问题
I have a byte[4] which contains a 32-bit unsigned integer (in big endian order) and I need to convert it to long (as int can't hold an unsigned number).
Also, how do I do it vice-versa (i.e. from long that contains a 32-bit unsigned integer to byte[4])?
回答1:
Sounds like a work for the ByteBuffer.
Somewhat like
public static void main(String[] args) {
byte[] payload = toArray(-1991249);
int number = fromArray(payload);
System.out.println(number);
}
public static int fromArray(byte[] payload){
ByteBuffer buffer = ByteBuffer.wrap(payload);
buffer.order(ByteOrder.BIG_ENDIAN);
return buffer.getInt();
}
public static byte[] toArray(int value){
ByteBuffer buffer = ByteBuffer.allocate(4);
buffer.order(ByteOrder.BIG_ENDIAN);
buffer.putInt(value);
buffer.flip();
return buffer.array();
}
回答2:
You can use ByteBuffer, or you can do it the old-fashioned way:
long result = 0x00FF & byteData[0];
result <<= 8;
result += 0x00FF & byteData[1];
result <<= 8;
result += 0x00FF & byteData[2];
result <<= 8;
result += 0x00FF & byteData[3];
回答3:
Guava has useful classes for dealing with unsigned numeric values.
http://docs.guava-libraries.googlecode.com/git/javadoc/com/google/common/primitives/UnsignedInts.html#toLong(int)
来源:https://stackoverflow.com/questions/9855087/converting-32-bit-unsigned-integer-big-endian-to-long-and-back