xor

问题 The following code System.out.println("1 0 0: " + (true ^ false ^ false)); System.out.println("1 0 1: " + (true ^ false ^ true)); System.out.println("1 1 0: " + (true ^ true ^ false)); System.out.println("1 1 1: " + (true ^ true ^ true)); System.out.println("0 0 0: " + (false ^ false ^ false)); System.out.println("0 0 1: " + (false ^ false ^ true)); System.out.println("0 1 0: " + (false ^ true ^ false)); System.out.println("0 1 1: " + (false ^ true ^ true)); outputs: 1 0 0: true 1 0 1: false

问题 I have two hex numbers, which, for the purposes of experimenting with OTP, I'm trying to XOR with C#. Unfortunately, both numbers are on the order of hundreds of digits - clearly far too large to store in an int or long . How do I store/XOR them? Right now, I'm storing as BigInteger s like so: public static string XOR(string string_1, string string_2){ BigInteger b1 = BigInteger.Parse(string_1, System.Globalization.NumberStyles.AllowHexSpecifier); BigInteger b2 = BigInteger.Parse(string_2,

问题 I am trying to xor two values which are like below: Variable 1 : 6463334891 Variable 2 : 1000212390 When i did xor with these values in php it gives me wrong answer. It should give me "7426059853" This is my code $numericValue = (int)$numericValue; $privateKey = (int)$privateKey; echo "Type of variable 1 ".gettype($numericValue)."<br />"; echo "Type of variable 2 ".gettype($privateKey)."<br />"; $xor_val = (int)$numericValue ^ (int)$privateKey; echo "XOR Value :".$xor_val."<br />"; 回答1: That

问题 Description: I am trying to write a simple program for fun that will read in a phrase then xor encrypt it then output the encrypted phrase to the terminal window. See code below for more info. code: #include #include using namespace std; int main () { string mystr; cout << "What's the phrase to be Encrypted? "; char key[11]="ABCDEFGHIJK"; //The Encryption Key, for now its generic getline(cin, mystr); string result; for (int i=0; i<10; i++) { result.push_back(mystr[i] ^ key[i]); cout << result

问题 Given an m*n binary matrix A, m*p binary matrix B, where n > m what is an efficient algorithm to compute X such that AX=B? For example: A = 1 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 0 1 0 1 1 0 1 0 1 0 1 0 1 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 1 1 1 0 B = 0 1 0 1 1 0 1 1 0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 1 0 1 0 0 0 1 1 0 0 0 1 1 0 0 1 1 0 0 0 0 1 1 1 1 0 0 0 1 1 0 0 0 1 0 0 1 0 0 1 0 1 0 0 1 1 0 Note, when I say binary matrix I mean matrix defined over the field Z_2, that is, where all arithmetic is mod 2.

问题 I'm having issues working with iOS Swift 2.0 to perform an XOR on a [UInt8] and convert the XORd result to a String. I'm having to interface with a crude server that wants to do simple XOR encryption with a predefined array of UInt8 values and return that result as a String. Using iOS Swift 2.0 Playground, create the following array: let xorResult : [UInt8] = [24, 48, 160, 212] // XORd result let result = NSString(bytes: xorResult, length: xorResult.count, encoding: NSUTF8StringEncoding) The

问题 I write a bash script who manipulate hex values and i need to do XOR operation between two hexa numbers. My problem is when i try in bash prompt it's work and return right value but in script this value is false. When XOR variable $ExtendAuthKey and $IPAD the result must be : 181ad673a5d94f0e12c8894ea26381b363636363636363636363636363636363636363636363636363636363636363636363636363636363636363636363636 But in fact i get this value : 3906369333256140342 I dont understand this behavior, if you

问题 I need to encrypt / decrypt a file using xor with Python 3, I have a code that works fine in Python 2, but when trying to adapt it to Python 3, gives me some errors that I can't solve. This code works fine in Python 2.7: from itertools import cycle def xore(data, key): return ''.join(chr(ord(a) ^ ord(b)) for (a, b) in zip(data, cycle(key))) with open('inputfile.jpg', 'rb') as encry, open('outputfile.jpg', 'wb') as decry: decry.write(xore(encry.read(), 'anykey')) The error when trying to run

问题 I a forum it is mentioned that given array of n numbers: arr[0........n-1] The following conditon holds, ^ is the xor operator ` f(l,r) = f(0,r) ^ f(0,l-1) where f(l,r) = arr[l]^arr[l+1]^........arr[r] I checked the above taking number of arrays and different values of l and r and YES , it is true. But I don't understand how? Can somebody please explain the logic behind this? 回答1: Use the simplest property of XOR f(0,r) ^ f(0,l-1) = f(l,r) => (f(0,r) ^ f(0,l-1)) ^ f(0,l-1) = f(0,l-1) ^ f(l,r)

问题 I have seen the solution for this problem in this thread -> How to find a duplicate element in an array of shuffled consecutive integers? But the problem I am now having is little varied from it. int arr[10] = {1,2,3,4,5,6,7,8,4,9}; int a= 0; for(int i=0;i<10;i++) { a= a^ arr[i] ^i; } cout<<a; Consider the above mentioned code snippet. Things work fine as it is. But when I add a 0 to the above mentioned array like, int arr[11] = {0,1,2,3,4,5,6,7,8,4,9}; I am not getting the proper duplicate