Java ByteBuffer BigEndian Double

Question

When I try to write in a file a binary files with value like this :

public static main(String[] args){
    ByteBuffer output = ByteBuffer.allocate(80);
    output.order(ByteOrder.BIG_ENDIAN);
    output.putDouble(545.5);
    appendByteArrayInFile("c:/myPath/", "test.bin", output.array());
}
    private static void appendByteArrayInFile(String exportDirectory, String fileName, byte[] toAppendInFile) {
    if (toAppendInFile != null) {
        File targetExport = createPathAndFile(exportDirectory + fileName);
        try (FileOutputStream output = new FileOutputStream(targetExport, true)) {
            output.write(toAppendInFile);
        } catch (Exception e) {
            // no
        }
    }
}
private static File createPathAndFile(String path) {
    File targetExport = new File(path);
    targetExport.getParentFile().mkdirs();
    return targetExport;
}

The thing is that when I look at the file generated, it seems that the double is putted in little endian style, and when I switch ByteOrder to little-endian, the double is written in big-endian ... But when I put an int, the endianness is correct.

Output with double in BigEndian :

01000000 10000001 00001100 00000000 00000000 00000000 00000000 00000000

Output with double in littleEndian :

00000000 00000000 00000000 00000000 00000000 00001100 10000001 01000000

Output with int in bigEndian:

00000000 00000000 00000010 00100001

VGR · Accepted Answer · 2017-05-22T16:58:37.580

2

As I understand it, the significand in an IEEE floating-point number is not “endian” at all, because it’s not a standalone numeric value. It’s a sequence of binary digits that would follow the decimal point if the significand were expressed as a base 2 number. (To the left of the decimal point, “1.” is always assumed.) The significand is represented as 0001 00001100 00000000 00000000 00000000 00000000 00000000, which means the actual significand is 1.00010000110…₂. The actual value, then, is 1.0001000011₂ × 2⁹, which is 545.5.

edited May 22 '17 at 16:58

answered May 22 '17 at 15:34

VGR

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If there is no endianness in a floating-point number, then why is there differences between little and big endian writings? – H. Saoud May 23 '17 at 07:41
Actually, I just noticed that your code contains `buffer.order(ByteOrder.BIG_ENDIAN);` … meaning it is setting the byte order of a *different* ByteBuffer (not `output`). – VGR May 23 '17 at 23:46
It's a mistake during copying my code, my bad t is actually output.order(ByteOrder.BIG_ENDIAN); – H. Saoud May 24 '17 at 08:39
Executing putDouble on a little-endian ByteBuffer just reverses the bytes, as you are seeing. The point of my answer is that the first value you printed is correct and is in fact big-endian; the bits represent the decimal digits in the order they would occur in the base 2 significand. – VGR May 24 '17 at 14:05

Java ByteBuffer BigEndian Double

1 Answers1