setBit java method using bit shifting and hexadecimal code - question
- by somewhat_confused
I am having trouble understanding what is happening in the two lines with the 0xFF7F and the one below it. There is a link here that explains it to some degree.
http://www.herongyang.com/java/Bit-String-Set-Bit-to-Byte-Array.html
I don't know if 0xFF7FposBit) & oldByte) & 0x00FF
are supposed to be 3 values 'AND'ed together or how this is supposed to be read. If anyone can clarify what is happening here a little better, I would greatly appreciate it.
private static void setBit(byte[] data,
final int pos,
final int val) {
int posByte = pos/8;
int posBit = pos%8;
byte oldByte = data[posByte];
oldByte = (byte) (((0xFF7F>>posBit) & oldByte) & 0x00FF);
byte newByte = (byte) ((val<<(8-(posBit+1))) | oldByte);
data[posByte] = newByte;
}
passed into this method as parameters from a selectBits method was setBit(out,i,val);
out = is byte[] out = new byte[numOfBytes]; (numOfBytes can be 7 in this situation)
i = which is number [57], the original number from the PC1 int array holding the 56-integers.
val = which is the bit taken from the byte array from the getBit() method.
