Convert double to byte[] array

Question

How can I convert double to byte array in Java? I looked at many other posts, but couldn't figure out the right way.

Input = 65.43 
byte[] size = 6
precision = 2   (this might change based on input)

expected output (byte[]) = 006543

Can I do it without using functions like doubleToLongBits()?

Are you trying to create a binary-encoded decimal? That's what your output is, but it isn't very clear from the question. — lynks
– lynks, Commented Oct 25, 2012 at 15:20
Not exactly, I want the decimal point to be removed. If I give input 9999.1234, my output should be 999912 — swap
– swap, Commented Oct 25, 2012 at 15:30

lynks · Accepted Answer · 2012-10-25 15:48:40Z

18

Real double to byte[] Conversion

double d = 65.43;
byte[] output = new byte[8];
long lng = Double.doubleToLongBits(d);
for(int i = 0; i < 8; i++) output[i] = (byte)((lng >> ((7 - i) * 8)) & 0xff);
//output in hex would be 40,50,5b,85,1e,b8,51,ec

double to BCD Conversion

double d = 65.43;
byte[b] output = new byte[OUTPUT_LENGTH];
String inputString = Double.toString(d);
inputString = inputString.substring(0, inputString.indexOf(".") + PRECISION);
inputString = inputString.replaceAll(".", "");
if(inputString.length() > OUTPUT_LENGTH) throw new DoubleValueTooLongException();
for(int i = inputString.length() - 1; i >= 0; i--) output[i] = (byte)inputString.charAt(i)
//output in decimal would be 0,0,0,0,6,5,4,3 for PRECISION=2, OUTPUT_LENGTH=8

answered Oct 25, 2012 at 15:48

lynks

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2 Comments

lynks Over a year ago

The second one is what you asked for, the first is what I started typing before your comment, thought I'd include it anyway.

Caelum Over a year ago

Good job, I'd imagine the first thing is what the large majority of the ~13,602 viewers came to see.

ope · Accepted Answer · 2021-04-13 09:30:48Z

9

ByteBuffer.allocate(8).putDouble(yourDouble).array()

edited Apr 13, 2021 at 9:30

ope

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answered Jul 1, 2019 at 7:33

Xuxin Jin

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Comments

durron597 · Accepted Answer · 2012-10-25 15:41:31Z

1

public static byte[] encode(double input, int size, int precision) {
    double tempInput = input;

    for (int i = 0; i < precision; i++) tempInput *= 10;

    int output = (int) tempInput;

    String strOut = String.format("%0"+size+"d", output);

    return strOut.getBytes();
}

answered Oct 25, 2012 at 15:41

durron597

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2 Comments

sataniccrow Over a year ago

it's perfect, but I would add also a parameter to indicate the locale to avoid problem with , or . in the decimal's part of the input

durron597 Over a year ago

Yeah but don't forget to do that job in a different function. The function I gave you takes a double as input, not a string... use NumberFormat format = NumberFormat.getInstance(Locale.FRANCE)) for example if you are getting a comma instead of a period as input. (and then use format.parse(inputString).doubleValue(); ) Don't forget to check this answer :)

sataniccrow · Accepted Answer · 2012-10-25 15:47:25Z

0

double doubleValue = 10.42123;
DecimalFormat df = new DecimalFormat("#.##");
String newDouble = df.format(doubleValue);
byte[] byteArray = (newDouble.replace(",", "")).getBytes();

for (byte b : byteArray) {
    System.out.println((char)b+"");
}

answered Oct 25, 2012 at 15:47

sataniccrow

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1 Comment

sataniccrow Over a year ago

italian locale... so , as decimal sign

swap · Accepted Answer · 2012-10-26 13:09:04Z

0

This is what I got based on your inputs and it serves my purpose. Thanks for helping out!

static int formatDoubleToAscii(double d, int bytesToUse, int minPrecision, byte in[], int startPos) {

        int d1 = (int)(d * Math.pow(10, minPrecision));

        String t = String.format("%0"+bytesToUse+"d", d1).toString();
        System.out.println("d1 = "+ d1 + " t="+ t + " t.length=" + t.length());

        for(int i=0 ; i<t.length() ; i++, startPos++) {
            System.out.println(t.charAt(i));
            in[startPos] = (byte) t.charAt(i);
        }           

        return startPos;
    }

answered Oct 26, 2012 at 13:09

swap

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