xor

问题 I'm trying to remember how the math is worked out to compute the remainder of an XOR algorithm in Cyclical Redundancy Checks to verify the remainder bits of a network message. I shouldn't have tossed that text book. This is easily done in code, but how is it worked out by hand? I know it looks something like a standard division algorithm, but I can't remember where to go from there to get the remainder. ___________ 1010 | 101101000 Note: I did google it, but wasn't able to find a place where

问题 Maybe this is just obvious to everyone but can someone explain where XOR (or Exclusive-OR) got its name from? What does the word Exclusive really mean? Not that it matters, but its just stuck in my head since morning. OR: 0 0 0 0 1 1 1 0 1 1 1 1 XOR: 0 0 0 0 1 1 1 0 1 1 1 0 Is it "exclusively 0 for inputs 1,1", "special version of OR" or something else? 回答1: XOR is an "exclusive OR" because it only returns a "true" value of 1 if the two values are exclusive, i.e. they are both different. 回答2:

问题 I saw this code from an answer on PPCG: echo BeeABBeeoBodBaBdOdPQBBgDQgDdp^"\n\n\t8b\n\n\t\nb&\nb b \n%%nb%%%\n%\nQ"; I know PHP casts undefined constants into strings, so the equivalent code is: echo 'BeeABBeeoBodBaBdOdPQBBgDQgDdp' ^ "\n\n\t8b\n\n\t\nb&\nb b \n%%nb%%%\n%\nQ"; The output of those are: Holy Hole In A Donut, Batman! Can someone explain to me how the XOR of those two strings produce that line of output? 回答1: According to this official example, using XOR on strings will operate

问题 I'm trying to to teach a neural net of 2 inputs, 4 hidden nodes (all in same layer) and 1 output node. The binary representation works fine, but I have problems with the Bipolar. I can't figure out why, but the total error will sometimes converge to the same number around 2.xx. My sigmoid is 2/(1+ exp(-x)) - 1. Perhaps I'm sigmoiding in the wrong place. For example to calculate the output error should I be comparing the sigmoided output with the expected value or with the sigmoided expected

问题 Basically to make this quick and simple, I'm looking to run an XOR conditional in django template. Before you ask why don't I just do it in the code, this isn't an option. Basically I need to check if a user is in one of two many-to-many objects. req.accepted.all and req.declined.all Now they can only be in one or the other (hence the XOR conditional). From the looking around on the docs the only thing I can figure out is the following {% if user.username in req.accepted.all or req.declined

问题 I came across a post How to find a duplicate element in an array of shuffled consecutive integers? but later realized that this fails for many input. For ex: arr[] = {601,602,603,604,605,605,606,607} #include <stdio.h> int main() { int arr[] = {2,3,4,5,5,7}; int i, dupe = 0; for (i = 0; i < 6; i++) { dupe = dupe ^ a[i] ^ i; } printf ("%d\n", dupe); return 0; } How can I modify this code so that the duplicate element can be found for all the cases ? 回答1: From original question: Suppose you

问题 An XOR linked list is a modified version of a normal doubly-linked list in which each node stores just one "pointer" instead of two. That "pointer" is composed of the XOR of the next and previous pointers. To traverse the list, two pointers are needed - one to the current node and one to the next or previous node. To traverse forward, the previous node's address is XORed with the "pointer" stored in the current node, revealing the true "next" pointer. The C++ standard causes a bunch of

问题 I'm trying to xor 2 binaries using python like this but my output is not in binary any help? a = "11011111101100110110011001011101000" b = "11001011101100111000011100001100001" y = int(a) ^ int(b) print y 回答1: a = "11011111101100110110011001011101000" b = "11001011101100111000011100001100001" y = int(a,2) ^ int(b,2) print '{0:b}'.format(y) 回答2: To get the Xor'd binary to the same length, as per the OP's request, do the following: a = "11011111101100110110011001011101000" b =

问题 I'm trying to xor 2 binaries using python like this but my output is not in binary any help? a = "11011111101100110110011001011101000" b = "11001011101100111000011100001100001" y = int(a) ^ int(b) print y 回答1: a = "11011111101100110110011001011101000" b = "11001011101100111000011100001100001" y = int(a,2) ^ int(b,2) print '{0:b}'.format(y) 回答2: To get the Xor'd binary to the same length, as per the OP's request, do the following: a = "11011111101100110110011001011101000" b =

问题 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance. Closed 6 years ago . I read on a site that using xor swaps is fast because it doesn't use a temporary variable. Here's an example: #include <stdio.h> int