I have some binary operations that are not working like I expect.
I have byte array with the first 2 bytes having these values : 0x5, and 0xE0.
I want to combine them into an integer value that should be 0x5E0.
I tried doing :
int val = (b[i]) << 8 | b[i+1];
but the value is coming out 0xFFFFFFEE0 and the first byte 0x5 is getting lost
I thought this would be easy? What am I doing wrong?
Try: int val = ((b[i] & 0xff) << 8) | (b[i + 1] & 0xff). Bytes are (unfortunately) signed in Java, so if the high bit is set, it gets sign-extended when converted to an integer.
The problem is that byte data type is signed. Therefore, b[i+1] gets sign-extended before performing the operation, becoming 0xFFFFFFE0. When it gets OR-ed with 0x0500 from b[i]<<8, the 0x0500 gets lost.
You can fix this by AND-ing with 0xFF before performing the operation:
public static int toInt16(byte high, byte low) {
int res = (high << 8);
res |= (low & 0xFF);
return res & 0xFFFF;
}
Demo.
Related
i have a java code to read a unsigned integer from high 4bit and another from lower 4bit
byte[] value = getBytes(1);
int first = (value[0] & 0xF0) >> 4;
int second = value[0] & 0x0F;
i have to write those 2 integers(first, second) back to new 1 byte using java.
please help me
Basically:
first << 4 | second
(assuming first and second were obtained with the original code, so they are in the range 0x0..0xF).
However, the result of bitwise operations is int for int operands, so this expression is of type int. You need to cast it to byte:
byte b = (byte) (first << 4 | second)
For a peer-to-peer audio client, I need to have the ability to change the output volume to a desired level. In my case, the volume is a floating point number between zero and one.
I modify the audio stream this way:
void play(byte[] buf)
{
for (int i = 0; i < buf.length; i += 2)
{
// sample size is 2 bytes, so convert to int and then back
int data = ((buf[i + 1] & 0xFF) << 8) | (buf[i] & 0xFF);
data = (int) (data * outputVolume);
buf[i] = (byte) (data & 0xFF);
buf[i + 1] = (byte) (data >> 8);
}
line.write(buf, 0, Math.min(buf.length, line.available()));
}
Now, when outputVolume is set to 0, the output is silent. When it is set to 1, it behaves normal and quality is fine (as it is not modified). But any numbers between 0 and 1 produce a horrible noise which is louder than the expected stream itself. At 0.5, the noise reaches it's loudest point.
I don't want to use the controls of the audio mixer itself (like gain control or volume control) because I had compatibility problems this way and later on, I want to modify the byte stream even more so I have to iterate through the stream anyway.
Assuming the audio data is signed (because I think it would be pretty unusual to have unsigned 16-bit samples), there is a mistake in that code, because you also need to sign extend the sample.
You can remove the & 0xFF from the high byte which will let sign extension happen automatically:
int data = (buf[i + 1] << 8) | (buf[i] & 0xFF);
If for some reason you couldn't modify the and-shift-or expression, you could do sign extension like this:
int data = ((buf[i + 1] & 0xFF) << 8) | (buf[i] & 0xFF);
data = (data << 16) >> 16;
The result of the shifting expression is equivalent to this:
if (data > 32767)
data -= 65536;
And this:
if ((i & 0x80_00) != 0)
i |= 0xFF_FF_00_00;
(Those would also work.)
However, in your case you can just remove the & 0xFF from the high byte.
For a quick explanation, if you had some 16-bit sample like this (which is -1):
11111111_11111111
If you just convert to 32-bit without sign extending, you would get:
00000000_00000000_11111111_11111111
But that's 65536, not -1. Sign extension fills the upper bits with 1s if the MSB in the 16-bit value was set:
11111111_11111111_11111111_11111111
Refer to the question How to Convert two 8 bit represented byte to single 16 bit represented integer value in Android.
I got an answer like this
short yourinteger16 = (short)(((bytes[0] & 0xFF) << 8) | (bytes[1] & 0xFF));
This answer is correct for the positive number. But in the case of the negative number, it's failing.
For example, I am sending the value from the BLE to the application as -10. The value will convert from the BLE as -10000 because of mAh/mV conversion of current and voltage. These values are split into two bytes and I am getting the byte value as -39 and -16 in my application. I am passing the byte to the method as like below.
short yourinteger16 = (short)(((-39 & 0xFF) << 8) | (-16 & 0xFF));
But I am getting the result as 9.77 as the float value of yourinteger16 .
Anyone has any idea about this?. Any solution please update me.
Full Code:
Integer ampValue = null;
if (mBleDataHashMap.containsKey(SuperMuttBleConst.RESP_I_HIGH) &&
mBleDataHashMap.containsKey(SuperMuttBleConst.RESP_I_LOW)) {
ampValue = get8ByteTo16Byte(mBleDataHashMap.get(SuperMuttBleConst.RESP_I_HIGH),
mBleDataHashMap.get(SuperMuttBleConst.RESP_I_LOW));
}
if (ampValue != null) {
float newAmp = ampValue.floatValue();
newAmp = newAmp/1000;
mAmpTextvw.setText("" + newAmp);
}
Method
protected Integer get8ByteTo16Byte(int firstValue, int secondValue) {
Short integerValue = (short)((((byte) firstValue & 0xFF) << 8) | ((byte) secondValue & 0xFF));
return new Integer(integerValue);
}
You're receiving -39 and -16 perfectly (the high and low bytes for -10000, respectively).
Use addition instead of ORing the high and low bytes.
Please try the following
short result = (short) (((short)(-39 & (byte)0xFF) << 8) + (short)(-16 & (byte)0xFF));
The negative low bytes is causing trouble for the high byte when dealing with 2's complement arithmetic.
I have to compress a list of short-values into a byte array, but only the last X bits of the value.
Given this method:
byte[] compress(int bitsPerWord, List<Short> input){
...
}
The BitsPerWorld will never be bigger than the given values in the input field.
Example: 10 bits per word => maximum value 1023
I also may not waste bits, I have to save X bits in the first Y bytes, and then append the next X bits directly to them.
Example:
Input(Short) [ 500, 150, 100 ]
Input(Binary):0000000111110100 0000000001101000 0000000001100100
Output (10 bits per short): 0111110100 0001101000 0001100100
Output (As byte array):0111 1101 0000 0110 1000 0001 1001 0000
What the result should look like
Any way to do this efficiently? BitSet seems not fitting for this task, because i would have to set every single bit explicit.
Thanks
Efficient in what way?
In terms of work required, extending BitSet adding a bulk put method and an index is super efficient; little work and thinking required.
The alternative, shifting and masking bits is moderately complicated in terms of programming effort if you know your ways with bitwise operations. It may be a major obstacle if you don't.
Considering you already use wrapper types and collections, indicating troughput is not your major concern, extending BitSet is probably all you need.
You need to perform some bit manipulations, and for that to work you need to find a repeatable pattern. In this case, you have a list of "short" values, but actually you just use the rightmost 10 bits. Since you want to pack those into bytes, the minimum repeatable pattern is 40 bits long (5 bytes, 4 10-bit values). That is the "block size" for processing.
You would then have a loop that would do the main parsing of full blocks, plus maybe a special case at the end for the final incomplete block.
byte[] pack10(List<Short> source) {
final int nBlk = source.size() / 4;
final int remBits = (source.size() % 4) * 10;
final int remBytes = (remBits / 8) + (remBits % 8 > 0 ? 1 : 0);
byte[] ret = new byte[nBlk*5 + remBytes];
final short bitPat = (short)0b0000001111111111;
for (int iBlk = 0; iBlk < nBlk; ++iBlk) {
// Parse full blocks
List<Short> curS = source.subList(iBlk*4, (iBlk+1)*4);
ret[iBlk*5 ] = (byte) ((curS.get(0) & bitPat) >> 2);
ret[iBlk*5+1] = (byte) ((curS.get(0) & bitPat) << 6
| (curS.get(1) & bitPat) >> 4);
ret[iBlk*5+2] = (byte) ((curS.get(1) & bitPat) << 4
| (curS.get(2) & bitPat) >> 6);
ret[iBlk*5+3] = (byte) ((curS.get(2) & bitPat) << 2
| (curS.get(3) & bitPat) >> 8);
ret[iBlk*5+4] = (byte) (curS.get(3) & bitPat);
}
// Parse final values
List<Short> remS = source.subList(nBlocks*4, source.size());
if (remS.size() >= 1) {
ret[nBlk*5 ] = (byte) ((remS.get(0) & bitPat) >> 2);
ret[nBlk*5+1] = (byte) ((remS.get(0) & bitPat) << 6);
}
if (remS.size() >= 2) { // The first byte is appended to
ret[nBlk*5+1] |= (byte) ((remS.get(1) & bitPat) >> 4);
ret[nBlk*5+2] = (byte) ((remS.get(1) & bitPat) << 4);
}
if (remS.size() == 3) { // The first byte is appended to
ret[iBlk*5+2] |= (byte) ((curS.get(2) & bitPat) >> 6);
ret[iBlk*5+3] = (byte) ((curS.get(2) & bitPat) << 2);
}
return ret;
}
That is a specific version for 10-bit values; if you want a version with a generic number of values you'd have to generalise from that. The bit pattern operations changes, and all the system becomes less efficient if the pattern is computed at runtime (i.e. if the number of bits is a variable like in your example).
There are several people who have already written a BitOutputStream in Java. Pick one of them, wrap it in a ByteArrayOutputStream, and you’re done.
All,
I have been practicing coding problems online. Currently I am working on a problem statement Problems where we need to convert Big Endian <-> little endian. But I am not able to jot down the steps considering the example given as:
123456789 converts to 365779719
The logic I am considering is :
1 > Get the integer value (Since I am on Windows x86, the input is Little endian)
2 > Generate the hex representation of the same.
3 > Reverse the representation and generate the big endian integer value
But I am obviously missing something here.
Can anyone please guide me. I am coding in Java 1.5
Since a great part of writing software is about reusing existing solutions, the first thing should always be a look into the documentation for your language/library.
reverse = Integer.reverseBytes(x);
I don't know how efficient this function is, but for toggling lots of numbers, a ByteBuffer should offer decent performance.
import java.nio.ByteBuffer;
import java.nio.ByteOrder;
...
int[] myArray = aFountOfIntegers();
ByteBuffer buffer = ByteBuffer.allocate(myArray.length*Integer.BYTES);
buffer.order(ByteOrder.LITTLE_ENDIAN);
for (int x:myArray) buffer.putInt(x);
buffer.order(ByteOrder.BIG_ENDIAN);
buffer.rewind();
int i=0;
for (int x:myArray) myArray[i++] = buffer.getInt(x);
As eversor pointed out in the comments, ByteBuffer.putInt() is an optional method, and may not be available on all Java implementations.
The DIY Approach
Stacker's answer is pretty neat, but it is possible to improve upon it.
reversed = (i&0xff)<<24 | (i&0xff00)<<8 | (i&0xff0000)>>8 | (i>>24)&0xff;
We can get rid of the parentheses by adapting the bitmasks. E. g., (a & 0xFF)<<8 is equivalent to a<<8 & 0xFF00. The rightmost parentheses were not necessary anyway.
reversed = i<<24 & 0xff000000 | i<<8 & 0xff0000 | i>>8 & 0xff00 | i>>24 & 0xff;
Since the left shift shifts in zero bits, the first mask is redundant. We can get rid of the rightmost mask by using the logical shift operator, which shifts in only zero bits.
reversed = i<<24 | i>>8 & 0xff00 | i<<8 & 0xff0000 | i>>>24;
Operator precedence here, the gritty details on shift operators are in the Java Language Specification
Check this out
int little2big(int i) {
return (i&0xff)<<24 | (i&0xff00)<<8 | (i&0xff0000)>>8 | (i>>24)&0xff;
}
The thing you need to realize is that endian swaps deal with the bytes that represent the integer. So the 4 byte number 27 looks like 0x0000001B. To convert that number, it needs to go to 0x1B000000... With your example, the hex representation of 123456789 is 0x075BCD15 which needs to go to 0x15CD5B07 or in decimal form 365779719.
The function Stacker posted is moving those bytes around by bit shifting them; more specifically, the statement i&0xff takes the lowest byte from i, the << 24 then moves it up 24 bits, so from positions 1-8 to 25-32. So on through each part of the expression.
For example code, take a look at this utility.
Java primitive wrapper classes support byte reversing since 1.5 using reverseBytes method.
Short.reverseBytes(short i)
Integer.reverseBytes(int i)
Long.reverseBytes(long i)
Just a contribution for those who are looking for this answer in 2018.
I think this can also help:
int littleToBig(int i)
{
int b0,b1,b2,b3;
b0 = (i&0x000000ff)>>0;
b1 = (i&0x0000ff00)>>8;
b2 = (i&0x00ff0000)>>16;
b3 = (i&0xff000000)>>24;
return ((b0<<24)|(b1<<16)|(b2<<8)|(b3<<0));
}
Just use the static function (reverseBytes(int i)) in java which is under Integer Wrapper class
Integer i=Integer.reverseBytes(123456789);
System.out.println(i);
output:
365779719
the following method reverses the order of bits in a byte value:
public static byte reverseBitOrder(byte b) {
int converted = 0x00;
converted ^= (b & 0b1000_0000) >> 7;
converted ^= (b & 0b0100_0000) >> 5;
converted ^= (b & 0b0010_0000) >> 3;
converted ^= (b & 0b0001_0000) >> 1;
converted ^= (b & 0b0000_1000) << 1;
converted ^= (b & 0b0000_0100) << 3;
converted ^= (b & 0b0000_0010) << 5;
converted ^= (b & 0b0000_0001) << 7;
return (byte) (converted & 0xFF);
}