virtual

navigation property should be virtual - not required in ef core?

由 本小妞迷上赌 提交于 2019-11-26 09:01:37
问题 As I remember in EF navigation property should be virtual: public class Blog { public int BlogId { get; set; } public string Name { get; set; } public string Url { get; set; } public string Tags { get; set; } public virtual ICollection<Post> Posts { get; set; } } But I look at EF Core and don\'t see it as virtual: public class Student { public int ID { get; set; } public string LastName { get; set; } public string FirstMidName { get; set; } public DateTime EnrollmentDate { get; set; } public

C++ What is the purpose of casting to void? [duplicate]

由 蹲街弑〆低调 提交于 2019-11-26 08:30:32
问题 This question already has answers here : Closed 9 years ago . Possible Duplicate: casting unused return values to void I read some source code, and in it many virtual functions in the interface classes are declared and default-implemented as such: virtual bool FunctionName(TypeName* pointer) { (void)pointer; return true; } May I ask what is the purpose of casting the pointer to void in the default implementation? 回答1: Multiple purposes depending on what you cast Marking your intention to the

Can you write virtual functions / methods in Java?

由 北城以北 提交于 2019-11-26 06:04:46
问题 Is it possible to write virtual methods in Java, as one would do in C++? Or, is there a proper Java approach which you can implement that produces similar behavior? Could I please have some examples? 回答1: From wikipedia In Java , all non-static methods are by default " virtual functions. " Only methods marked with the keyword final , which cannot be overridden, along with private methods , which are not inherited, are non-virtual . 回答2: Can you write virtual functions in Java? Yes. In fact,

C++ 虚析构函数

由 牧云@^-^@ 提交于 2019-11-26 06:03:27
C++ 虚析构函数 flyfish 2015-3-12 Effective C++ 条款07 为多态基类声明virtual析构函数 Declare destructors virtual in polymorphic base classes. 1 polymorphic(带多态性质的)base classes应该声明一个virtual析构函数。如果class带有任何virtual函数，它就应该拥有一个virtual析构函数。 2 Classes的设计目的如果不是作为base classes使用，或不是为了具备多态性(polymorphically)，就不应该声明virtual析构函数。 理由： C++中含有纯虚函数的类称为抽象类，它不能生成对象.只能用作一个基类，由派生类提供 纯虚函数 的实现。 如下 parent 是一个抽象基类，由派生类实现function. class parent { parent(); ~parent(); virtual void function()=0; } class child: public parent { child(); ~child(); void function();//抽象基类的函数实现 } 调用 parent* p=new child(); delete p; 这时候 delete 操作只调用~parent();

Making operator<< virtual?

由 ⅰ亾dé卋堺 提交于 2019-11-26 04:44:16
问题 I need to use a virtual << operator. However, when I try to write: virtual friend ostream & operator<<(ostream& os,const Advertising& add); I get the compiler error Error 1 error C2575: \'operator <<\' : only member functions and bases can be virtual How can I turn this operator virtual? 回答1: The problem with this setup is that the operator<< you defined above is a free function, which can't be virtual (it has no receiver object). In order to make the function virtual, it must be defined as a

Why use &#39;virtual&#39; for class properties in Entity Framework model definitions?

由 房东的猫 提交于 2019-11-26 02:39:57
问题 In the following blog: http://weblogs.asp.net/scottgu/archive/2010/07/16/code-first-development-with-entity-framework-4.aspx The blog contains the following code sample: public class Dinner { public int DinnerID { get; set; } public string Title { get; set; } public DateTime EventDate { get; set; } public string Address { get; set; } public string HostedBy { get; set; } public virtual ICollection<RSVP> RSVPs { get; set; } } public class RSVP { public int RsvpID { get; set; } public int

virtual assignment operator C++

由 孤街醉人 提交于 2019-11-26 02:29:41
问题 Assignment Operator in C++ can be made virtual. Why is it required? Can we make other operators virtual too? 回答1: The assignment operator is not required to be made virtual. The discussion below is about operator= , but it also applies to any operator overloading that takes in the type in question, and any function that takes in the type in question. The below discussion shows that the virtual keyword does not know about a parameter's inheritance in regards to finding a matching function

C++ virtual function from constructor [duplicate]

由 浪子不回头ぞ 提交于 2019-11-26 01:29:53
问题 This question already has an answer here: Calling virtual functions inside constructors 13 answers Why the following example prints \"0\" and what must change for it to print \"1\" as I expected ? #include <iostream> struct base { virtual const int value() const { return 0; } base() { std::cout << value() << std::endl; } virtual ~base() {} }; struct derived : public base { virtual const int value() const { return 1; } }; int main(void) { derived example; } 回答1: Because base is constructed

Accessing class members on a NULL pointer

由 倖福魔咒の 提交于 2019-11-26 01:29:11
问题 I was experimenting with C++ and found the below code as very strange. class Foo{ public: virtual void say_virtual_hi(){ std::cout << \"Virtual Hi\"; } void say_hi() { std::cout << \"Hi\"; } }; int main(int argc, char** argv) { Foo* foo = 0; foo->say_hi(); // works well foo->say_virtual_hi(); // will crash the app return 0; } I know that the virtual method call crashes because it requires a vtable lookup and can only work with valid objects. I have the following questions How does the non

Virtual/pure virtual explained

由 孤人 提交于 2019-11-25 23:59:47
问题 What exactly does it mean if a function is defined as virtual and is that the same as pure virtual? 回答1: From Wikipedia's Virtual function ... In object-oriented programming, in languages such as C++, and Object Pascal, a virtual function or virtual method is an inheritable and overridable function or method for which dynamic dispatch is facilitated. This concept is an important part of the (runtime) polymorphism portion of object-oriented programming (OOP). In short, a virtual function
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