bit-manipulation

We have a database of images where I have calculated the PHASH using Dr. Neal Krawetz's method as implemented by David Oftedal . Part of the sample code calculates the difference between these longs is here: ulong hash1 = AverageHash(theImage); ulong hash2 = AverageHash(theOtherImage); uint BitCount(ulong theNumber) { uint count = 0; for (; theNumber > 0; theNumber >>= 8) { count += bitCounts[(theNumber & 0xFF)]; } return count; } Console.WriteLine("Similarity: " + ((64 - BitCount(hash1 ^ hash2)) * 100.0) / 64.0 + "%"); The challenge is that I only know one of these hashes and I want to query

Why does Java return -2147483648 when I bit shift 1 << 63 ? The expected result is 9 223 372 036 854 775 808 , tested with Wolfram Alpha and my calculator. I tested: System.out.print((long)(1 << (63))); Adam Mihalcin There's an important thing to note about the line System.out.print((long)(1 << (63))); You first take (1 << 63) , and then you cast to long. As a result, you are actually left-shifting in integers, so the long cast doesn't have any effect. That's why shifting 63 bits left gives the min integer rather than the min long. But there's another, more important point. Java longs are

Cut and dry... while I never have enough logical operations for it to be a performance bottleneck - I wonder, would I be better off using bitwise and (&) and bitwise or (|) as opposed to the same-named logical operators (&& and ||) if possible? Perhaps the question can be prefaced by the fact that I don't know of a library to convert Java to assembly to see the # of operations. Bitwise operators avoid branching instructions, even in Java code execution. As a result you have no expensive branch prediction misses and no jumps at all. From my experience, they can be measurably faster when used in

How do I set, clear and toggle a bit in Rust? Like many other languages, the bitwise operators & (bitwise AND), | (bitwise OR), ^ (bitwise XOR) exist: fn main() { let mut byte: u8 = 0b0000_0000; byte |= 0b0000_1000; // Set a bit println!("0b{:08b}", byte); byte &= 0b1111_0111; // Unset a bit println!("0b{:08b}", byte); byte ^= 0b0000_1000; // Toggle a bit println!("0b{:08b}", byte); } The main difference from other languages is in bitwise NOT, which uses ! instead of ~ : fn main() { let mut byte: u8 = 0b0000_0000; byte = !byte; // Flip all bits println!("0b{:08b}", byte); } You can also shift

I'm parsing unsigned bits from a DatagramSocket. I have a total of 24bits (or 3 bytes) coming in - they are: 1 unsigned 8bit integer followed by a 16bit signed integer. But java never stores anything more than a signed byte into a byte/byte array? When java takes in these values, do you lose that last 8th bit? DatagramSocket serverSocket = new DatagramSocket(666); byte[] receiveData = new byte[3]; <--Now at this moment I lost my 8th bit System.out.println("Binary Server Listing on Port: "+port); while (true) { DatagramPacket receivePacket = new DatagramPacket(receiveData, receiveData.length);

I have a uint8 (unsigned 8 bit integer) coming in from a UDP packet. Java only uses signed primitives. How do I parse this data structure correctly with java? Simply read it as as a byte and then convert to an int. byte in = udppacket.getByte(0); // whatever goes here int uint8 = in & 0xFF; The bitmask is needed, because otherwise, values with bit 8 set to 1 will be converted to a negative int. Example: This: 10000000 Will result in: 11111111111111111111111110000000 So when you afterwards apply the bitmask 0xFF to it, the leading 1's are getting cancelled out. For your information: 0xFF ==

I have found one example in Data and Communication Networking book written by Behrouza Forouzan regarding upper- and lowercase letters which differ by only one bit in the 7 bit code. For example, character A is 1000001 (0x41) and character a is 1100001 (0x61).The difference is in bit 6, which is 0 in uppercase letters and 1 in lowercase letters. If we know the code for one case, we can easily find the code for the other by adding or subtracting 32 in decimal, or we can just flip the sixth bit. What does all this mean? I have found myself very confused with all these things. Could someone

Is there a straightforward way to extracting the exponent from a power of 2 using bitwise operations only? EDIT: Although the question was originally about bitwise operations, the thread is a good read also if you are wondering "What's the fastest way to find X given Y = 2 X in Python ?"** I am currently trying to optimize a routine ( Rabin-Miller primality test ) that reduces an even number N in the forms 2**s * d . I can get the 2**s part by: two_power_s = N & -N but I can't find a way to extract just " s " with a bitwise operation. Workarounds I am currently testing without too much