bit-manipulation

How can the XOR operation (on two 32 bit ints) be implemented using only basic arithmetic operations? Do you have to do it bitwise after dividing by each power of 2 in turn, or is there a shortcut? I don't care about execution speed so much as about the simplest, shortest code. Edit: This is not homework, but a riddle posed on a hacker.org . The point is to implement XOR on a stack-based virtual machine with very limited operations (similar to the brainfuck language and yes - no shift or mod). Using that VM is the difficult part, though of course made easier by an algorithm that is short and

I'm trying to write kotlin code like: for (byte b : hash) stringBuilder.append(String.format("%02x", b&0xff)); but I have nothing to do with the "&". I'm trying to use "b and 0xff" but it doesn't work. The bitwise "and" seems to work on Int, not byte. java.lang.String.format("%02x", (b and 0xff)) it's ok to use 1 and 0xff Kolin provides bitwise operator-like infix functions available for Int and Long only. So it's necessary to convert bytes to ints to perform bitwise ops: val b : Byte = 127 val res = (b.toInt() and 0x0f).toByte() // evaluates to 15 UPDATE: Since Kotlin 1.1 these operations are

I'm writing some code for a very limited system where the mod operator is very slow. In my code a modulo needs to be used about 180 times per second and I figured that removing it as much as possible would significantly increase the speed of my code, as of now one cycle of my mainloop does not run in 1/60 of a second as it should. I was wondering if it was possible to re-implement the modulo using only bit shifts like is possible with multiplication and division. So here is my code so far in c++ (if i can perform a modulo using assembly it would be even better). How can I remove the modulo

I'm building a PowerPC interpreter, and it works quite well. In the Power architecture the condition register CR0 (EFLAGS on x86) is updated on almost any instruction. It is set like this. The value of CR0 is 1, if the last result was negative, 2 if the last result was positive, 4 otherwise. My first naive method to interpret this is: if (n < 0) cr0 = 1 else if (n > 0) cr0 = 2; else cr0 = 4; However I understand that all those branches won't be optimal, being run millions of times per second. I've seen some bit hacking on SO, but none seemed adeguate. For example I found many examples to

I have 1 bit in a byte (always in the lowest order position) that I'd like to invert. ie given 00000001 I'd like to get 00000000 and with 00000000 I'd like 00000001. I solved it like this: bit > 0 ? 0 : 1; I'm curious to see how else it could be done. How about: bit ^= 1; This simply XOR's the first bit with 1, which toggles it. If you want to flip bit #N, counting from 0 on the right towards 7 on the left (for a byte), you can use this expression: bit ^= (1 << N); This won't disturb any other bits, but if the value is only ever going to be 0 or 1 in decimal value (ie. all other bits are 0),

I'd like to get the sign of a float value as an int value of -1 or 1. Avoiding conditionals is always a good idea in reducing computational cost. For instance, one way I can think of would be to use a fast bit-shift to get the sign: float a = ...; int sign = a >> 31; //0 for pos, 1 for neg sign = ~sign; //1 for pos, 0 for neg sign = sign << 1; //2 for pos, 0 for neg sign -= 1; //-1 for pos, 1 for neg -- perfect. Or more concisely: int sign = (~(a >> 31) << 1) - 1; Does this seem like a good approach? Will this work for all platforms, given endianness concerns (as MSB holds sign)? assylias Any