xor

I feel silly for not being able to figure this out, but I am lost. I am trying to XOR two C strings. #include <stdio.h> #include <memory.h> #include <stdlib.h> int main() { char plainone[16]; char plaintwo[16]; char xor[17]; strcpy(plainone, "PlainOne"); strcpy(plaintwo, "PlainTwo"); int i=0; for(i=0; i<strlen(plainone);i++) xor[i] ^= (char)(plainone[i] ^ plaintwo[i]); printf("PlainText One: %s\nPlainText Two: %s\n\none^two: %s\n", plainone, plaintwo, xor); return 0; } My output is: $ ./a.out PlainText One: PlainOne PlainText Two: PlainTwo one^two: Why doesn't the xor array read as anything?

Does Ruby have a plain-English keyword for exclusive or, like they have "and" and "or"? If not, is this because exclusive or doesn't allow evaluation short-cutting? Firstly, I don't think shortcircuiting can sensibly apply to XOR: whatever the value of the first operand, the second needs to be examined. Secondly, and, &&, or and || use shortcircuiting in all cases; the only difference between the "word" and "symbol" versions is precedence. I believe that and and or are present to provide the same function as perl has in lines like process_without_error or die I think the reason for not having

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 operations on pointers and integers to result in undefined behavior - for example, you cannot guarantee that

Here's a question which deals with algorithm and bitwise xor operation. We are given x1*x2*x3*....*xn=P , where star( * ) operation represents XOR(bitwise) operation and x1 to xn are positive integers . P is also a positive integer. We need to find min(a1+a2+a3+.....an) such that this relation holds--> (x1+a1)*(x2+a2)*(x3+a3)*....*(xn+an)=0 . '+' represents normal addition operation. Restatement as a Bounded Integer Linear Programming Problem The problem can be restated as the following bounded ILP (integer linear programming) problem. Let x1,...,xN be as in the original problem, i.e. the goal