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CRC16 CCITT Java implementation

There is a function written in C that calculates CRC16 CCITT. It helps reading data from RFID reader - and basically works fine. I would like to write a function in Java that would do similar thing.
I tried online converter page to do this, but the code I got is garbage.
Could you please take a look at this and advice why Java code that should do the same generates different crc?
Please find attached original C function:
void CRC16(unsigned char * Data, unsigned short * CRC, unsigned char Bytes)
{
int i, byte;
unsigned short C;
*CRC = 0;
for (byte = 1; byte <= Bytes; byte++, Data++)
{
C = ((*CRC >> 8) ^ *Data) << 8;
for (i = 0; i < 8; i++)
{
if (C & 0x8000)
C = (C << 1) ^ 0x1021;
else
C = C << 1;
}
*CRC = C ^ (*CRC << 8);
}
}
And here is the different CRC function written in JAVA that should calculate the same checksum, but it does not:
public static int CRC16_CCITT_Test(byte[] buffer) {
int wCRCin = 0x0000;
int wCPoly = 0x1021;
for (byte b : buffer) {
for (int i = 0; i < 8; i++) {
boolean bit = ((b >> (7 - i) & 1) == 1);
boolean c15 = ((wCRCin >> 15 & 1) == 1);
wCRCin <<= 1;
if (c15 ^ bit)
wCRCin ^= wCPoly;
}
}
wCRCin &= 0xffff;
return wCRCin;
}
When I try calculating 0,2,3 numbers in both functions I got different results:
for C function it is (DEC): 22017
for JAVA function it is (DEC): 28888
OK. I have converter C into Java code and got it partially working.
public static int CRC16_Test(byte[] data, byte bytes) {
int dataIndex = 0;
short c;
short [] crc= {0};
crc[0] = (short)0;
for(int j = 1; j <= Byte.toUnsignedInt(bytes); j++, dataIndex++) {
c = (short)((Short.toUnsignedInt(crc[0]) >> 8 ^ Byte.toUnsignedInt(data[dataIndex])) << 8);
for(int i = 0; i < 8; i++) {
if((Short.toUnsignedInt(c) & 0x8000) != 0) {
c = (short)(Short.toUnsignedInt(c) << 1 ^ 0x1021);
} else {
c = (short)(Short.toUnsignedInt(c) << 1);
}
}
crc[0] = (short)(Short.toUnsignedInt(c) ^ Short.toUnsignedInt(crc[0]) << 8);
}
return crc[0];
}
It gives the same CRC values as C code for 0,2,3 numbers, but i.e. for numbers 255, 216, 228 C code crc is 60999 while JAVA crc is -4537.
OK. Finally thanks to your pointers I got this working.
The last change required was changing 'return crc[0]' to:
return (int) crc[0] & 0xffff;
... and it works...
Many thanks to all :)

There is nothing wrong. For a 16 bit value, –4537 is represented as the exact same 16 bits as 60999 is. If you would like for your routine to return the positive version, convert to int (which is 32 bits) and do an & 0xffff.

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Here my code in C.
unsigned int crc32b(unsigned char *message) {
int i, j;
unsigned int byte, crc, mask;
i = 0;
crc = 0xFFFFFFFF;
while (message[i] != 0) {
byte = message[i]; // Get next byte.
crc = crc ^ byte;
for (j = 7; j >= 0; j--) { // Do eight times.
mask = -(crc & 1);
crc = (crc >> 1) ^ (0xEDB88320 & mask);
}
i = i + 1;
}
return ~crc;
}
But in Java, unsign int is not supported. so I try to get unsign by &0xFFFFFF
But and it wrong. How to fix it in java

You can easily emulate most unsigned arithmetic with signed integers in Java (some things are more annoying, such as division, but it is not used here).
This in particular is very easy, 0xFFFFFFFF is just itself, also known as -1.
The >> should be replaced by its unsigned counterpart, >>>.
And the byte from the message has to be masked to undo the sign-extension.
In total (not tested)
int crc32b(byte[] message) {
int i, j;
int b, crc, mask;
i = 0;
crc = -1;
while (i < message.length) {
b = message[i] & 0xFF; // Get next byte.
crc = crc ^ b;
for (j = 7; j >= 0; j--) { // Do eight times.
mask = -(crc & 1);
crc = (crc >>> 1) ^ (0xEDB88320 & mask);
}
i = i + 1;
}
return ~crc;
}
That's still quite C-ish though.

IBM to IEEE floating point conv

Is there any standard method in java to convert IBM 370(in the form of bytes) to IEEE format.?Any algorithm for the conversion would help..
I tried writing a java code..But i fail to understand where do i go wrong. When i give the input as -2.000000000000000E+02, i'm getting the value as -140.0 in IEEE format. and in othercase when i give the input as 3.140000000000000E+00 i'm getting the value as 3.1712502374909226 in IEEE format Any help on this would be highly appreciated
private void conversion() {
byte[] buffer = //bytes to be read(8 bytes);
int sign = (buffer[0] & 0x80);
// Extract exponent.
int exp = ((buffer[0] & 0x7f) - 64) * 4 - 1;
//Normalize the mantissa.
for (int i = 0; i < 4; i++) {//since 4 bits per hex digit
if ((buffer[1] & 0x80) == 0) {
buffer = leftShift(buffer);
exp = exp - 1;
}
}
// Put sign and mantissa back in 8-byte number
buffer = rightShift(buffer);// make room for longer exponent
buffer = rightShift(buffer);
buffer = rightShift(buffer);
exp = exp + 1023;//Excess 1023 format
int temp = exp & 0x0f;//Low 4 bits go into B(1)
buffer[1]= (byte)((buffer[1]&0xf) | (temp *16));
buffer[0]= (byte)(sign | ((exp/16) & 0x7f));
}
private byte[] rightShift(byte[] buf) {
int newCarry = 0;
int oldCarry = 0;
for(int i = 1; i<buf.length; i++) {
newCarry = buf[i] & 1;
buf[i] = (byte)((buf[i] & 0xFE)/2 + (oldCarry != 0 ? 0x80 : 0));
oldCarry = newCarry;
}
return buf;
}
private byte[] leftShift(byte[] buf) {
int newCarry = 0;
int oldCarry = 0;
for(int i = buf.length-1; i>0; i--) {
newCarry = buf[i] & 1;
buf[i] = (byte)((buf[i] & 0x7F)*2 + (oldCarry != 0 ? 1 : 0));
oldCarry = newCarry;
}
return buf;
}

I can see a couple different solutions to your question:
Use the text representation as an intermediary reference
Do a straight conversion C code

This IBM Technical Article includes algorithms for converting from IBM floating point formats to IEE floating point.

There is a bug in the leftShift() function, where you should mask with 0x80 instead of 1. Here is the corrected function.
private byte[] leftShift(byte[] buf) {
int newCarry = 0;
int oldCarry = 0;
for(int i = buf.length-1; i>0; i--) {
newCarry = buf[i] & 0x80;
buf[i] = (byte)((buf[i] & 0x7F)*2 + (oldCarry != 0 ? 1 : 0));
oldCarry = newCarry;
}
return buf;
}
I tested with the wiki example -118.625 If I understand correctly, the bias for IBM double is also 64, so the binary will be 11000010 01110110 10100000 00000000 00000000 00000000 00000000 00000000. After the fix, the program can produce -118.625 correctly.
I know it is an old post, but I currently ran into the same situation too.

Bitwise operator for simply flipping all bits in an integer?

I have to flip all bits in a binary representation of an integer. Given:
10101
The output should be
01010
What is the bitwise operator to accomplish this when used with an integer? For example, if I were writing a method like int flipBits(int n);, what would go in the body? I need to flip only what's already present in the number, not all 32 bits in the integer.

The ~ unary operator is bitwise negation. If you need fewer bits than what fits in an int then you'll need to mask it with & after the fact.

Simply use the bitwise not operator ~.
int flipBits(int n) {
return ~n;
}
To use the k least significant bits, convert it to the right mask.
(I assume you want at least 1 bit of course, that's why mask starts at 1)
int flipBits(int n, int k) {
int mask = 1;
for (int i = 1; i < k; ++i)
mask |= mask << 1;
return ~n & mask;
}
As suggested by Lưu Vĩnh Phúc, one can create the mask as (1 << k) - 1 instead of using a loop.
int flipBits2(int n, int k) {
int mask = (1 << k) - 1;
return ~n & mask;
}

There is a number of ways to flip all the bit using operations
x = ~x; // has been mentioned and the most obvious solution.
x = -x - 1; or x = -1 * (x + 1);
x ^= -1; or x = x ^ ~0;

Well since so far there's only one solution that gives the "correct" result and that's.. really not a nice solution (using a string to count leading zeros? that'll haunt me in my dreams ;) )
So here we go with a nice clean solution that should work - haven't tested it thorough though, but you get the gist. Really, java not having an unsigned type is extremely annoying for this kind of problems, but it should be quite efficient nonetheless (and if I may say so MUCH more elegant than creating a string out of the number)
private static int invert(int x) {
if (x == 0) return 0; // edge case; otherwise returns -1 here
int nlz = nlz(x);
return ~x & (0xFFFFFFFF >>> nlz);
}
private static int nlz(int x) {
// Replace with whatever number leading zero algorithm you want - I can think
// of a whole list and this one here isn't that great (large immediates)
if (x < 0) return 0;
if (x == 0) return 32;
int n = 0;
if ((x & 0xFFFF0000) == 0) {
n += 16;
x <<= 16;
}
if ((x & 0xFF000000) == 0) {
n += 8;
x <<= 8;
}
if ((x & 0xF0000000) == 0) {
n += 4;
x <<= 4;
}
if ((x & 0xC0000000) == 0) {
n += 2;
x <<= 2;
}
if ((x & 0x80000000) == 0) {
n++;
}
return n;
}

faster and simpler solution :
/* inverts all bits of n, with a binary length of the return equal to the length of n
k is the number of bits in n, eg k=(int)Math.floor(Math.log(n)/Math.log(2))+1
if n is a BigInteger : k= n.bitLength();
*/
int flipBits2(int n, int k) {
int mask = (1 << k) - 1;
return n ^ mask;
}

One Line Solution
int flippingBits(int n) {
return n ^ ((1 << 31) - 1);
}

I'd have to see some examples to be sure, but you may be getting unexpected values because of two's complement arithmetic. If the number has leading zeros (as it would in the case of 26), the ~ operator would flip these to make them leading ones - resulting in a negative number.
One possible workaround would be to use the Integer class:
int flipBits(int n){
String bitString = Integer.toBinaryString(n);
int i = 0;
while (bitString.charAt(i) != '1'){
i++;
}
bitString = bitString.substring(i, bitString.length());
for(i = 0; i < bitString.length(); i++){
if (bitString.charAt(i) == '0')
bitString.charAt(i) = '1';
else
bitString.charAt(i) = '0';
}
int result = 0, factor = 1;
for (int j = bitString.length()-1; j > -1; j--){
result += factor * bitString.charAt(j);
factor *= 2;
}
return result;
}
I don't have a java environment set up right now to test it on, but that's the general idea. Basically just convert the number to a string, cut off the leading zeros, flip the bits, and convert it back to a number. The Integer class may even have some way to parse a string into a binary number. I don't know if that's how the problem needs to be done, and it probably isn't the most efficient way to do it, but it would produce the correct result.
Edit: polygenlubricants' answer to this question may also be helpful

I have another way to solve this case,
public static int complementIt(int c){
return c ^ (int)(Math.pow(2, Math.ceil(Math.log(c)/Math.log(2))) -1);
}
It is using XOR to get the complement bit, to complement it we need to XOR the data with 1, for example :
101 XOR 111 = 010
(111 is the 'key', it generated by searching the 'n' square root of the data)
if you are using ~ (complement) the result will depend on its variable type, if you are using int then it will be process as 32bit.

As we are only required to flip the minimum bits required for the integer (say 50 is 110010 and when inverted, it becomes 001101 which is 13), we can invert individual bits one at a time from the LSB to MSB, and keep shifting the bits to the right and accordingly apply the power of 2. The code below does the required job:
int invertBits (int n) {
int pow2=1, int bit=0;
int newnum=0;
while(n>0) {
bit = (n & 1);
if(bit==0)
newnum+= pow2;
n=n>>1;
pow2*=2;
}
return newnum;
}

import java.math.BigInteger;
import java.util.Scanner;
public class CodeRace1 {
public static void main(String[] s) {
long input;
BigInteger num,bits = new BigInteger("4294967295");
Scanner sc = new Scanner(System.in);
input = sc.nextInt();
sc.nextLine();
while (input-- > 0) {
num = new BigInteger(sc.nextLine().trim());
System.out.println(num.xor(bits));
}
}
}

The implementation from openJDK, Integer.reverse():
public static int More ...reverse(int i) {
i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
i = (i << 24) | ((i & 0xff00) << 8) |
((i >>> 8) & 0xff00) | (i >>> 24);
return i;
}
Base on my experiments on my laptop, the implementation below was faster:
public static int reverse2(int i) {
i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555;
i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333;
i = (i & 0x0f0f0f0f) << 4 | (i >>> 4) & 0x0f0f0f0f;
i = (i & 0x00ff00ff) << 8 | (i >>> 8) & 0x00ff00ff;
i = (i & 0x0000ffff) << 16 | (i >>> 16) & 0x0000ffff;
return i;
}
Not sure what's the reason behind it - as it may depends on how the java code is interpreted into machine code...

If you just want to flip the bits which are "used" in the integer, try this:
public int flipBits(int n) {
int mask = (Integer.highestOneBit(n) << 1) - 1;
return n ^ mask;
}

public static int findComplement(int num) {
return (~num & (Integer.highestOneBit(num) - 1));
}

int findComplement(int num) {
int i = 0, ans = 0;
while(num) {
if(not (num & 1)) {
ans += (1 << i);
}
i += 1;
num >>= 1;
}
return ans;
}

Binary 10101 == Decimal 21
Flipped Binary 01010 == Decimal 10
One liner (in Javascript - You could convert to your favorite programming language )
10 == ~21 & (1 << (Math.floor(Math.log2(21))+1)) - 1
Explanation:
10 == ~21 & mask
mask : For filtering out all the leading bits before the significant bits count (nBits - see below)
How to calculate the significant bit counts ?
Math.floor(Math.log2(21))+1 => Returns how many significant bits are there (nBits)
Ex:
0000000001 returns 1
0001000001 returns 7
0000010101 returns 5
(1 << nBits) - 1 => 1111111111.....nBits times = mask

It can be done by a simple way, just simply subtract the number from the value
obtained when all the bits are equal to 1 .
For example:
Number: Given Number
Value : A number with all bits set in a given number.
Flipped number = Value – Number.
Example :
Number = 23,
Binary form: 10111
After flipping digits number will be: 01000
Value: 11111 = 31
We can find the most significant set bit in O(1) time for a fixed size integer. For
example below code is for a 32-bit integer.
int setBitNumber(int n)
{
n |= n>>1;
n |= n>>2;
n |= n>>4;
n |= n>>8;
n |= n>>16;
n = n + 1;
return (n >> 1);
}

Extract bit sequences of arbitrary length from byte[] array efficiently

I'm looking for the most efficient way of extracting (unsigned) bit sequences of arbitrary length (0 <= length <= 16) at arbitrary position. The skeleton class show how my current implementation essentially handles the problem:
public abstract class BitArray {
byte[] bytes = new byte[2048];
int bitGet;
public BitArray() {
}
public void readNextBlock(int initialBitGet, int count) {
// substitute for reading from an input stream
for (int i=(initialBitGet>>3); i<=count; ++i) {
bytes[i] = (byte) i;
}
prepareBitGet(initialBitGet, count);
}
public abstract void prepareBitGet(int initialBitGet, int count);
public abstract int getBits(int count);
static class Version0 extends BitArray {
public void prepareBitGet(int initialBitGet, int count) {
bitGet = initialBitGet;
}
public int getBits(int len) {
// intentionally gives meaningless result
bitGet += len;
return 0;
}
}
static class Version1 extends BitArray {
public void prepareBitGet(int initialBitGet, int count) {
bitGet = initialBitGet - 1;
}
public int getBits(int len) {
int byteIndex = bitGet;
bitGet = byteIndex + len;
int shift = 23 - (byteIndex & 7) - len;
int mask = (1 << len) - 1;
byteIndex >>= 3;
return (((bytes[byteIndex] << 16) |
((bytes[++byteIndex] & 0xFF) << 8) |
(bytes[++byteIndex] & 0xFF)) >> shift) & mask;
}
}
static class Version2 extends BitArray {
static final int[] mask = { 0x0, 0x1, 0x3, 0x7, 0xF, 0x1F, 0x3F, 0x7F, 0xFF,
0x1FF, 0x3FF, 0x7FF, 0xFFF, 0x1FFF, 0x3FFF, 0x7FFF, 0xFFFF };
public void prepareBitGet(int initialBitGet, int count) {
bitGet = initialBitGet;
}
public int getBits(int len) {
int offset = bitGet;
bitGet = offset + len;
int byteIndex = offset >> 3; // originally used /8
int bitIndex = offset & 7; // originally used %8
if ((bitIndex + len) > 16) {
return ((bytes[byteIndex] << 16 |
(bytes[byteIndex + 1] & 0xFF) << 8 |
(bytes[byteIndex + 2] & 0xFF)) >> (24 - bitIndex - len)) & mask[len];
} else if ((offset + len) > 8) {
return ((bytes[byteIndex] << 8 |
(bytes[byteIndex + 1] & 0xFF)) >> (16 - bitIndex - len)) & mask[len];
} else {
return (bytes[byteIndex] >> (8 - offset - len)) & mask[len];
}
}
}
static class Version3 extends BitArray {
int[] ints = new int[2048];
public void prepareBitGet(int initialBitGet, int count) {
bitGet = initialBitGet;
int put_i = (initialBitGet >> 3) - 1;
int get_i = put_i;
int buf;
buf = ((bytes[++get_i] & 0xFF) << 16) |
((bytes[++get_i] & 0xFF) << 8) |
(bytes[++get_i] & 0xFF);
do {
buf = (buf << 8) | (bytes[++get_i] & 0xFF);
ints[++put_i] = buf;
} while (get_i < count);
}
public int getBits(int len) {
int bit_idx = bitGet;
bitGet = bit_idx + len;
int shift = 32 - (bit_idx & 7) - len;
int mask = (1 << len) - 1;
int int_idx = bit_idx >> 3;
return (ints[int_idx] >> shift) & mask;
}
}
static class Version4 extends BitArray {
int[] ints = new int[1024];
public void prepareBitGet(int initialBitGet, int count) {
bitGet = initialBitGet;
int g = initialBitGet >> 3;
int p = (initialBitGet >> 4) - 1;
final byte[] b = bytes;
int t = (b[g] << 8) | (b[++g] & 0xFF);
final int[] i = ints;
do {
i[++p] = (t = (t << 16) | ((b[++g] & 0xFF) <<8) | (b[++g] & 0xFF));
} while (g < count);
}
public int getBits(final int len) {
final int i;
bitGet = (i = bitGet) + len;
return (ints[i >> 4] >> (32 - len - (i & 15))) & ((1 << len) - 1);
}
}
public void benchmark(String label) {
int checksum = 0;
readNextBlock(32, 1927);
long time = System.nanoTime();
for (int pass=1<<18; pass>0; --pass) {
prepareBitGet(32, 1927);
for (int i=2047; i>=0; --i) {
checksum += getBits(i & 15);
}
}
time = System.nanoTime() - time;
System.out.println(label+" took "+Math.round(time/1E6D)+" ms, checksum="+checksum);
try { // avoid having the console interfere with our next measurement
Thread.sleep(369);
} catch (InterruptedException e) {}
}
public static void main(String[] argv) {
BitArray test;
// for the sake of getting a little less influence from the OS for stable measurement
Thread.currentThread().setPriority(Thread.MAX_PRIORITY);
while (true) {
test = new Version0();
test.benchmark("no implementaion");
test = new Version1();
test.benchmark("Durandal's (original)");
test = new Version2();
test.benchmark("blitzpasta's (adapted)");
test = new Version3();
test.benchmark("MSN's (posted)");
test = new Version4();
test.benchmark("MSN's (half-buffer modification)");
System.out.println("--- next pass ---");
}
}
}
This works, but I'm looking for a more efficient solution (performance wise). The byte array is guaranteed to be relatively small, between a few bytes up to a max of ~1800 bytes. The array is read exactly once (completely) between each call to the read method. There is no need for any error checking in getBits(), such as exceeding the array etc.
It seems my initial question above isn't clear enough. A "bit sequence" of N bits forms an integer of N bits, and I need to extract those integers with minimal overhead. I have no use for strings, as the values are either used as lookup indices or are directly fed into some computation. So basically, the skeleton shown above is a real class and getBits() signature shows how the rest of the code interacts with it.
Extendet the example code into a microbenchmark, included blitzpasta's solution (fixed missing byte masking). On my old AMD box it turns out as ~11400ms vs ~38000ms. FYI: Its the divide and modulo operations that kill the performance. If you replace /8 with >>3 and %8 with &7, both solutions are pretty close to each other (jdk1.7.0ea104).
There seemed to be a bit confusion about how and what to work on. The first, original post of the example code included a read() method to indicate where and when the byte buffer was filled. This got lost when the code was turned into the microbench. I re-introduced it to make this a little clearer.
The idea is to beat all existing versions by adding another subclass of BitArray which need to implement getBits() and prepareBitGet(), the latter may be empty. Do not change the benchmarking to give your solution an advantage, the same could be done for all the existing solutions, making this a completely moot optimization! (really!!)
I added a Version0, which does nothing but increment the bitGet state. It always returns 0 to get a rough idea how big the benchmark overhead is. Its only there for comparison.
Also, an adaption on MSN's idea was added (Version3). To keep things fair and comparable for all competitors, the byte array filling is now part of the benchmark, as well as a preparatory step (see above). Originally MSN's solution did not do so well, there was lots of overhead in preparing the int[] buffer. I took the liberty of optimizing the step a little, which turned it into a fierce competitor :)
You might also find that I de-convoluted your code a little. Your getBit() could be condensed into a 3-liner, probably shaving off one or two percent. I deliberately did this to keep the code readable and because the other versions aren't as condensed as possible either (again for readability).
Conclusion (code example above update to include versions based on all applicable contributions). On my old AMD box (Sun JRE 1.6.0_21), they come out as:
V0 no implementaion took 5384 ms
V1 Durandal's (original) took 10283 ms
V2 blitzpasta's (adapted) took 12212 ms
V3 MSN's (posted) took 11030 ms
V4 MSN's (half-buffer modification) took 9700 ms
Notes: In this benchmark an average of 7.5 bits is fetched per call to getBits(), and each bit is only read once. Since V3/V4 have to pay a high initialization cost, they tend to show better runtime behavior with more, shorter fetches (and consequently worse the closer to the maximum of 16 the average fetch size gets). Still, V4 stays slightly ahead of all others in all scenarios.
In an actual application, the cache contention must be taken into account, since the extra space needed for V3/v4 may increase cache misses to a point where V0 would be a better choice. If the array is to be traversed more than once, V4 should be favored, since it fetches faster than every other and the costly initialization is amortized after the fist pass.

If you just want the unsigned bit sequence as an int.
static final int[] lookup = {0x0, 0x1, 0x3, 0x7, 0xF, 0x1F, 0x3F, 0x7F, 0xFF, 0x1FF, 0x3FF, 0x7FF, 0xFFF, 0x1FFF, 0x3FFF, 0x7FFF, 0xFFFF };
/*
* bytes: byte array, with the bits indexed from 0 (MSB) to (bytes.length * 8 - 1) (LSB)
* offset: index of the MSB of the bit sequence.
* len: length of bit sequence, must from range [0,16].
* Not checked for overflow
*/
static int getBitSeqAsInt(byte[] bytes, int offset, int len){
int byteIndex = offset / 8;
int bitIndex = offset % 8;
int val;
if ((bitIndex + len) > 16) {
val = ((bytes[byteIndex] << 16 | bytes[byteIndex + 1] << 8 | bytes[byteIndex + 2]) >> (24 - bitIndex - len)) & lookup[len];
} else if ((offset + len) > 8) {
val = ((bytes[byteIndex] << 8 | bytes[byteIndex + 1]) >> (16 - bitIndex - len)) & lookup[len];
} else {
val = (bytes[byteIndex] >> (8 - offset - len)) & lookup[len];
}
return val;
}
If you want it as a String (modification of Margus' answer).
static String getBitSequence(byte[] bytes, int offset, int len){
int byteIndex = offset / 8;
int bitIndex = offset % 8;
int count = 0;
StringBuilder result = new StringBuilder();
outer:
for(int i = byteIndex; i < bytes.length; ++i) {
for(int j = (1 << (7 - bitIndex)); j > 0; j >>= 1) {
if(count == len) {
break outer;
}
if((bytes[byteIndex] & j) == 0) {
result.append('0');
} else {
result.append('1');
}
++count;
}
bitIndex = 0;
}
return result.toString();
}

Well, depending on how far you want to go down the time vs. memory see-saw, you can allocate a side table of every 32-bits at every 16-bit offset and then do a mask and shift based on the 16-bit offset:
byte[] bytes = new byte[2048];
int bitGet;
unsigned int dwords[] = new unsigned int[2046];
public BitArray() {
for (int i=0; i<bytes.length; ++i) {
bytes[i] = (byte) i;
}
for (int i= 0; i<dwords.length; ++i) {
dwords[i]=
(bytes[i ] << 24) |
(bytes[i + 1] << 16) |
(bytes[i + 2] << 8) |
(bytes[i + 3]);
}
}
int getBits(int len)
{
int offset= bitGet;
int offset_index= offset>>4;
int offset_offset= offset & 15;
return (dwords[offset_index] >> offset_offset) & ((1 << len) - 1);
}
You avoid the branching (at the cost of quadrupling your memory footprint). And is looking up the mask really that much faster than (1 << len) - 1?

Just wondering why can't you use java.util.BitSet;
Basically what you can do, is to read the whole data as byte[], convert it to binary in string format and use string utilities like .substring() to do the work. This will also work bit sequences > 16.
Lets say you have 3 bytes: 1, 2, 3 and you want to extract bit sequence from 5th to 16th bit.
Number Binary
1 00000001
2 00000010
3 00000011
Code example:
public static String getRealBinary(byte[] input){
StringBuilder sb = new StringBuilder();
for (byte c : input) {
for (int n = 128; n > 0; n >>= 1){
if ((c & n) == 0)
sb.append('0');
else sb.append('1');
}
}
return sb.toString();
}
public static void main(String[] args) {
byte bytes[] = new byte[]{1,2,3};
String sbytes = getRealBinary(bytes);
System.out.println(sbytes);
System.out.println(sbytes.substring(5,16));
}
Output:
000000010000001000000011
00100000010
Speed:
I did a testrun for 1m times and on my computer it took 0.995s, so its reasonably very fast:
Code to repeat the test yourself:
public static void main(String[] args) {
Random r = new Random();
byte bytes[] = new byte[4];
long start, time, total=0;
for (int i = 0; i < 1000000; i++) {
r.nextBytes(bytes);
start = System.currentTimeMillis();
getRealBinary(bytes).substring(5,16);
time = System.currentTimeMillis() - start;
total+=time;
}
System.out.println("It took " +total + "ms");
}

You want at most 16 bits, taken from an array of bytes. 16 bits can span at most 3 bytes.
Here's a possible solution:
int GetBits(int bit_index, int bit_length) {
int byte_offset = bit_index >> 3;
return ((((((byte_array[byte_offset]<<8)
+byte_array[byte_offset+1])<<8)
+byte_array[byte_offset+2]))
>>(24-(bit_index&7)+bit_length))))
&((1<<bit_length)-1);
}
[Untested]
If you call this a lot you should precompute the 24-bit values for the 3 concatenated bytes, and store those into an int array.
I'll observe that if you are coding this in C on an x86, you don't even need to precompute the 24 bit array; simply access the by te array at the desire offset as a 32 bit value. The x86 will do unaligned fetches just fine. [commenter noted that endianess mucks this up, so it isn't an answer, OK, do the 24 bit version.]

Since Java 7 BitSet has the toLongArray method, which I believe will do exactly what the question asks for:
int subBits = (int) bitSet.get(lowBit, highBit).toLongArray()[0];
This has the advantage that it works with sequences larger than ints or longs. It has the performance disadvantage that a new BitSet object must be allocated, and a new array object to hold the result.
It would be really interesting to see how this compares with the other methods in the benchmark.