JavaScript traces its ancestry back to C, and C does not have a logical XOR operator. Mainly because it’s not useful. Bitwise XOR is extremely useful, but in all my years of programming I have never needed a logical XOR. If you have two boolean variables you can mimic XOR with: if (a != b) … Read more
and has higher precedence than or, so you can just write the decomposed version: {% if user.username in req.accepted.all and user.username not in req.declined.all or user.username not in req.accepted.all and user.username in req.declined.all %} For efficiency, using with to skip reevaluating the querysets: {% with accepted=req.accepted.all declined=req.declined.all username=user.username %} {% if username in accepted and … Read more
As others have pointed out, logical XOR is the same as not-equal for booleans, so you can do this: // XOR validation if( isEmptyString(firstStr) != isEmptyString(secondStr) ) { alert(SOME_VALIDATION_MSG); return; }
Note that A + X == (A xor X) + ((A and X)<<1). So: A xor X = A + X – ((A and X)<<1) = B + X A – B = (A and X)<<1 And we have: (A – B) and not (A<<1) = 0 (All bits in (A – B) are also … Read more
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 native support for hexadecimal integer literals, with the 0x prefix. “Hexadecimal” is just a presentation detail; the arithmetic is done in binary, and then the result is … Read more
I think I have a O(n lg U) algorithm for this, where U is the largest number. The idea is similar to user949300‘s, but with a bit more detail. The intuition is as follows. When you’re XORing two numbers together, to get the maximum value, you want to have a 1 at the highest possible … Read more
First Try Using scipy.weave and SSE2 intrinsics gives a marginal improvement. The first invocation is a bit slower since the code needs to be loaded from the disk and cached, subsequent invocations are faster: import numpy import time from os import urandom from scipy import weave SIZE = 2**20 def faster_slow_xor(aa,bb): b = numpy.fromstring(bb, dtype=numpy.uint64) … Read more
bool xorValue = bool1 ^ bool2;
A XOR B in english would be translated as “are A and B not equal”. So xor ax, ax will set ax to zero since ax is always equal to itself. A B | A XOR B 0 0 | 0 1 0 | 1 0 1 | 1 1 1 | 0
You want something like this: import sun.misc.BASE64Decoder; import sun.misc.BASE64Encoder; import java.io.IOException; public class StringXORer { public String encode(String s, String key) { return base64Encode(xorWithKey(s.getBytes(), key.getBytes())); } public String decode(String s, String key) { return new String(xorWithKey(base64Decode(s), key.getBytes())); } private byte[] xorWithKey(byte[] a, byte[] key) { byte[] out = new byte[a.length]; for (int i = 0; … Read more