virtual

How do I know the the complete virtual path that my application is currently hosted? For example: http://www.mysite.com/myApp or http://www.mysite.com/myApp/mySubApp I know the application path of HttpRequest but it only returns the folder name that my application is currently hosted, but how do I get the initial part? The domain name part of the path is not really a property of the application itself, but depends on the requesting URL. You might be able to reach a single Web site from many different host names. To get the domain name associated with the current request , along with the

I read that , "When a program executes an instruction like : MOV REG,1000 , it does so to copy the contents of the memory address 1000 to REG. Address can be generated using indexing,base registers,segment registers and other ways. These program generated address are called virtual address and form the virtual address space." Can anyone please explain me,what does it (These program generated address are called virtual address) mean ? Programs and data are stored as numbers in memory cells. Each memory cell has a unique number, called its address . The range of numbers representing valid

　　工厂模式：定义一个创建对象的接口，让其子类自己决定实例化哪一个工厂类，工厂模式使其创建过程延迟到子类进行。工厂模式属于创建型模式，其主要解决的是计划在 不同条件下创建不同实例 时接口选择的问题。 实现： 　　1、创建接口 1 class IShape 2 { 3 public: 4 virtual void draw() = 0; 5 virtual ~IShape() {}; 6 }; 　　2、创建接口实体类 1 class Rectangle :public IShape 2 { 3 virtual void draw(); 4 }; 5 6 class Square : public IShape 7 { 8 virtual void draw(); 9 }; 10 11 void Rectangle::draw() 12 { 13 cout << "this is Rectangle" << endl; 14 } 15 16 void Square::draw() 17 { 18 cout << "this is Square" << endl; 19 } 　　3、创建工厂类接口 1 class IShapeFactory 2 { 3 public: 4 virtual IShape* createShape() = 0; 5 virtual

I'm trying to create a Windows Virtual Drive ( like c:\ ) to map a remote storage. The main purpose is to do it in a clear way to the user. Therefore the user wouldn't know that he is writing/reading from another site. I was searching for available products, and i find that FUSE is not an option in Windows and WebDAV maps directly the drive, and i would like to build a middle layer between windows and remote storage to implement some kind of services. Another alternatives exists, such as Dokan, that is very expensive, and System.IO.IsolatedStorage Namespace, that doesn't seem to explicity

What are the advantages and disadvantages of each of these? interface methods virtual methods abstract methods When one should choose what? What are the points one should keep in mind when making this decision? Virtual and abstract are almost the same. A virtual method has an implementation in the base class that can optionally be overridden, while an abstract method hasn't and must be overridden in a child class. Otherwise they are the same. Choosing between them depends on the situation. If you got a base implementation, you use virtual. If you don't, and you need every descendant to

这一节实例主要实现使用服务器和客户端进行通信。 设计流程图如下： 具体操作步骤如下： 创建服务器基本对话框： 项目------MFC App----Dialog base---Windows Sockets 添加控件： 添加成员变量及按键消息处理函数 1 ///////////////////////////////////////////////////////////////////////////// 2 // CServerDlg dialog 3 4 class CServerDlg : public CDialog 5 { 6 public: 7 CSocket serSock; 8 CSocket ChatSendSock; 9 // Construction 10 public: 11 CServerDlg(CWnd* pParent = NULL); // standard constructor 12 13 // Dialog Data 14 //{{AFX_DATA(CServerDlg) 15 enum { IDD = IDD_SERVER_DIALOG }; 16 CString m_edit; 17 //}}AFX_DATA 18 19 // ClassWizard generated virtual function overrides 20 //{

模拟 #include<iostream> #include<algorithm> #include<string.h> #include<stdio.h> #include<map> using namespace std; int vis[100000000]; int n,flag; long long a[100005]; char s[400]; int chang(char x) { if (x>='0'&&x<='9') { int tmp=x-'0'; return tmp; } if (x>='A'&&x<='C') return 2; if (x>='D'&&x<='F') return 3; if (x>='G'&&x<='I') return 4; if (x>='J'&&x<='L') return 5; if (x>='M'&&x<='O') return 6; if (x>='P'&&x<='S') return 7; if (x>='T'&&x<='V') return 8; if (x>='W'&&x<='Y') return 9; //返回对应的数值 } int main() { scanf("%d",&n); flag=0; for (int i=1; i<=n; i++) { scanf("%s",s); //输入字符串 long long

八方向 深搜 #include <iostream> #include<cstdio> #include<cstdlib> #include<algorithm> using namespace std; char maps[1050][1050]; int n,m,d[8]= {-1,-1,-1,0, 0, 1,1,1}; int z[8]= {-1, 0, 1,-1,1,-1,0,1}; //8个方向 void dfs(int x, int y) { maps[x][y] = '*'; //变为*，相当于走过 for (int i = 0; i < 8; i++) { int nx = x + d[i], ny = y + z[i]; if (0<=nx&&nx<m&&0<=ny&&ny<n) { if (maps[nx][ny]=='@') { dfs(nx, ny); } } } } int main() { while(~scanf("%d%d", &m, &n)&&(m || n)) { int count = 0; for(int i = 0; i < m; i++) scanf("%s", maps[i]); for(int i = 0; i < m; i++) for(int j = 0; j < n; j++) { if(maps[i][j]=='@') {