Java decimal to binary int array

Question

For this exercise I'm making I want a decimal < 4096 to be written in binary form in an int array.

So for example, 4 would be {0,0,0,0,0,0,0,0,0,1,0,0}. I need this for (nearly) all integers up to 4096, so I've written this piece of code:

for(int k=0; k<4096; k++){
    int[] myNumber =  { (k / 2048) % 2, (k / 1024) % 2, (k / 512) % 2, (k / 256) % 2, (k / 128) % 2, (k / 64) % 2, (k / 32) % 2, (k / 16) % 2, (k / 8) % 2, (k / 4) % 2, (k / 2) % 2, (k / 1) % 2 }
    /* Some processing */
}

This looks kind of ugly, so that's why I'm curious to find out if there is a more elegant way of achieving this?

For the interested reader:
I chose for the array approach of storing the binary numbers, because I need to perform some shifting and addition modulo 2. I'm using an LFSR, and this is my implementation for that:

public class LFSR {

    private int[] polynomial;

    public LFSR(int[] polynomial) {
        this.polynomial = polynomial;
    }

    public int[] shiftLeft(int[] input) {
        int[] result = new int[input.length];

        int out = input[0];
        result[input.length - 1] = out;
        for (int i = input.length - 1; i > 0; i--) {
            result[i - 1] = (input[i] + polynomial[i - 1] * out) % 2;
        }

        return result;
    }

}

Any suggestions?

Bit operators: vipan.com/htdocs/bitwisehelp.html leepoint.net/notes-java/data/expressions/bitops.html. Maybe you don't even need the int arrays. — madth3
– madth3, Commented Mar 22, 2012 at 20:21

Jochen · Accepted Answer · 2012-03-22 20:21:14Z

6

Some pseudo code:

While (int i = 0; i < 12; i++) { 
   bitarray[i] = numericalValue & 0x1;
   numericalValue = numericalValue >> 1;
}

So, shifting right by one bit is division by 2, and ANDing with 1 always leaves you only with the lowest bit which is what you want.

answered Mar 22, 2012 at 20:21

Jochen

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6 Comments

nicktalbot Over a year ago

This isn't quite right as in the example the highest bit comes first where as your code would put it last.

nhaarman Over a year ago

Ah, the teacher hinted to this indeed, but I didn't realize how easy it can be. Thanks!

nhaarman Over a year ago

@nickt but that is just a simple addition. This is for the whole exercise a more elegant way.

StriplingWarrior Over a year ago

@Niek: If this is homework-related, will you please tag it as such?

nicktalbot Over a year ago

Yes you just need to reverse the order. I did give the correct answer below but my tablet messed up the formatting. :-(

|

Cory Dolphin · Accepted Answer · 2012-03-22 20:26:57Z

0

One quick suggestion would be to switch to a byte array instead of an int array, simply to save space, as they will only be 'bits'.

With regards to improving the elegance of your solution, it is perhaps easier to use subcomputations:

int[] intToBinaryArray(int dec){

int[] res = int[12]
for(int i =0; i < 12; i++)
    bitarray[i] = numericalValue & 0x1; //grab first bit only
    dec  /= 2;
}

return res;
}

answered Mar 22, 2012 at 20:26

Cory Dolphin

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Comments

Chandra Sekhar · Accepted Answer · 2012-03-22 20:29:43Z

0

String s = Integer.toBinaryString(int value);

Now convert the String to int[]

int[] intArray = new int[s.length()];
for (int i = 0; i < s.length(); i++) {
intArray[i] = Character.digit(s.charAt(i), 10);
}

answered Mar 22, 2012 at 20:29

Chandra Sekhar

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Comments

Andreas Dolk · Accepted Answer · 2012-03-22 20:30:30Z

0

This is somewhat shorter ;)

    int[] bits = new int[13];
    String bin = Integer.toBinaryString(8192 + value);
    for(int i = 1; i < bin.length(); i++) {
        bits[i-1] = bin.charAt(i)-'0';
    }

answered Mar 22, 2012 at 20:30

Andreas Dolk

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Comments

StriplingWarrior · Accepted Answer · 2012-03-22 20:33:40Z

"Elegance" is in the eye of the beholder, but I'd start by creating a method to avoid repetition and improve clarity:

int[] myNumber =  { getBit(k, 12), getBit(k, 11), ... };

I personally feel this is the "most elegant" way to get a particular bit:

int getBit(int v, int i)
{
    return v >> i & 1;
}

Then you have to decide whether you want to keep repeating yourself with the calls to getBit or whether you'd rather just use a single while or for loop to populate the whole array. You'd think it'd be quicker the way you've got it written, but there's good chance the JIT compiler automatically unrolls your loop for you if you use a loop like Jochen suggests.

Since this particular operation is entirely self contained, you might even want to create a special method for it:

int[] getBits(int v, int num)
{
    int[] arr = new int[num];
    for(int i=0; i<num; i++) {
        arr[i] = getBit(v, num - i - 1);
    }
    return arr;
}

That makes it easier to unit test, and you can reuse it in a variety of situations.

bomber8013 · Accepted Answer · 2012-03-22 20:36:45Z

0

public int[] toBin (int num)
{
int[] ret = new int[8];
for (int i = 7, p = 0; i>=0; i--, p++)
{
ret[i] = (num/2**p) % 2;
}
return ret;
}

answered Mar 22, 2012 at 20:36

bomber8013

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Comments

StriplingWarrior · Accepted Answer · 2012-03-22 20:46:47Z

0

You could use the bit shift operator together withbthe bitwise AND operator as follows. Note bitCount - i - 1 is needed to get the high bit first.

    final int bitCount =12; // Increase to support greater than 4095
    int[] number = new int[bitCount];
    for( int i = 0; i < bitCount; i++ )
    {
      number[i] = ( k >>> ( bitCount - i - 1 ) ) & 1;
    }

edited Mar 22, 2012 at 20:46

StriplingWarrior

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answered Mar 22, 2012 at 20:31

nicktalbot

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3 Comments

nicktalbot Over a year ago

Thanks! :-) The formatting doesn't seem to work on my tablet.

StriplingWarrior Over a year ago

All you needed was an empty line between your paragraph and the code.

nicktalbot Over a year ago

@StriplingWarrior Thanks I will try that next time.

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