xor

问题 What operation does the following ‘C’ statement perform? star = star ^ 0b00100100; (A) Toggles bits 2 and 5 of the variable star. (B) Clears all bits except bits 2 and 5 of the variable star. (C) Sets all bits except bits 2 and 5 of the variable star. (D) Multiplies value in the variable star with 0b00100100. I'm still clueless about this. Can someone help me out? 回答1: XOR operator (also called "logical addition") is defined like this: a b a^b ----------- 0 0 0 0 1 1 1 0 1 1 1 0 So a^0 leaves

问题 What operation does the following ‘C’ statement perform? star = star ^ 0b00100100; (A) Toggles bits 2 and 5 of the variable star. (B) Clears all bits except bits 2 and 5 of the variable star. (C) Sets all bits except bits 2 and 5 of the variable star. (D) Multiplies value in the variable star with 0b00100100. I'm still clueless about this. Can someone help me out? 回答1: XOR operator (also called "logical addition") is defined like this: a b a^b ----------- 0 0 0 0 1 1 1 0 1 1 1 0 So a^0 leaves

问题 What is the simplest way to do a three-way exclusive OR? In other words, I have three values, and I want a statement that evaluates to true IFF only one of the three values is true. So far, this is what I've come up with: ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) Is there something simpler to do the same thing? Here's the proof that the above accomplishes the task: a = true; b = true; c = true ((a ^ b) && (a ^ c) && !(b && c))

问题 I found this code to swap two numbers without using a third variable, using the XOR ^ operator. Code: int i = 25; int j = 36; j ^= i; i ^= j; j ^= i; Console.WriteLine("i:" + i + " j:" + j); //numbers Swapped correctly //Output: i:36 j:25 Now I changed the above code to this equivalent code. My Code: int i = 25; int j = 36; j ^= i ^= j ^= i; // I have changed to this equivalent (???). Console.WriteLine("i:" + i + " j:" + j); //Not Swapped correctly //Output: i:36 j:0 Now, I want to know, Why

问题 I have a quick question about using XOR two swap two string literals. so I have the following: #include <stdio.h> #include <stdlib.h> #include <string.h> void intSwap(int *a, int *b){ *a=*a^*b; *b=*a^*b; *a=*a^*b; } void swapString(char **a, char **b){ char *temp=*a; *a=*b; *b=temp; } void main(){ char *s= "ha"; char *t= "Oh"; printf("%s and %s \n",s,t); // prints ha Oh swapString(&s,&t); printf("%s and %s \n",s,t); // prints Oh ha int a=10; int b=5; printf("%d %d\n",a,b); //print 10 5

问题 Hi i have an issue while doing xoring between objects in c#. Suppose i have a class having below properties - public string UserName { get; set; } public bool IsUserEmployed { get; set; } public bool IsUserValid { get; set; }` I have 2 lists of this classes: List<Class1> List1 = new List<Class1>(); List<Class1> List2 = new List<Class1>(); I tried the basic xoring as follows: FinalList.AddRange(List1.Except(List2)); FinalList.AddRange(List2.Except(List1));` But i didnt got the results in the

问题 How can I XOR images with BMP extension using C# built in methods or any other image processing technique? I'm now doing this, but I want much more efficient method. using System; using System.Drawing; using System.Drawing.Imaging; namespace IProcessing { /// <summary> /// Summary description for LogicalOperator. /// </summary> public class LogicalOperator { Bitmap bmpimg; public LogicalOperator() { // // TODO: Add constructor logic here // } public Bitmap XORing( Bitmap bmp ) { BitmapData

问题 Apparently this used to work on ruby 1.8.7 but unfortunately not on 1.9.2 class String def xor(key) text = dup text.length.times {|n| text[n] ^= key[n.modulo key.size] } text end end def encode(_original, _pass = 'testvendor') _original.xor(_pass) end puts encode('Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat.

问题 I need to port some JS code which involves Math.random()*2147483648)^(new Date).getTime() . While it looks like for smaller numbers, the python function and the JS function are equivalent in function, but with large numbers like this, the values end up entirely different. Python: >>> 2147483647 ^ 1257628307380 1257075044427 Javascript: > 2147483647 ^ 1257628307380 -1350373301 How can I get the Javascript value from python? 回答1: Python has unlimited-precision integers, while Javascript is

问题 I would need to do horizontal xor of two 128bit integers (by 32bit integers) and combine the results to one 64bit integer. So operation like this: uint32_t x0[4]; uint32_t x1[4]; uint32_t xor0 = x0[0]; uint32_t xor1 = x1[0]; for (int i = 1; i < 4; ++i) { xor0 ^= x0[i]; xor1 ^= x1[i]; } uint64_t xor = uint64_t(xor1) << 32 | xor0; I finally found following code, that seems to work: __m128i x0 = ...; __m128i x1 = ...; __m128i xor64_0 = _mm_unpackhi_epi64(x0, x1); __m128i xor64_1 = _mm_unpacklo