bitwise-operators

问题 How do you do the XOR bitwise operation if you only have available the AND and the OR operations? 回答1: Creating my own scripting language - ChrisScript - you just need something like: #!/bin/chrish bit XOR (bit A, bit B) { bit notA; bit notB; IF (A == 0) notA = 1 ELSE notA = 0; IF (B == 0) notB = 1 ELSE notB = 0; F = ((A && notB) || (notA && B)); RETURN F; } Even without NOT, it can be emulated like this. But this is the best solution you're going to get without having some form of inverter.

问题 I am trying to use a Boolean mask to get a match from 2 different dataframes. U Using the logical OR operator: x = df[(df['A'].isin(df2['B'])) or df['A'].isin(df2['C'])] Output: ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all(). However using the bitwise OR operator, the results are returned successfully. x = df[(df['A'].isin(df2['B'])) | df['A'].isin(df2['C'])] Output: x Is there a difference in both and would bitwise OR be the best

问题 I am trying to make a test to tell whether my PC performs arithmetic or logical right shift by right-shifting hexadecimal FFFFFFFF by 1 . I know that an integer -1 reads as FFFFFFFF in hexadecimal since it is the two's complement of 1 . Right-shifting -1 by 1 results in FFFFFFFF and shows the PC performed arithmetic right shift. But if I just type in 0xFFFFFFFF >> 1 , it resulted in 7FFFFFFF and shows that the PC performed logical right shift instead. Why did that happen? See for the code

问题 I have two input integer numbers and an output list< int> myoutputlist. My inputs are lets say A=0x 110101 B=0x 101100 then I have calculated a C integer number depending on A and B numbers.I already coded its algorithm, I can calculate C integer. C integer shows which bits should be changed. 1 value represents changing bits, 0 values represents unchanging bits. Only one bit should be changed in each time. As C integer depends on A and B inputs, sometimes 1 bit, sometimes 3 bits, sometimes 8

问题 I understand that: int bit = (number >> 3) & 1; Will give me the bit 3 places from the left, so lets say 8 is 1000 so that would be 0001. What I don't understand is how "& 1" will remove everything but the last bit to display an output of simply "1". I know that this works, I know how to get a bit from an int but how is it the code is extracting the single bit? Code... int number = 8; int bit = (number >> 3) & 1; Console.WriteLine(bit); 回答1: Unless my boolean algebra from school fails me,

问题 It looks like my previous question update haven't been noticed, therefore a new question. #dump1 var_dump('two identical strings' | 'two identical strings'); # mind the | // string(21) "two identical strings" #dump2 var_dump('two identical strings' ^ 'two identical strings'); # mind the ^ // string(21) "" Why #dump2 shows that length == 21, but outputs none/invisible symbols? Plus, when pasted in Notepad++ there are no signs of 21 symbols inside that string either, well actually, not even 1

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 8 years ago . How do I overload the &, | and ^ operators in C# and how does overloading work? I have been trying to find some good answers for some time but have not found any. I will share any helpful articles. 回答1: By

问题 I'm a bit baffled by this. Shouldn't the values truncate after the shift? Does anyone know why this happens? long a, b, c, n; //assign any value to a, set b and c to 0x000...0 n = 128; //any number works; b = a << n; c = b >> n; a == (b >> n); // True a == c; //True; Postscript I've always understood that if you shift a buffer in any direction, the values that "fall" outside the buffer size get truncated, and that they're essentially lost unless you get them from the original buffer. Thats

问题 I'm trying to find a way to make PHP Bitwise XOR results match the results of JavaScript Bitwise XOR . I came across different questions of this issue, and all without answers. Here are a few of them: Javascript & PHP Xor equivalent php bitwise XOR and js bitwise XOR producing different results JS bitwise XOR operator acts differently from PHP’s counterpart. How to get the same result as PHP returns? I know that PHP is using 64 bit compared to 32 bit JavaScript, but my question is, is there

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 years ago . How to write unsigned right shift operator in sql server? The expression is like value >>> 0 Here is the e.g. -5381>>>0 = 4294961915 回答1: T-SQL has no bit-shift operators, so you'd have to implement one yourself. There's an implementation of a bitwise shifts here: http://dataeducation.com/bitmask-handling-part-4