Converting 32-bit unsigned integer (big endian) to long and back

余生长醉 提交于 2019-11-27 16:33:35

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

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