Bitconverter for Java
Following the advice provided in the question https://stackoverflow.com/questions/1738244/what-is-the-java-equivalent-of-net-bitconverter I have begun implementing my own bitconverter for Java but am not getting equivalent results.
Could someone please guide me on what I might be doing incorrectly?
public static byte[] GetBytes(Integer value) {
ByteArrayOutputStream byteStream = new ByteArrayOutputStream();
DataOutputStream stream = new DataOutputStream(byteStream);
try {
stream.writeInt(value);
} catch (IOException e) {
return new byte[4];
}
return byteStream.toByteArray();
}
byte[] result = BitConverter.GetBytes(1234); //JAVA: [0, 0, 4, -4开发者_Go百科6]
byte[] result = BitConverter.GetBytes(1234); //C#: [210, 4, 0, 0]
That is just endianness (-46 and 210 is because of Java's signed bytes, but that is just a UI thing). Either reverse the array contents, or use shift operations to write the int.
Note: the endianness that .NET emits depends on the platform. I would suggest using KNOWN ENDIANNESS in both cases; most likely by using shift operations from both. Or perhaps a better idea: just use a pre-canned, platform independent serialization format (for example: protocol buffers, which has good support on both Java and .NET/C#).
For example; if I was writing an int value
to a byte[] buffer
(starting at offset
), I might use:
buffer[offset++] = (byte)value;
buffer[offset++] = (byte)(value>>8);
buffer[offset++] = (byte)(value>>16);
buffer[offset++] = (byte)(value>>24);
this is guaranteed little-endian, and similar code should work on any framework.
The C# BitConverter
will use the endianness of the underlying achitecture. In most environments, it will be little-endian (as it is in your case). Java's DataOutputStream
however will always write in big-endian ("the portable way"). You'll need to check the endianness of the machine and write accordingly if you want to match the behavior.
Also, bytes in java are signed so the output is just a cosmetic difference. The bit representation is the same so you don't need to worry about that.
To check the endianness of your machine, use the java.nio.ByteOrder.nativeOrder()
method. Then use the java.nio.ByteBuffer
instead where you may specify the byte order()
and write the data.
You could then implement your method like this:
public static byte[] GetBytes(int value)
{
ByteBuffer buffer = ByteBuffer.allocate(4).order(ByteOrder.nativeOrder());
buffer.putInt(value);
return buffer.array();
}
if any body need..C# to JAVA BitConverter.ToInt32
public static int toInt32_2(byte[] bytes, int index)
{
int a = (int)((int)(0xff & bytes[index]) << 32 | (int)(0xff & bytes[index + 1]) << 40 | (int)(0xff & bytes[index + 2]) << 48 | (int)(0xff & bytes[index + 3]) << 56);
// int a = (int)((int)(0xff & bytes[index]) << 56 | (int)(0xff & bytes[index + 1]) << 48 | (int)(0xff & bytes[index + 2]) << 40 | (int)(0xff & bytes[index + 3]) << 32);
//Array.Resize;
return a;
}
Also Int16
public static short toInt16(byte[] bytes, int index) //throws Exception
{
return (short)((bytes[index + 1] & 0xFF) | ((bytes[index] & 0xFF) << 0));
//return (short)(
// (0xff & bytes[index]) << 8 |
// (0xff & bytes[index + 1]) << 0
//);
}
BitConverter.getBytes
public static byte[] GetBytesU16(long value)
{
ByteBuffer buffer = ByteBuffer.allocate(8).order(ByteOrder.nativeOrder());
buffer.putLong(value);
return buffer.array();
}
Building on Jeff's answer, you can use a single ByteBuffer
to convert both to and from int
and byte[]
. Here is code you can drop into a class to convert to/from Little Endian:
ByteBuffer _intShifter = ByteBuffer.allocate(Integer.SIZE / Byte.SIZE)
.order(ByteOrder.LITTLE_ENDIAN);
public byte[] intToByte(int value) {
_intShifter.clear();
_intShifter.putInt(value);
return _intShifter.array();
}
public int byteToInt(byte[] data)
{
_intShifter.clear();
_intShifter.put(data, 0, Integer.SIZE / Byte.SIZE);
_intShifter.flip();
return _intShifter.getInt();
}
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