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
precisi
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
ByteBuffer.allocate(8).putDouble()).array()
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();
}
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;
}
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+"");
}