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I have recently come across the problem of creating arrays with values that have a specified bit length. Say an array with 13bits instead of 8,16,32 etc. I tried to look for a good tutorial/article about it as I am new to bit operations. Though I am not really sure of what to search for.
I presume the array would work with a backing array of bytes or longs...

My ultimate question is if you can show me if there is a duplicate question or tutorial out there.
If not perhaps show me an example. AND if you got the time write a short explanation.

Thank you.

EDIT: The purpose is not to make an array of say longs but only use 40% of it. I want it to be packed together to save space to be compatible with the thing im making.

Caspar Noree
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1 Answers1

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It's not possible to "create your own primitive types" in java. Also I don't think there is any library around here to do what you want. I think most people would go with the overhead of losing some memory, especially at bit level. Maybe C or Cpp would have been a wiser choice (and I'm not even sure).

You'll have to create your own bit manipulation library. There are many ways to do it, I'll give you one. I began using a byte[] but it's more complex. As a rule, use the biggest normal type (ex: for a 48bit elements, use 32 bit types as storage). so let's go with an int array (16 bits) for 100 of your 13bits types. I'll use big-endian-style storage.

int intArraySize = 100 * 16 / 13 + 1; // + 1 is just to be sure...
int[] intArray = new int[byteArraySize];

Now, how do you access the sixth value for example. You'll always need at least and at most two int of your array and an integer to store it.

int pos = 6;
int buffer = 0;
int firstPart = int Array[ (pos  * 13) /16]; // 1010 0110 1100 0011
int secondPart = int Array[ (pos * 13) /16 + 1]; // 1001 1110 0101 1111
int begin = pos * 13 % 16;

The variable begin = 14 is the bit at which your number begins. So that means on your 13bits elements there are (16-14) 3 bits in the first (left) int and the rest (13-3 = 10) in the second (right). The number you want is 1010 0110 1100 0{011 and 1001 1110} 0101 1111.

You're gonna put these two ints into one now. Right shift the secondPart 3 times (so it's the right part of your final number), and left shift the firstPart 10 times, add them in the buffer. Because it's a 13bits elements, you'll need to clean ( with a bitmask ) the 3 first elements of your 16 bit in the buffer, and voila !

I'll let you guess how to insert a value in the array (try doing the same step, but in reverse) and be carefull not to erase other values. And if you haven't looked yet: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/op3.html

Disclaimer: I didn't try the code, but you get the general idea. There might be some errors, maybe you'll have to add or remove 1 to begin. But you get the general idea. The first thing you should do is make a function that prints/log any integer (or byte, or whatever) into it's binary representation. Multiple possibilities here: Print an integer in binary format in Java because you're gonne need them to test every step of your code.

I still think it's a bad idea to store your special number this way, (seriously memory is rarely gonna be an issue), but I found the exercise interesting, and maybe you really need taht kind of storage. If your curious, take a look at the ByteArrayOutputStream, I'm not sure you'll ever need this for what you're doing but who knows.

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Asoub
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