xor

& has && . | has || . Why doesn't ^ have ^^ ? I understand that it wouldn't be short-circuiting, but it would have different semantics. In C, true is really any non-zero value. Bitwise XOR is not always the same thing as logical XOR: int a=strcmp(str1,str2);// evaluates to 1, which is "true" int b=strcmp(str1,str3);// evaluates to 2, which is also "true" int c=a ^^ b; // this would be false, since true ^ true = false int d=a ^ b; //oops, this is true again, it is 3 (^ is bitwise) Since you can't always rely on a true value being 1 or -1 , wouldn't a ^^ operator be very helpful? I often have to

I'm getting into assembly and I keep running into xor, for example: xor ax, ax Does it just clear the register's value? 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 xor reg, reg is often used to clear register. It can be an alternative to mov reg, 0 AFAIR, it was faster (or shorter) in some cases. And of course, XOR itself is eXclusive OR (a.k.a.: exclusive disjunction) operation (but it's a shame to describe here such basics - use Wikipedia) xor ax, ax is

I'm trying to get into C programming, and I'm having trouble writing a bitwise XOR function with only ~ and & operators. Example: bitXor(4, 5) = 1 . How can I achieve this? So far I have this: int bitXor(int x, int y) { return z; } Well, let's think about this. What does XOR do? x y XOR ------------ 0 0 0 1 0 1 0 1 1 1 1 0 So how do we turn that into a function? Let's think about AND, and the inverse order of AND (~x&~y) (this happens to be NOR): (~x&~y) x y AND NOR --------------------- 0 & 0 = 0 1 1 & 0 = 0 0 0 & 1 = 0 0 1 & 1 = 1 0 Looking at those two outputs, it's pretty close, all we

There is XOR function in Java - a^b For exemple: 5^3 = 6 Can you tell me inverse function? If I have 6 and 3 can i get range of numbers which include number 5 ? The inverse is XOR! If you have: c = a^b; You can get a or b back if you have the other value available: a = c^b; // or b^c (order is not important) b = c^a; // or a^c For example if a = 5 , b = 3 (and thus c = 6 as you mentioned) you get: b=0011 (3) a=0101 (5) c=0110 (6) XOR or c=0110 (6) XOR ---------- ---------- a=0101 (5) b=0011 (3) The inverse of XOR is XOR..... 来源： https://stackoverflow.com/questions/14279866/what-is-inverse

I personally like the exclusive or , ^ , operator when it makes sense in the context of boolean checks because of its conciseness. I much prefer to write if (boolean1 ^ boolean2) { //do it } than if((boolean1 && !boolean2) || (boolean2 && !boolean1)) { //do it } but I often get confused looks from other experienced Java developers (not just the newbies), and sometimes comments about how it should only be used for bitwise operations. I'm curious as to the best practices regarding the usage of the ^ operator. You can simply use != instead. I think you've answered your own question - if you get

Here is what I read about it but cant understand exactly what it does: One way to implement rubber-banding is to draw in XOR mode. You set XOR mode by calling the setXORMode() method for a graphics context and passing a color to it — usually the background color. In this mode the pixels are not written directly to the screen. The color in which you are drawing is combined with the color of the pixel currently displayed together with a third color that you specify, by exclusive ORing them together, and the resultant pixel color is written to the screen. The third color is usually set to be the

This question already has an answer here: Comparing two lists and only printing the differences? (XORing two lists) 6 answers I'm a beginner in Python, and I have to do the XOR between two lists (the first one with the length : 600 and the other 60) I really don't know how to do that, if somebody can explain me how, it will be a pleasure. I have to do that to find the BPSK signal module, and I'm wondering on how doing that with two lists that haven't the same length. I saw this post : Comparing two lists and only printing the differences? (XORing two lists) but the length of lists is the same

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 main(void) { int a=234,b=789; b=b^a; a=b^a; b=b^a; printf("a=%d,b=%d",a,b); return 0; } Why don't people use this technique in real life code? Is it just poor style? Is there something not well defined about it? Is it an optimisation that my compiler might produce from more clear code, automatically? Because readability is preferred over performance. Because tmp = a; a = b; b = tmp; is not that slow. Because the compiler will optimize it anyway. Because it works