bit-shift

问题 I was asked to implement integer division with logarithmic time complexity using only bit shifts, additions and subtractions. I can see how I can deal with a divisor which is a power of 2, but how can I deal with an odd divisor, such that the time remains logarithmic? Is it even possible? EDIT: a way to do it in a time complexity that isn't logarithmic but still better than linear will also be welcomed. Thanks 回答1: It's just like doing long division on paper but in binary. You shift bits from

问题 Is the difference between integer multiply(temporarily forgetting about division) still in favor of shifting and if so how big is the difference? It simply seems such a low level optimization, even if you wanted it the shouldn't the (C#/Java) to bytecode compiler or the jit catch it in most cases? Note: I tested the compiled output for C#(with gmcs Mono C# compiler version 2.6.7.0) and the multiply examples didn't use shift for multiplying even when multiplying by a multiple of 2. C# http:/

问题 Is there any efficient algorithm that allows to insert bit bit to position index when working with uint16_t ? I've tried reading bit-by-bit after index , storing all such bits into array of char , changing bit at index , increasing index , and then looping again, inserting bits from array, but could be there a better way? So I know how to get, set, unset or toggle specific bit, but I suppose there could be better algorithm than processing bit-by-bit. uint16_t bit_insert(uint16_t word, int bit

问题 I have a school assignment that requires me to convert a word from little endian to big endian in three different ways. One of them is by using multiplication and division. I know that shifting to the left multiplies the number by 2 but i still cant figure out how to utilise that. Here is me doing it using rotate. Can someone help me step on this and do it with division and multiplication? .data .text .globl main main: li $t0,0x11223344 #number to be converted in t0 rol $t1,$t0,8 li $t2

问题 I am trying to replicate the behavior of javascript bit-shift and bit-wise operations in java. Did you ever try to do this before, and how can you do it reliably and consistently even with longs? var i=[some array with large integers]; for(var x=0;x<100;x++) { var a=a large integer; var z=some 'long'>2.1 billion; //EDIT: z=i[x]+=(z>>>5)^(a<<2))+((z<<4)^(a<<5)); } what would you do to put this into java? 回答1: Yes. Java has bit-wise operators and shift operators. Is there something in

问题 This question already has answers here : what does it mean to bitwise left shift an unsigned char with 16 (2 answers) Closed 12 months ago . I am trying to understand shift operators in C/C++, but they are giving me a tough time. I have an unsigned 8-bit integer initialized to a value, for the example, say 1. uint8_t x = 1; From my understanding, it is represented in the memory like |0|0|0|0|0||0||0||1| . Now, when I am trying to left shit the variable x by 16 bit, I am hoping to get output 0

问题 Suppose I have the following input: 1234 How can I get the following output? 3412 This is obtained by circularly shifting (or rotating) the digits of the input twice. I have tried the following code: number = 1234 bin(number >> 1) but it is not producing the results I was expecting. 回答1: The >> operator does a binary bitshift. It moves the binary representation of 1234 on place to the right, discarding the rightmost (least significant) bit. Therefore you code does not result in 3412 . You

问题 I just discovered livebindings with Delphi. And created my first components for handling a control-word for a frequency converter. The component it self seems to work well testing it in the form designer. However, compiling and running the application things doesn't work. Screenshot from livbindings like this: And here is the code for the component unit cBits2Byte; interface uses System.SysUtils, System.Classes; type TBits2Byte = class(TComponent) private { Private declarations } fBit00,

问题 Many lossless algorithms in signal processing require an evaluation of the expression of the form ⌊ a / 2 b ⌋, where a , b are signed ( a possibly negative, b non-negative) integers and ⌊·⌋ is the floor function. This usually leads to the following implementation. int floor_div_pow2(int numerator, int log2_denominator) { return numerator >> log2_denominator; } Unfortunately, the C standard states that the result of the >> operator is implementation-defined if the left operand has a signed

问题 I want to sort floats in C with radix. Below is the code I have. However, my output is not correct. For example, if I run the code with 3.1, -5, and 1 my sorted values are printed as 3.000000, -5.000000, and 1.000000. I know to cast correctly from a float to an int back to a float, I need to apply the following logic but I am not sure how to integrate this into rfloat() because I tried and am getting many errors. How would I be able to correctly apply a bitwise radix sort to floats? float x =