virtual

简介 SSI是英文Server Side Includes的缩写，翻译成中文就是服务器端包含的意思。从技术角度上说，SSI就是在HTML文件中，可以通过注释行调用的命令或指针。SSI具有强大的功能，只要使用一条简单的SSI命令就可以实现整个网站的内容更新，时间和日期的动态显示，以及执行shell和CGI脚本程序等复杂的功能。SSI可以称得上是那些资金短缺、时间紧张、工作量大的网站开发人员的最佳帮手。 (Server-side Includes)服务器端包含提供了一种对现有HTML文档增加动态内容的方法。apache和iis都可以通过配置支持SSI(默认Apache不开启SSI，SSI这种技术已经比较少用了)，在网页内容被返回给用户之前，服务器会执行网页内容中的SSI标签。在很多场景中，用户输入的内容可以显示在页面中，比如一个存在反射XSS漏洞的页面，如果输入的payload不是xss代码而是ssi的标签，服务器又开启了ssi支持的话就会存在SSI漏洞。 环境配置 修改nginx.conf文件来启用SSI： ssi on; ssi_silent_errors off; ssi_types text/shtml; 　　 SSI语法 显示服务器端环境变量<#echo> 1、获取本文档名称： <!–#echo var="DOCUMENT_NAME"–> 2、获取当前时间： <!–#echo

先按住右边的Alt键，然后按一下(右边)ctrl键 来源： https://www.cnblogs.com/wangwenhui/p/11810872.html

方法一：dmesg 举例如下： ###这是阿里云的云主机 [root@xxx ~]# dmesg |grep -i virtual [ 0.000000] Booting paravirtualized kernel on KVM [ 0.707486] KVM setup paravirtual spinlock [ 1.811087] input: VirtualPS/2 VMware VMMouse as /devices/platform/i8042/serio1/input/input2 [ 1.811257] input: VirtualPS/2 VMware VMMouse as /devices/platform/i8042/serio1/input/input3 [ 1.824106] systemd[1]: Detected virtualization kvm. [ 1.853583] systemd[1]: Starting Setup Virtual Console... ###这是一台物理机 [root@xxx]# dmesg |grep -i virtual Booting paravirtualized kernel on bare hardware input: Macintosh mouse button emulation as

This question already has answers here : Closed 7 years ago . Possible Duplicate: g++ undefined reference to typeinfo Undefined symbols “vtable for …” and “typeinfo for…”? I can't use my class. class Accel { public: virtual void initialize(void); virtual void measure(void); virtual void calibrate(void); virtual const int getFlightData(byte); }; class Accel_ad : public Accel { public: Accel_ad() : Accel(){} void initialize(void) {/*code code code...*/} void measure(void) {/*measure code*/} const int getFlightData(byte axis){/*getting data*/} void calibrate(void) { int findZero[FINDZERO]; int

编译器到底做了什么实现的虚函数的晚绑定呢？我们来探个究竟。 编译器对每个包含虚函数的类创建一个表（称为V TA B L E）。在V TA B LE中，编译器放置特定类的虚函数地址。在每个带有虚函数的类 中，编译器秘密地置一指针，称为v p o i n t e r（缩写为V PT R），指向这个对象的V TA B L E。通过基类指针做虚函数调用时（也就是做多态调用时），编译器静态地插入取得这个V P TR，并在V TA B L E表中查找函数地址的代码，这样就能调用正确的函数使晚捆绑发生。为每个类设置V TA B L E、初始化V PTR、为虚函数调用插入代码，所有这些都是自动发生的，所以我们不必担心这些。利用虚函数，这个对象的合适的函数就能被调用，哪怕在编译器还不知道这个对象的特定类型的情况下。（《C++编程思想》） 在任何类中不存在显示的类型信息，可对象中必须存放类信息，否则类型不可能在运行时建立。那这个类信息是什么呢？我们来看下面几个类： class no_virtual { public: void fun1() const{} int fun2() const { return a; } private: int a; } class one_virtual { public: virtual void fun1() const{} int fun2() const {

Possible Duplicate: C++ static virtual members? Can we have a virtual static method (in C++) ? I've tried to compile the following code : #include <iostream> using namespace std; class A { public: virtual static void f() {cout << "A's static method" << endl;} }; class B :public A { public: static void f() {cout << "B's static method" << endl;} }; int main() { /* some code */ return 0; } but the compiler says that : member 'f' cannot be declared both virtual and static so I guess the answer is no , but why ? thanks , Ron No. static on a function in a class means that the function doesn't need

In CRTP to avoid dynamic polymorphism , the following solution is proposed to avoid the overhead of virtual member functions and impose a specific interface: template <class Derived> struct base { void foo() { static_cast<Derived *>(this)->foo(); }; }; struct my_type : base<my_type> { void foo() {}; // required to compile. < Don't see why }; struct your_type : base<your_type> { void foo() {}; // required to compile. < Don't see why }; However it seems that the derived class does not require a definition to compile as it inherits one (the code compiles fine without defining a my_type::foo). In

I'm in need of some code which fakes the actual file system to a fake one. So, when I start it it converts /home/user/Documents/fake_fs to / , so every Dir or File call goes to that directory. An example: I want to make a file on /some_file , so I use: File.open('/some_file', 'w') do |f| f.puts 'something on this file' end And it would write it on /home/user/Documents/fake_fs/some_file instead of /some_file . Is there any way of doing this? Thanks! You've got two options: Option 1 - Use a Gem to Fake it out FakeFS will do exactly what you want, with the caveat that some file system operations