bitwise-operators

问题 Just played around with unusual bitwise operations in JavaScript and I got some weird results in some cases: Usual cases 1 << 0 // returns 1, makes sense 100 << 0 // returns 100, makes sense 100 >> 0 // returns 100, definitely makes sense But these, when shift by 0 bits, all yield zero 9E99 << 0 // returns 0 ..... Why all bits are cleared? 9E99 >> 0 // returns 0 also ..... All bits cleared? Infinity >> 0 // returns 0 Infinity << 0 // returns 0 -Infinity << 0 // returns 0 .... Can't explain

问题 I was recently given a quiz in one of my classes. The question is below: Write a function (called cmp ) in C that accepts two integers ( x and y ) and returns: -1 if x < y , 0 if x = y , 1 if x > y . Write cmp as concise as possible. The most concise function that I could think of was: int cmp(int x, int y) { return ((x < y) ? (-1) : ((x == y) ? (0) : (1))); } But I have a feeling there could be a bit manipulation that I could use to do this more concisely. Perhaps a combination of & and ^ ?

问题 When I try to run a LINQ query of the form: MongoCollection<MyEntity> collection; collection.AsQueryable().Where(entity => (entity.Flags & MyFlags.AFlag) != MyFlags.None); I get an ArgumentException with the message Unsupported where clause: ((Int32)((Int32)entity.Flags & 4) != 0). Is this a known bug/feature? Is there any workaround? From the documentation it seems like MongoDB has a bitwise update, but not a bitwise query. For comparison, the same query runs smoothly above Redis using

问题 I have the following division that I need to do often: int index = pos / 64; Division can be expensive in the cpu level. I am hoping there is a way to do that with bitwise shift. I would also like to understand how you can go from division to shift, in other words, I don't want to just memorize the bitwise expression. 回答1: int index = pos >> 6 will do it, but this is unnecessary. Any reasonable compiler will do this sort of thing for you. Certainly the Sun/Oracle compiler will. The general

问题 I wrote following code in Eclipse: byte b = 10; /* some other operations */ b = ~b; Eclipse wanted a cast to byte in the line of the bitwise complement. It said: "Type mismatch: cannot convert from int to byte". I also tried this with other bitwise operations and on other integral types. It was with short and char the same. Only long and integer could use bitwise operations. Is there a reason for this? 回答1: Unary (such as ~ ) and binary operators in Java subject their operands to "unary

问题 I've recently been writing some code for a research project that I'm working on, where efficiency is very important. I've been considering scraping some of the regular methods I do things in and using bitwise XORs instead. What I'm wondering is if this will make if a difference (if I'm performing this operation say several million times) or if it's the same after I use 03 in g++. The two examples that come to mind: I had an instance where (I'm working with purely positive ints) I needed to

问题 It's a well known fact that multiplication, integer division, and modulo by powers of two can be rewritten more efficiently as bitwise operations: >>> x = randint(50000, 100000) >>> x << 2 == x * 4 True >>> x >> 2 == x // 4 True >>> x & 3 == x % 4 True In compiled languages such as C/C++ and Java, tests have shown that bitwise operations are generally faster than arithmetic operations. (See here and here). However, when I test these in Python, I am getting contrary results: In [1]: from

问题 I want to be able to access the sign bit of a number in C++. My current code looks something like this: int sign bit = number >> 31; That appears to work, giving me 0 for positive numbers and -1 for negative numbers. However, I don't see how I get -1 for negative numbers: if 12 is 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 then -12 is 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0011 and shifting it 31 bits would make 0000 0000 0000 0000 0000 0000 0000 0000 0000

问题 In a nutshell I have two images I want to overlay one over the other using a mask so that only parts of the second image show up. This is part of a real time image processing routine so I need the operation to happen as fast as possible. Specifically, I have two 32 bit image BGR byte arrays. In addition, I have a byte array that represents an image mask. I want to generate a new byte array where byte array A is overlayed on top of byte array B using the mask array to decide which byte is used

问题 I'm implementing a quantization algorithm from a textbook. I'm at a point where things pretty much work, except I get off-by-one errors when rounding. This is what the textbook has to say about that: Rounded division by 2^p may be carried out by adding an offset and right-shifting by p bit positions Now, I get the bit about the right shift, but what offset are they talking about? Here's my sample code: def scale(x, power2=16): if x < 0: return -((-x) >> power2) else: return x >> power2 def