xor

I found the following XOR encryption function on the internet: void xor_encrypt(char *key, char *string) { int i, string_length = strlen(string); for(i=0; i<string_length; i++) { string[i]=string[i]^key[i]; printf("%i", string[i]); } } It works perfect, but I would like to decrypt the string also. For example: void xor_decrypt(char *key, char *encrypted_string) { //decrypt method goes here } So basically after I encrypt the string, I would use the same encryption key to decrypt the previously encrypted string. I'm pretty new to programming and I would just like to know how to decrypt the

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)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => false a = true; b = true;

I often see code like int hashCode(){ return a^b; } Why XOR? Of all bit-operations XOR has the best bit shuffling properties. This truth-table explains why: A B AND 0 0 0 0 1 0 1 0 0 1 1 1 A B OR 0 0 0 0 1 1 1 0 1 1 1 1 A B XOR 0 0 0 0 1 1 1 0 1 1 1 0 As you can see for AND and OR do a poor job at mixing bits. OR will on average produce 3/4 one-bits. AND on the other hand will produce on average 3/4 null-bits. Only XOR has an even one-bit vs. null-bit distribution. That makes it so valuable for hash-code generation. Remember that for a hash-code you want to use as much information of the key

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 does my code give incorrect output? EDIT: Okay, got it. The first point to make is that obviously you

Is there an XOR operator or equivalent function in SQL Server (T-SQL)? There is a bitwise XOR operator - the caret (^), i.e. for: SELECT 170 ^ 75 The result is 225. For logical XOR, use the ANY keyword and NOT ALL, i.e. WHERE 5 > ANY (SELECT foo) AND NOT (5 > ALL (SELECT foo)) Using boolean algebra, it is easy to show that: A xor B = (not A and B) or (A and not B) A B | f = notA and B | g = A and notB | f or g | A xor B ----+----------------+----------------+--------+-------- 0 0 | 0 | 0 | 0 | 0 0 1 | 1 | 0 | 1 | 1 1 0 | 0 | 1 | 1 | 1 1 1 | 0 | 0 | 0 | 0 As clarified in your comment,

how can we XOR hex numbers in python eg. I want to xor 'ABCD' to '12EF'. answer should be B922. i used below code but it is returning garbage value def strxor(a, b): # xor two strings of different lengths if len(a) > len(b): return "".join(["%s" % (ord(x) ^ ord(y)) for (x, y) in zip(a[:len(b)], b)]) else: return "".join(["%s" % (ord(x) ^ ord(y)) for (x, y) in zip(a, b[:len(a)])]) key ='12ef' m1='abcd' print strxor(key,m1) Whoa. You're really over-complicating it by a very long distance. Try: >>> print hex(0x12ef ^ 0xabcd) 0xb922 You seem to be ignoring these handy facts, at least: Python has

I wrote a short C++ program to do XOR encryption on a file, which I may use for some personal files (if it gets cracked it's no big deal - I'm just protecting against casual viewers). Basically, I take an ASCII password and repeatedly XOR the password with the data in the file. Now I'm curious, though: if someone wanted to crack this, how would they go about it? Would it take a long time? Does it depend on the length of the password (i.e., what's the big-O)? The problem with XOR encryption is that for long runs of the same characters, it is very easy to see the password. Such long runs are