
}
return out.toString();
}
And I'm following along and I get to the line:
b = (in[i] & 0xFC) >> 2;
and I don't get it...why would you bitwise and 252 to a number then shift it right 2...wouldn't it be the same if you just shifted the byte itself without doing the bitwise operation? example:
b = in[i] >> 2;
Say my in[i] was the letter e...represented as 101 or in binary 01100101. If I shift that 2 to the right I get 011001 or 25. If I bitwise & it I get
01100101
11111100
--------
01100100
but then the shift is going to chop off the last 2 anyway...so why bother doing it?
Can somebody clarify for me please. Thanks.

IN in[i] >> 2, in[i] is converted to an int first. If it was a negative byte (with the high bit set) it will be converted to a negative int (with the now-highest 24 bits set as well).
In (in[i] & 0xFC) >> 2, in[i] is converted to an int as above, and then & 0xFC makes sure the extra bits are all reset to 0.
You're partially right, in that (in[i] & 0xFF) >> 2 would give the same result. & 0xFF is a common way to convert a byte to a non-negative int in the range 0 to 255.
The only way to know for sure why the original developer used 0xFC, and not 0xFF, is to ask them - but I speculate that it's to make it more obvious which bits are being used.

How to change 37 bits in a byte array to all complement of existing value

I have a byte array myByteArray[82]
I want to change exactly 37 bits in this array to the complement value. ie. if bit0 has a '1' I want to change it to '0'. I need to change first 37 such bits to introduce error in this byte array.
Please suggest how to do this

It's not entirely clear what you are trying to do. My best understanding is that you have an array of 82 bytes and you want to invert the lowest 37 bits of the array. Since a byte is 8 bits, so you can do this:
byte[] myByteArray = new byte[82];
// invert lowest 32 bits, 8 at a time
for (int i = 0; i < 4; ++i) {
myByteArray[i] = (byte)(~myByteArray[i]);
}
// invert next five bits
myByteArray[4] = (byte) (
(myByteArray[4] & 0xE0) // top 3 bits unchanged
|
((~myByteArray[4)) & 0x1F) // bottom 5 bits inverted
);

try
byte[] a82 = ...
Set<Integer> set = new HashSet<Integer>();
while (set.size() < 37) {
set.add((int) (Math.random() * 82));
}
for (int i : set) {
int ibyte = i / 8;
int ibit = i % 8;
int m = 1 << ibit;
a[ibyte] ^= m;
}

This works:
int nBits = 37;
int i = 0;
for (; i<nBits / 8; ++i)
myByteArray[i] = (byte)((byte) myByteArray[i] ^ 0xFF);
myByteArray[i] = (byte)(myByteArray[i] ^ ((0xFF >>> 5) ^ 0xFF));
Whether or not you do the final XOR with the 0xFF on the last line depends on whether you consider the most significant bit to be the first bit (then use it) or the last bit (then omit it);

To invert a specific 37 bits, randomly chosen:
// array of 82 bytes with 37 selected bits set to 1, all the rest zero
// you could generate this programmatically as well if you need a different
// set of bits each time, but your question implies you don't
byte[] mask = { 0x00, 0x01, 0x02, 0x80, .... 0x00 };
for (int i=0; i<myByteArray.length; i++)
{
myByteArray[i] ^= mask[i];
}
Use the bit-exclusive-or operator ^. The truth table for xor is
M a s k
| 0 | 1
D -+---+---
a 0| 0 | 1
t -+---+---
a 1| 1 | 0
Wherever there's a 1 in the mask the corresponding bit in the data will be flipped.