bit-manipulation

问题 I'm a bit naive when it comes to bitwise logic and I have what is probably a simple question... basically if I have this (is ActionScript but can apply in many languages): var color:uint = myObject.color; var red:uint = color >>> 16; var green:uint = color >>> 8 & 0xFF; var blue:uint = color & 0xFF; I was wondering what exactly the `& 0xFF' is doing to green and blue. I understand what an AND operation does, but why is it needed (or a good idea) here? The source for this code was here: http:/

问题 I'd like a way to calculate (x + y)/2 for any two integers x, y in Java. The naive way suffers from issues if x+y > Integer.MAX_VALUE, or < Integer.MIN_VALUE. Guava IntMath uses this technique: public static int mean(int x, int y) { // Efficient method for computing the arithmetic mean. // The alternative (x + y) / 2 fails for large values. // The alternative (x + y) >>> 1 fails for negative values. return (x & y) + ((x ^ y) >> 1); } ... but this rounds towards negative infinity, meaning the

问题 I want to get the values of INT_MIN and INT_MAX . I've tried ~0 and ~0 >> 1 since the leftmost bit is a sign bit but I got -1 for both of them. It's so confused that why ~0 doesn't turn out to be 0xffffffff and ~0 >> 1 to be 0x7fffffff ? 回答1: Use: ~0U >> 1 Suffix 'U' for unsigned shift behavior. so, confused that why not ~0 turns out to be 0xffffffff ? See, what is 0 say in four bytes representation: BIT NUMBER 31 0 ▼ ▼ number bits 0000 0000 0000 0000 0000 0000 0000 0000 ▲ ▲ MSB LSB LSB -

问题 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. 回答1: Bitwise operators avoid branching instructions, even in Java code execution. As a result you have no expensive

问题 short permissions = 0755; short requested = 0700; short result = permissions & requested; I get a compiler error: error possible loss of precision found : int required: short If I'm not totally wrong the result of binary AND is as long as the longest operand. Why is the result an integer? Would there be a performance hit, if I would cast to short? (short) permissions & requested 回答1: If I'm not totally wrong the result of binary AND is as long as the longest operand. Why is the result an

问题 Let's say that data is 1011 1001 and the mask is 0111 0110 , then you have: input data: 1011 1001 input mask: 0111 0110 apply mask: 0011 0000 (based on `input mask`) bits selected: -011 -00- (based on `input mask`) right packed: ---0 1100 expected result: 0000 1100 (set left `8 - popcount('input mask')` bits to zero) So the final output is 0000 1100 (note that the 3 positions on the left which are unspecified are zero-filled). You can see that wherever the bits in input mask is 1 the

问题 Could you provide a convincing explanation, or a mathematical proof, to why the following function calculates the negabinary representation of a given number? function quickNegabinary(number) { var mask = 0xAAAAAAAA; return ((number + mask) ^ mask).toString(2); } 回答1: Negabinary notation Negabinary notation uses a radix of -2. This means that, as in every number system with a negative base, every other bit has a negative value: position: ... 7 6 5 4 3 2 1 0 binary: ... 128 64 32 16 8 4 2 1

问题 I'm trying to packing 10 bit pixels in to a continuous byte stream, using SIMD instructions. The code below does it "in principle" but the SIMD version is slower than the scalar version. The problem seem to be that I can't find good gather/scatter operations that load the register efficiently. Any suggestions for improvement? // SIMD_test.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include "Windows.h" #include <tmmintrin.h> #include <stdint.h> #include

问题 I'm looking at the entry Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup from Bit Twiddling hacks. I can easily see how the second algorithm in that entry works static const int MultiplyDeBruijnBitPosition2[32] = { 0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8, 31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9 }; r = MultiplyDeBruijnBitPosition2[(uint32_t)(v * 0x077CB531U) >> 27]; which calculates n = log2 v where v is known to be a

问题 I was reading about flag enums and bitwise operators, and came across this code: enum file{ read = 1, write = 2, readandwrite = read | write } I read somewhere about why there is a inclusive or statement and how there can't be an &, but can't find the article. Can someone please refresh my memory and explain the reasoning? Also, how can I say and/or? Eg. if dropdown1="hello" and/or dropdown2="hello".... Thanks 回答1: First Question: A | does a bitwise or; a bit will be set in the result if it