bit | 易学教程

Arithmetic Left shift time complexity

阅读更多关于 Arithmetic Left shift time complexity

问题 What is the time complexity of * Arithmetic left shift* / * Arithmetic right shift* operators of a n bit operand for example doing x = y << 2; whow much time will it take ? 回答1: Complexity, with the O(…) notation, is an asymptotic characterization of the time an algorithm takes when the input size becomes larger and larger. It is meaningless for algorithms that can take only a finite number of inputs. << can take 2^32 * 32 different inputs, hence a finite number of inputs, therefore it is

BIT: Unable to understand update operation in Binary index Tree

阅读更多关于 BIT: Unable to understand update operation in Binary index Tree

问题 I have just read answer on this question and was very satisfied and it is indeed a fantastic answer. It taught me the working of BIT. But at the end, the second last paragraph is where I am struggling. It says, Similarly, let's think about how we would do an update step. To do this, we would want to follow the access path back up to the root, updating all nodes where we followed a left link upward. We can do this by essentially doing the above algorithm, but switching all 1's to 0's and 0's

OpenCV Mat读取和显示图片

阅读更多关于 OpenCV Mat读取和显示图片

一、从文件读取图像并显示 1. 程序在基于VS2013搭建OpenCV开发环境这篇文章的最后给出了一个简单的Demo，这个例子跟本篇使用的例子是一样的。打开C++ IDE并创建一个新的项目，新建一个源文件，粘贴下面的代码： 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 #include <opencv2\highgui\highgui.hpp> #include <iostream> using namespace cv; using namespace std; int main( int argc, const char ** argv) { Mat img = imread( "earth.jpg" , CV_LOAD_IMAGE_UNCHANGED); if (img.empty()) { cout << "图像加载失败！" << endl; //system("pause"); return -1; } //创建一个名字为MyWindow的窗口 namedWindow( "MyWindow" , CV_WINDOW_AUTOSIZE); //在MyWindow的窗中中显示存储在img中的图片 imshow( "MyWindow" , img); //等待直到有键按下 waitKey

Does the type of bitfield affect structure alignement

阅读更多关于 Does the type of bitfield affect structure alignement

问题 I have the following structures: struct bf_struct1 { uint64_t bf1 : 1; uint64_t bf2 : 6; uint64_t bf3 : 2; uint64_t bf4 : 55; } struct bf_struct2 { uint8_t bf1 : 1; uint8_t bf2 : 6; uint8_t bf3 : 2; uint64_t bf4 : 55; } Does the structure member alignment depend on type of a bitfield members? 回答1: From the horse's mouth: 6.7.2.1 Structure and union specifiers ... 5 A bit-field shall have a type that is a qualified or unqualified version of _Bool, signed int , unsigned int , or some other

Bitwise Logic in C

阅读更多关于 Bitwise Logic in C

问题 I have some familiarity with bitwise operations, but this function just went over my head. void binary_print(unsigned int value) { unsigned int mask = 0xff000000; // Start with a mask for the highest byte. unsigned int shift = 256*256*256; // Start with a shift for the highest byte. unsigned int byte, byte_iterator, bit_iterator; for (byte_iterator=0; byte_iterator < 4; byte_iterator++) { byte = (value & mask) / shift; // Isolate each byte. printf(" "); for (bit_iterator=0; bit_iterator < 8;

Looping through Bits C

阅读更多关于 Looping through Bits C

问题 I'm trying to loop through the bits of an unsigned char, but I'm not sure where to start, eventually, I'll perform other bitwise operating on the bits, such as ~ and xor..etc. 回答1: Looping over bits can be done in several ways: You can do a destructive loop, when you shift the value, and test the initial or the final bit, depending on the order in which you would like to enumerate bits, or You can do a non-destructive loop, when you use bitwise AND to test the number with a single-bit mask ,

How can I get first n bits of floating point number as integer in python

阅读更多关于 How can I get first n bits of floating point number as integer in python

问题 Suppose I have 0.625 as a floating point is 0b.101 , so if I want the first two bits of that as an integer i.e. 0b10 = 2 , how can I achieve this in python? I've tried taking the number to a power of 2 and casting to an int, so if I want n bits I do int(0.625*(2**n)) . But that is not working for me. The problem occurs when I have a number greater than 1 so 24.548838022726972 will give me 392 rather than 12 for the first four bits. ( 24.548838022726972 = 0b11000.100011001... ) 回答1: If you

Getting essid via ioctl in ruby

阅读更多关于 Getting essid via ioctl in ruby

问题 To avoid relying on the wireless tools I want to get the essid directly from the device with ioctl, in C this wouldn't be a problem, but in Ruby it's quite different. The problem is following struct from wireless.h that is used as input/reply of ioctl: struct iw_point { void __user *pointer; /* Pointer to the data (in user space) */ __u16 length; /* number of fields or size in bytes */ __u16 flags; /* Optional params */ }; The pointer part must be a valid address of a memory area, followed by

解决Python下pip install MySQL-python失败的问题

阅读更多关于解决Python下pip install MySQL-python失败的问题

命令： pip install MySQL-python 报错： _mysql.c(42) : fatal error C1083: Cannot open include file: 'config-win.h': No s uch file or directory error: command 'C:\\Program Files (x86)\\Common Files\\Microsoft\\Visual C++ for Python\\9.0\\VC\\Bin\\cl.exe' failed with exit status 2 环境： OS：Win7-64bit Python:2.7.9 解决方案：在这里下载对应文件 http://www.codegood.com/downloads There are 11 files , weighing 7.9 MiB in total . Displaying 1 to 11 of 11 files. MySQL-python for Windows Distributions MySQL-python-1.2.2.win-amd64-py2.5.rar » 862.5 KiB - May 2, 2009 MySQLdb 1.2.2 for Windows and Python 2.5 64 bit MySQL-python

Best way to store long binary (up to 512 bit) in C#

阅读更多关于 Best way to store long binary (up to 512 bit) in C#

问题 I'm trying to figure out the best way to store large binary (more than 96 bit) numbers in C# I'm building application that will automatically allocate workers for shifts. Shifts can be as short as 15 minutes (but this might be even smaller in the future). To avoid double-booking of workers, I plan to have binary map of their daily time: 24 hours separated in equal chunks (15 minutes) and every chunk has a flag (0 for free, 1 for busy) So when we try to give another shift to a worker, we can