bit-manipulation

问题 #define SwapByte4(ldata) \ (((ldata & 0x000000FF) << 24) | \ ((ldata & 0x0000FF00) << 8) | \ ((ldata & 0x00FF0000) >> 8) | \ ((ldata & 0xFF000000) >> 24)) What does that 0x000000FF represent? I know that decimal 15 is represented in hex as F, but why is it << 24? 回答1: Here is a hex value, 0x12345678, written as binary, and annotated with some bit positions: |31 24|23 16|15 8|7 bit 0| +---------------+---------------+---------------+---------------+ |0 0 0 1 0 0 1 0|0 0 1 1 0 1 0 0|0 1 0 1 0 1

问题 #define SwapByte4(ldata) \ (((ldata & 0x000000FF) << 24) | \ ((ldata & 0x0000FF00) << 8) | \ ((ldata & 0x00FF0000) >> 8) | \ ((ldata & 0xFF000000) >> 24)) What does that 0x000000FF represent? I know that decimal 15 is represented in hex as F, but why is it << 24? 回答1: Here is a hex value, 0x12345678, written as binary, and annotated with some bit positions: |31 24|23 16|15 8|7 bit 0| +---------------+---------------+---------------+---------------+ |0 0 0 1 0 0 1 0|0 0 1 1 0 1 0 0|0 1 0 1 0 1

问题 I want to set first three bytes of an integer to 0 in C++. I tried this code, but my integer variable a is not changing, output is always -63. What am I doing wrong? #include <iostream> #include <string> int main() { int a = 4294967233; std::cout << a << std::endl; for(int i = 0; i< 24; i++) { a |= (0 << (i+8)); std::cout << a << std::endl; } } 回答1: Just use bitwise and ( & ) with a mask, there is no reason for loop: a &= 0xFF000000; // Drops all but the third lowest byte a &= 0x000000FF; //

问题 I need to check the value of the least significant bit (LSB) and most significant bit (MSB) of an integer in C/C++. How would I do this? 回答1: //int value; int LSB = value & 1; Alternatively (which is not theoretically portable, but practically it is - see Steve's comment) //int value; int LSB = value % 2; Details: The second formula is simpler. The % operator is the remainder operator. A number's LSB is 1 iff it is an odd number and 0 otherwise. So we check the remainder of dividing with 2.

问题 I'm messing with assembly language programming and I'm curious how I could tell if a number is a multiple of 4 using the logic operator AND? I know how to do it using "div" or "remainder" instructions but I'm trying to do this with bit manipulation of number/word. Can anyone point me in the right direction? I'm using MIPs but a Language agnostic answer is fine. 回答1: Well, to detect if a number is a multiple of another, you simply need to do x MOD y . If the result is 0 , then it is an even

问题 I got some numbers and I need to get how much they should be shifted for their lower bit to be at position 0. ex: 0x40000000 => 30 because 0x40000000 >> 30 = 1 768 = 512+256 => 8 This works if (Math.log2(x) == 31) return 31; if (Math.log2(x) > 31) x = x & 0x7FFFFFFF; return Math.log2(x & -x) Is there any more efficient or elegant way (builtin ?) to do this in javascript ? 回答1: You cannot get that result immediately with a builtin function, but you can avoid using Math.log2 . There is a little

问题 I write a algorithm (taken from "The C Programming Language") that counts the number of 1-bits very fast: int countBit1Fast(int n) { int c = 0; for (; n; ++c) n &= n - 1; return c; } But a friend told me that __builtin__popcount(int) is a lot faster, but less portable. I give it a try and was MANY times faster! Why it's so fast? I want to count bits as fast as possible, but without stick to a particular compiler. EDIT: I may use it on PIC micro-controllers and maybe on non-intel processors,

问题 I have the following question: If asked whether to use a shift vs a multiply or divide for example the answer would be, let the JVM optimize. Example here: is-shifting-bits-faster-than-multiplying Now I was looking at the jdk source, for example Priority Queue and the code uses only shifting for both multiplication and division (signed and unsigned). Taking for granted that the post in SO is the valid answer I was wondering why in jdk they prefer to do it by shifting? Is it some subtle detail

问题 If you have an enum that is used for bit flags, i.e., [Flags] internal enum _flagsEnum : byte { None = 0, //00000000 Option1 = 1, //00000001 Option2 = 1 << 1, //00000010 Option3 = 1 << 2, //00000100 Option4 = 1 << 3, //00001000 Option5 = 1 << 4, //00010000 Option6 = 1 << 5, //00100000 Option7 = 1 << 6, //01000000 Option8 = 1 << 7, //10000000 All = Byte.MaxValue,//11111111 } _flagsEnum myFlagsEnum = _flagsEnum.None; Is it faster to do.. bool hasFlag = myFlagsEnum.HasFlag(_flagsEnum.Option1);

问题 How can I perform bitwise queries on the DB with Django? I haven't found anything about it in the docs. Should I retrieve a queryset and then filter programmically? If you're interested, I use bitwise ops as an alternative to IN() statements in very large and complex queries, in order to improve performance. I have a DB containing millions of items (records). Some fields use binary representation of an item property. For example: the Color field can have multiple values, so it is structured