bitwise-operators

问题 I show my question by an example : int a = 1 << 0; // = 1 int flag = 1; bool b = flag & a; // = 1 < In c++ this line has no error but in c# error is like this : Cannot be applied to operands of type 'bool' and 'int' When b variable is true and when b variable is false in c#? How fix the error? When does c++ recognize that b variable is true? The other side should be (flag & a) != 0 or (flag & a) == 1 or something else? 回答1: In C# you write it like so: bool b = (flag & a) != 0; You can't

问题 Is it possible to extract the month from date represented as int (format YYYYMMDD, e.g. 20110401) using some bitwise operators? If so, how can it be done? edit: I am currently using 20110401 % 10000 / 100. I thought bit-wise could be faster. DateTime.Parse etc. are too slow for what I am trying to do. 回答1: You could efficiently extract the month with bitwise operations if you represented the date in a binary format, e.g., 5 bits for the day of the month, 4 bits for the month number, and the

问题 I'm struggling to learn how to pack four seperate values into a single byte. I'm trying to get a hex output of 0x91 and the binary representation is supposed to be 10010001 but instead I'm getting outputs of: 0x1010001 and 16842753 respectively. Or is there a better way to do this? uint8_t globalColorTableFlag = 1; uint8_t colorResolution = 001; uint8_t sortFlag = 0; uint8_t sizeOfGlobalColorTable = 001; uint32_t packed = ((globalColorTableFlag << 24) | (colorResolution << 16) | (sortFlag <<

问题 I want to read the 2nd, 5th and the 6th bit from a 32-bit register. I have decided to use struct bit fields to store them. Is the following data structure correct? struct readData { int unwanted:1; int reqbit1:1; int unwanted1:2; int reqbit2:2; int unwanted2:26; }; I'm not sure about how the bit fields are created. I'm going to use an API that will copy bytes from the h/w register to this structure directly. In that case, will reqbit1 contain the 2nd bit? As per my understanding, the compiler

问题 Why a positive number operated with bitwise or 0 not always positive in Javascript For example: 3391700000|0 -903267296 4260919000|0 -34048296 2884900000|0 -1410067296 I'm using chrome 64-bit on Linux related to: https://stackoverflow.com/a/12837315/1620210 回答1: Because JavaScript uses 32bit integers at most, but keep in mind every number is kind of a float in this language If you want to truncate them to an unsigned 32bit value: (3391700000|0) >>> 0 回答2: In JavaScript, the operands of

问题 I am bit confused about a expression I found in some C++ code: if (m & 1) pmm=-pmm; I am not a C/C++ coder, so Google gives me two things: & is bitwise AND if syntax is if (condition) statement So, how does the above statement work? Should I not require a if ((m & 1)==0) ? 回答1: Just to add to the more technical explanations, a simpler way of viewing if (m & 1) is that is tests whether m is odd (i.e. not an exact multiple of 2): if (m & 1) // m is odd (1, 3, 5, 7, 9, ...) - do something else /

问题 Can you please explain the below lines, with some good examples. A left arithmetic shift by n is equivalent to multiplying by 2 n (provided the value does not overflow). And: A right arithmetic shift by n of a two's complement value is equivalent to dividing by 2 n and rounding toward negative infinity. If the binary number is treated as ones' complement, then the same right-shift operation results in division by 2 n and rounding toward zero. 回答1: I will explain what happens in a base that we

问题 In C++, a bool is guaranteed to be 0 or 1 C++ (§4.5/4): An rvalue of type bool can be converted to an rvalue of type int, with false becoming zero and true becoming one. Consider the following function and what g++5.2 generates with -O3 int foo(bool a, bool b) { if(a&b) return 3; else return 5; } 0000000000000000 <_Z3foobb>: 0: 40 84 ff test %dil,%dil 3: 74 13 je 18 <_Z3foobb+0x18> 5: 40 84 f6 test %sil,%sil 8: b8 03 00 00 00 mov $0x3,%eax d: 74 09 je 18 <_Z3foobb+0x18> f: f3 c3 repz retq 11:

问题 I'm trying to recreate this function: int test(int x) { int i; for (i = 0; i < 32; i+=2) if ((x & (1<<i)) == 0) return 0; return 1; } But only using these bit-wise operators: !, ~, &, ^, |, +, <<, and >> (Meaning no loops or if statements either) I am so confused with this question I have been staring at it for like a hour and am still not sure where to start. I understand that basically it is taking x comparing it with 2^i where i is 0-31 and then returning 0 if x and 2^i do not share any of

问题 so if I have x = 0x01 and y = 0xff and I do x & y , I would get 0x01 . If I do x && y do I still get 0x01 , but the computer just says its true if its anything than 0x00 ? My question is are the bits the same after the operation regardless of bit-wise or logical AND/OR, but the computer just interprets the final result differently? In other words are the hex values of the result of & and && (likewise | and ||) operations the same? edit: talking about C here 回答1: The answer depends on the