Develop Reference

Where do we use BitSet and why do we use it in java?

I just found out that there is BitSet in java. There are already arrays and similar data structures. Where can BitSet be used?

As the above answer only explains what a BitSet is, I am providing here an answer of how I use BitSet and why. At first, I did not knew that the BitSet construct exists. I have a QR Code generator in C++ and for flexible reasons I don't want to use a specific Bitmap structures in returning this QR Code back to the caller. The QR Code is just black and white and can be represented as a series of bits. The problem was that in the JNI C++, I have to return the byte array that represents these series of bits and then I have to return the count of bits. Note that the size of the bytes array alone could not tell the count of bits. In effect, I am face with a scenario wherein my JNI C++ has to return two values:
the byte[] array
the count of bits
My first solution, was to return an array of boolean. The content of this array are the QR Code pixels, and the square root of the length of the array is the length of the side. Of course this worked but I felt wasted because it is supposed to be a series of bits. My next attempt was to return Pair<int, byte[]> object which, after lots of hair pulling i am not able to make it work in C++. Here comes the BitSet(145) construct. By returning this BitSet object, I am conveying two types of information i listed above. But there is minor trick. If QR Code pixel has total 144 pixels, because one side is 12, then you have to allocate BitSet(145) and do obj.set(144). That is, we introduce an artificial last bit that we then set, but this last bit is not part of the QR Code pixels. This ensures that, BitSet::length() correctly returns the bit count. So in Kotlin:
var pixels:BitSet = getqrpixels(inputdata)
var pixels_len = pixels.length() - 1
var side = sqrt(pixels_len.toFloat()).toInt()
drawSquareBitmap(pixels, side)
And thus, is my unexpected use case of this mysterious BitSet.

Take a look at this:
https://docs.oracle.com/javase/7/docs/api/java/util/BitSet.html
A BitSet is a vector of bits. Each entry in the list is either true (1) or false (0). The BitSet class comes with methods that resemble the bitwise operators. It is a little bit more flexible then a normal binary type.

BitSet, unlike a boolean[], is actually a dynamically sized bitmask. Essentially, instead of using booleans to store values, it uses longs, where each of the longs 64 bits are used to store a single bit.

Arbitrary array length in java

Is there a way to get an array in java which is longer than the length supported by an integer data type?
I'm looking for something that may be indexable using a big integer in Java because the natively supported array length is nowhere near as big as I need it to be for an algorithm I am implementing.

This library should be useful for you: http://fastutil.dsi.unimi.it/
It says:
"...provides also big (64-bit) arrays..."

Int32 gives you 17 8 gigabytes of storage. Do you have so much memory?
I think you should use sparse arrays, i.e. index elements by hash. For example, with just HashMap<BigInteger,YourValueType> or with some libs like BigMemory and alternatives http://terracotta.org/products/bigmemory

Are you sure that you don't have to change the algorithm? Integer.MAX is equal to 2^31-1, which is 2147483647, each int has 4 bytes what gives us: 8589934588 bytes of memory (8GB!!!).

How to use this FFT code with a variable sine wave size and integers instead of shorts?

http://www.koders.com/java/fid168F68D8D019CF9A4F17CA8AFEE102F8BE3B2C28.aspx?s=FFT#L21
2 part question, really. My understanding of FFT's is pretty general.
1) The code seems straight forward enough. But the solution I'm looking for requires a variable data array size (still will be a power of 2). So how should I change the sine wave that's used here? I see it's a static 1024 size sinewave that is clearly defined. How could I use a dynamic size of the real data array to generate a proper sinewave?
2) Also, the solution I'm looking for requires the use of ints instead of shorts like the example linked. As in the input data array will be an array of ints. What modifications of this code would be needed?
Thanks in advance for all your help.

If you need a variable size array, just use an ArrayList to insert your values. The code should be easy to adjust to accomodate this.
To change from a short to an int, simply change them all to ints. A short is just a 2 byte integer, instead of int which is 4 bytes.

How to manage and manipulate extremely large binary values

I need to read in a couple of extremely large strings which are comprised of binary digits. These strings can be extremely large (up to 100,000 digits) which I need to store, be able to manipulate (flip bits) and add together. My first though was to split the string in to 8 character chunks, convert them to bytes and store them in an array. This would allow me to flip bits with relative ease given an index of the bit needed to be flipped, but with this approach I'm unsure how I would go about adding the entirety of the two values together.
Can anyone see a way of storing these values in a memory efficient manner which would allow me to be able to still be able to perform calculations on them?
EDIT:
"add together" (concatenate? arithmetic addition?) - arithmetic addition
My problem is that in the hardest case I have two 100,000 bit numbers (stored in an array of 12,500 bytes). Storing and manually flipping bits isn't an issue, but I need the sum of both numbers and then to be able to find out what the xth bit of this is.

"Strings of binary digits" definitely sound like byte arrays to me. To "add" two such byte arrays together, you'd just allocate a new byte array which is big enough to hold everything, and copy the contents using System.arraycopy.
However that assumes each "string" is a multiple of 8 bits. If you want to "add" a string of 15 bits to another string of 15 bits, you'll need to do bit-shifting. Is that likely to be a problem for you? Depending on what operations you need, you may even want to just keep an object which knows about two byte arrays and can find an arbitrary bit in the logically joined "string".
Either way, byte[] is going to be the way forward - or possibly BitSet.

What about
// Addition
byte[] resArr = new byte[byteArr1.length];
for (int i=0; i<byteArr1.length; i++)
{
res = byteArr1[i]+byteArr2[i];
}
?
Is it something like this you are trying to do?

Why no readUnsignedInt in RandomAccessFile class?

I just found there is no readUnsignedInt() method in the RandomAccessFile class. Why? Is there any workaround to read an unsigned int out from the file?
Edit:
I want to read an unsigned int from file and put it into a long space.
Edit2:
Cannot use readLong(). it will read 8 bytes not 4 bytes. the data in the file have unsigned ints in 4 bytes range.
Edit3:
Found answer here: http://www.petefreitag.com/item/183.cfm
Edit4:
how about if the data file is little-endian? we need to bits swap first?

I'd do it like this:
long l = file.readInt() & 0xFFFFFFFFL;
The bit operation is necessary because the upcast will extend a negative sign.
Concerning the endianness. To the best of my knowledge all I/O in Java is done in big endian fashion. Of course, often it doesn't matter (byte arrays, UTF-8 encoding, etc. are not affected by endianness) but many methods of DataInput are. If your number is stored in little endian, you have to convert it yourself. The only facility in standard Java I know of that allows configuration of endianness is ByteBuffer via the order() method but then you open the gate to NIO and I don't have a lot of experience with that.

Edited to remove readLong():
You could use readFully(byte[] b, int off, int len) and then convert to Long with the methods here: How to convert a byte array to its numeric value (Java)?

Because there is no unsigned int type in java?
Why not readLong() ?
You can readLong and then take first 32 bits.
Edit
You can try
long value = Long.parseLong(Integer.toHexString(file.readInt()), 16);

Depending on what you are doing with the int, you may not need to turn it into a long. You just need to be aware of the operations you are performing. After all its just 32-bits and you can treat it as signed or unsigned as you wish.
If you want to play with the ByteOrder, the simplest thing to do may be to use ByteBuffer which allows you to set a byte order. If your file is less than 2 GB, you can map the entire file into memory and access the ByteBuffer randomly.