How can I convert a short (2 bytes) to a byte array in Java, e.g.
short x = 233;
byte[] ret = new byte[2];
...
it should be something like this. But not sure.
((0xFF << 8) & x) >> 0;
EDIT:
Also you can use:
java.nio.ByteOrder.nativeOrder();
To discover to get whether the native bit order is big or small. In addition the following code is taken from java.io.Bits which does:
byte (array/offset) to boolean
byte array to char
byte array to short
byte array to int
byte array to float
byte array to long
byte array to double
And visa versa.
ret[0] = (byte)(x & 0xff);
ret[1] = (byte)((x >> 8) & 0xff);
A cleaner, albeit far less efficient solution is:
ByteBuffer buffer = ByteBuffer.allocate(2);
buffer.putShort(value);
return buffer.array();
Keep this in mind when you have to do more complex byte transformations in the future. ByteBuffers are very powerful.
An alternative that is more efficient:
// Little Endian
ret[0] = (byte) x;
ret[1] = (byte) (x >> 8);
// Big Endian
ret[0] = (byte) (x >> 8);
ret[1] = (byte) x;
Figured it out, its:
public static byte[] toBytes(short s) {
return new byte[]{(byte)(s & 0x00FF),(byte)((s & 0xFF00)>>8)};
}
Short to bytes convert method In Kotlin works for me:
fun toBytes(s: Short): ByteArray {
return byteArrayOf((s.toInt() and 0x00FF).toByte(), ((s.toInt() and 0xFF00) shr (8)).toByte())
}
Several methods have been mentioned here. But which one is the best? Here follows some proof that the following 3 approaches result in the same output for all values of a short
// loops through all the values of a Short
short i = Short.MIN_VALUE;
do
{
// method 1: A SIMPLE SHIFT
byte a1 = (byte) (i >> 8);
byte a2 = (byte) i;
// method 2: AN UNSIGNED SHIFT
byte b1 = (byte) (i >>> 8);
byte b2 = (byte) i;
// method 3: SHIFT AND MASK
byte c1 = (byte) (i >> 8 & 0xFF);
byte c2 = (byte) (i & 0xFF);
if (a1 != b1 || a1 != c1 ||
a2 != b2 || a2 != c2)
{
// this point is never reached !!
}
} while (i++ != Short.MAX_VALUE);
Conclusion: less is more ?
byte b1 = (byte) (s >> 8);
byte b2 = (byte) s;
(As other answers have mentioned, watch out for LE/BE).
It depends how you want to represent it:
big endian or little endian? That will determine which order you put the bytes in.
Do you want to use 2's complement or some other way of representing a negative number? You should use a scheme that has the same range as the short in java to have a 1-to-1 mapping.
For big endian, the transformation should be along the lines of:
ret[0] = x/256; ret[1] = x%256;
public short bytesToShort(byte[] bytes) {
return ByteBuffer.wrap(bytes).order(ByteOrder.LITTLE_ENDIAN).getShort();
}
public byte[] shortToBytes(short value) {
byte[] returnByteArray = new byte[2];
returnByteArray[0] = (byte) (value & 0xff);
returnByteArray[1] = (byte) ((value >>> 8) & 0xff);
return returnByteArray;
}
short to byte
short x=17000;
byte res[]=new byte[2];
res[i]= (byte)(((short)(x>>7)) & ((short)0x7f) | 0x80 );
res[i+1]= (byte)((x & ((short)0x7f)));
byte to short
short x=(short)(128*((byte)(res[i] &(byte)0x7f))+res[i+1]);
Related
This question already has answers here:
How do I convert Long to byte[] and back in java
(18 answers)
Closed 6 years ago.
I have made long value from byte array using this code
byte[] by = {07, 53 -70, 74};
long value = 0;
for (int i = 0; i < by.length; i++) {
value = ((value << 8) + (by[i] & 0xff));
}
System.out.println(value);
out put is 520010
now I want reverse process on this and I tried it this way
long ts = 520010;
tm_stp[0] = (byte) ((byte) ts>>24);
tm_stp[1] = (byte) ((byte) ts>>16);
tm_stp[2] = (byte) ((byte) ts>>8);
tm_stp[3] = (byte) ts;
for (byte b : tm_stp) {
System.out.println(b);
}
and output is 0 0 0 74
what is wrong in my second part of code please help me, Thanks!
The problem is the fact that you cast to byte too early:
tm_stp[0] = (byte) ((byte) ts>>24);
tm_stp[1] = (byte) ((byte) ts>>16);
tm_stp[2] = (byte) ((byte) ts>>8);
//^^^^
This causes the ts value to be truncated, replacing the first 24 bytes with 0s. After that, shifting by anything greater than or equal to 8 will return 0. To avoid incorrect behaviour with negative values, you should also use a bitmask. The correct code should look like this:
tm_stp[0] = (byte) ((ts >> 24) & 0xFF);
tm_stp[1] = (byte) ((ts >> 16) & 0xFF);
tm_stp[2] = (byte) ((ts >> 8) & 0xFF);
tm_stp[3] = (byte) ((ts >> 0) & 0xFF); // >> 0 not actually required, only for symmetry
Don't convert ts to byte before shifting.
tm_stp[0] = (byte) (ts >> 24);
tm_stp[1] = (byte) (ts >> 16);
tm_stp[2] = (byte) (ts >> 8);
tm_stp[3] = (byte) ts;
I am trying to convert a byte to integer. All the searches I have done use byte[] which I assume is any array. I want to convert F byte (not b as show below) but it gives error to change: not applicable to the argument.
byte F;
mmInStream.read(packetBytes);
b [counter]= packetBytes[0];
F=b [counter];
counter++;
temp = byteToInt(b); //Convert byte to int
Here is a byte To Int I found on one of the sites.
private int byteToInt(byte[] b) {
int value= 0;
for(int i=0;i<b.length;i++){
int n=(b[i]<0?(int)b[i]+256:(int)b[i])<<(8*i);
value+=n;
}
return value;
}
Simply do:
byte b = ...;
int signedInt = b; // For negative bytes, resulting in negative ints
int unsignedInt = 0xFF & b; // For negative bytes, resulting in positive ints
FYI: An int is 4 bytes. So, that is the reason why the methods you found on the internet are using an array of bytes. They assume you pass an array of 4 bytes, which will be stitched together to make an int.
you can use this:
int i = 234;
byte b = (byte) i;
System.out.println(b); // -22
int i2 = b & 0xFF;
System.out.println(i2); // 234
or this one also:
public static byte[] intToByteArray(int a)
{
byte[] ret = new byte[4];
ret[3] = (byte) (a & 0xFF);
ret[2] = (byte) ((a >> 8) & 0xFF);
ret[1] = (byte) ((a >> 16) & 0xFF);
ret[0] = (byte) ((a >> 24) & 0xFF);
return ret;
}
and
public static int byteArrayToInt(byte[] b)
{
return (b[3] & 0xFF) + ((b[2] & 0xFF) << 8) + ((b[1] & 0xFF) << 16) + ((b[0] & 0xFF) << 24);
}
if b is unsigned
int i = b & 0xff;
Actually I need to transfer the integer value along with the bitmap via bluetooth.. Now my problem is I need to transfer the integer as single byte value..
Is tat possible to convert int as single byte value.. and retrieve it as a integer there... I tried byteValue() and the casting thing but its not usefull.. If my approach is right just help me out with this or say some other way.
(Each time when I am using casting then it's returning as 65535)
What about this?
public static byte[] intToByteArray(int a)
{
byte[] ret = new byte[4];
ret[3] = (byte) (a & 0xFF);
ret[2] = (byte) ((a >> 8) & 0xFF);
ret[1] = (byte) ((a >> 16) & 0xFF);
ret[0] = (byte) ((a >> 24) & 0xFF);
return ret;
}
and
public static int byteArrayToInt(byte[] b)
{
return (b[3] & 0xFF) + ((b[2] & 0xFF) << 8) + ((b[1] & 0xFF) << 16) + ((b[0] & 0xFF) << 24);
}
If you're completely sure, that your int variable contains a byte value [-128; 127] then it should be as simple as:
int i = 100; // your int variable
byte b = (byte) i;
A single byte (8 bits) can only contain 2^8 unsigned integers, i.e [0, 255]. For signed you loose the first bit and the range becomes [-128, 127]. If your integer fits then a simple cast should work.
for 0-255 numbers.
int i = 200; // your int variable
byte b = (byte)(i & 0xFF);
Given
private int width = 400;
private byte [] data = new byte [2];
I want to split the integer "width" into two bytes and load data[0] with the high byte and data[1] with the low byte.
That is binary value of 400 = 1 1001 0000
so data[0] should contain 0000 0001
and data[1] should contain 1001 0000
Using simple bitwise operations:
data[0] = (byte) (width & 0xFF);
data[1] = (byte) ((width >> 8) & 0xFF);
How it works:
& 0xFF masks all but the lowest eight bits.
>> 8 discards the lowest 8 bits by moving all bits 8 places to the right.
The cast to byte is necessary because these bitwise operations work on an int and return an int, which is a bigger data type than byte. The case is safe, since all non-zero bits will fit in the byte. For more information, see Conversions and Promotions.
Edit: Taylor L correctly remarks that though >> works in this case, it may yield incorrect results if you generalize this code to four bytes (since in Java an int is 32 bits). In that case, it's better to use >>> instead of >>. For more information, see the Java tutorial on Bitwise and Bit Shift Operators.
For converting two bytes the cleanest solution is
data[0] = (byte) width;
data[1] = (byte) (width >>> 8);
For converting an integer to four bytes the code would be
data[0] = (byte) width;
data[1] = (byte) (width >>> 8);
data[2] = (byte) (width >>> 16);
data[3] = (byte) (width >>> 24);
It doesn't matter whether >> or >>> is used for shifting, any one bits created by sign extension will not end up in the resulting bytes.
See also this answer.
This should do what you want for a 4 byte int. Note, it stores the low byte at offset 0. I'll leave it as an exercise to the reader to order them as needed.
public static byte[] intToBytes(int x) {
byte[] bytes = new byte[4];
for (int i = 0; x != 0; i++, x >>>= 8) {
bytes[i] = (byte) (x & 0xFF);
}
return bytes;
}
Integer is 32 bits (=4 bytes) in java, you know?
width & 0xff will give you the first byte,
width & 0xff00 >> 8 will give you the second, etc.
To get the high byte, shift right by 8 bits then mask off the top bytes. Similarly, to get the low byte just mask off the top bytes.
data[0] = (width >> 8) & 0xff;
data[1] = width & 0xff;
int width = 400;
byte [] data = new byte [2];
data[0] = (byte) ((width & 0xFF00) >> 8);
data[1] = (byte) (width & 0xFF);
for(int b = 0; b < 2; b++) {
System.out.println("printing byte " + b);
for(int i = 7; i >= 0; i--) {
System.out.println(data[b] & 1);
data[b] = (byte) (data[b] >> 1);
}
}
I suggest you have a look at the source for HeapByteBuffer. It has the conversion code for all primitive data types. (In fact you could just use a ByteBuffer ;)
I am reading 8 bytes of data in from a hardware device. I need to convert them into a numeric value. I think I want to convert them to a long as that should fit 8 bytes. I am not very familiar with Java and low level data type operations. I seem to have two problems (apart from the fact there is almost no documentation for the hardware in question), The bytes are expecting to be unsigned, so I can't do a straight integer conversion. I am not sure what endianness they are.
Any advice would be appreciated.
Ended up with this (taken from some source code I probably should have read a week ago):
public static final long toLong (byte[] byteArray, int offset, int len)
{
long val = 0;
len = Math.min(len, 8);
for (int i = (len - 1); i >= 0; i--)
{
val <<= 8;
val |= (byteArray [offset + i] & 0x00FF);
}
return val;
}
Shifting bytes according to the endianness of the data is fairly straightforward. There is a small trick with the long datatype, however, because a binary operation on integral types of int or smaller promotes the operands to int. For left shifts larger than 31 bits, this will result in zero, since all of the bits have been shifted out of the int range.
So, force promotion to long by including a long operand in the calculation. Below, I do this by masking each byte with the value 0xFFL, which is a long, forcing the result to be a long.
byte[] buf = new byte[8];
/* Fill buf somehow... */
long l = ((buf[0] & 0xFFL) << 56) |
((buf[1] & 0xFFL) << 48) |
((buf[2] & 0xFFL) << 40) |
((buf[3] & 0xFFL) << 32) |
((buf[4] & 0xFFL) << 24) |
((buf[5] & 0xFFL) << 16) |
((buf[6] & 0xFFL) << 8) |
((buf[7] & 0xFFL) << 0) ;
Byte#longValue() should do it
And if not (thanks for the source example) you can use java.nio.ByteBuffer such as in
public static long toLong(byte[] b) {
ByteBuffer bb = ByteBuffer.allocate(b.length);
bb.put(b);
return bb.getLong();
}
The initial order is BIG_ENDIAN you can reed more here
I believe that you could benefit from using java.nio. This is how you can store 8 bytes in a long:
// Byte Array TO Long
public static long batol(byte[] buff) {
return batol(buff, false);
}
public static long batol(byte[] buff, boolean littleEndian) {
assert(buff.length == 8);
ByteBuffer bb = ByteBuffer.wrap(buff);
if (littleEndian) bb.order(ByteOrder.LITTLE_ENDIAN);
return bb.getLong();
}
Of course, the resulting longs will have signed representation, but they will have identical binary values to the source data. For a 64 bit+ unsigned representation and arithmatic, you'll need to use BigInteger. Here's how to convert from "unsigned" data stored in a long to a correct BigInteger:
// "Unsigned" Long TO Big Integer
public static BigInteger ultobi(long ul) {
byte[] buff = new byte[8];
ByteBuffer.wrap(buff).asLongBuffer().put(ul);
return new BigInteger(+1, buff);
}
ByteBuffer buffer = ByteBuffer.wrap(bytes);
buffer.getLong();
For the endianness, test with some numbers you know, and then you will be using a byte shifting to move them into the long.
You may find this to be a starting point.
http://www.janeg.ca/scjp/oper/shift.html
The difficulty is that depending on the endianess will change how you do it, but you will shift by 24, 16, 8 then add the last one, basically, if doing 32 bits, but you are going longer, so just do extra shifting.
Take a look at BigInteger(byte[]). It is almost what you want except that it is a signed one. So you may add one more byte to it before you pass it on to BigInteger.
Another thing is that you should be aware of what endian your bytes are.
Hope this helps.
If you're reading from an InputStream, you may also want to look at DataInputStream.readLong(). Java 1.5 introduced Long.reverseBytes(long) which may help you with endianness.
You can use:
byte bVal = 127;
Long longVar = Long.valueOf(bVal);
public static long convertToLong(byte[] array)
{
ByteBuffer buffer = ByteBuffer.wrap(array);
return buffer.getLong();
}
The next code parses bytes as a signed number of arbitrary length ≤ 8.
static long bytesToSignedNumber(boolean reverseOrder, byte... bytes) {
if (bytes.length > 8) bytes = Arrays.copyOfRange(bytes, 0, 8); //delete this line to ignore the higher excess bytes instead of the lower ones
int l = bytes.length;
long number = 0;
for (int i = 0; i < l; i++) {
long current;
if (l==3 || l==5 || l==6 || l==7 //delete this line if you want to operate with 3,5,6,7-byte signed numbers (unlikely)
|| !reverseOrder && (i > 0)
|| reverseOrder && (i < l-1)) {
current = ((long) bytes[i]) & 0x0000_0000_0000_00ff; //unsigned
} else {
current = (long) bytes[i]; //signed
}
if (reverseOrder) number += current << (8 * i);
else number = (number << 8) + current;
}
return number;
}
It parses an byte array as a number, of minimum existing type, converted to long. A few examples:
bytesToSignedNumber(false, 0x01, 0x04) returns 260 (2 bytes as short)
bytesToSignedNumber(false, 0xF1, 0x04) returns -3836 (2 bytes as short)
bytesToSignedNumber(false, 0x01, 0x01, 0x04) returns 65796 (3 bytes as int)
bytesToSignedNumber(false, 0xF1, 0x01, 0x04) returns 15794436 (3 bytes as int)
bytesToSignedNumber(false, 0xF1, 0x01, 0x01, 0x04) returns -251592444 (4 bytes as int)
bytesToSignedNumber(false, 0x0F, 0x01, 0x01, 0x01, 0x01, 0x01, 0x01, 0x01, 0x04) returns 1081146489369067777 (8 of 9 bytes as long)
Another Alternative
From Google
com.google.common.primitives
Longs.fromByteArray(bytes);