I'm trying to extract 6 bytes from a byte array and convert them into a 48 bit signed integer value (I. E. a Java long). How can this be accomplished?
Edit: For example, if a byte array has one of the following:
byte[] minInt48 = new byte[] { (byte)0x80, 0, 0, 0, 0, 0 };
byte[] maxInt48 = new byte[] { (byte)0x7F, (byte)0xFF, (byte)0xFF, (byte)0xFF, (byte)0xFF, (byte)0xFF };
How can I parse this into a Java primitive (i.e. Java long) such that the sign value is preserved?
First you need to understand sign extension. You need to treat each byte as an unsigned value.
long v = 0;
byte q = -2; // unsigned value of 254
v = v + q;
System.out.println(v); // prints -2 which is not what you want.
v = 0;
v = v + (q & 0xFF); // mask off sign extension from q.
System.out.println(v); // prints 254 which is correct.
This is one way of doing it.
long val = 0;
byte[] bytes = { -1, 12, 99, -121, -3, 123
};
for (int i = 0; i < bytes.length; i++) {
// shift the val left by 8 bits first.
// then add b. You need to mask it with 0xFF to
// eliminate sign extension to a long which will
// result in an incorrect conversion.
val = (val << 8) | ((i == 0 ? bytes[i]
: bytes[i] & 0xFF));
}
System.out.println(val);
new BigInteger(by).longValue()