pure-virtual

问题 This question already has answers here : Pure virtual function with implementation (8 answers) Closed 6 days ago . Pure virtual functions (when we set = 0 ) can also have a function body. What is the use to provide a function body for pure virtual functions, if they are not going get called at all? 回答1: Your assumption that pure virtual function cannot be called is absolutely incorrect. When a function is declared pure virtual, it simply means that this function cannot get called dynamically

问题 Nearly the final step but still some strange erros.... bash-3.2$ make g++ -Wall -c -g Myworld.cc g++ -Wall -g solvePlanningProblem.o Position.o AStarNode.o PRM.o PRMNode.o World.o SingleCircleWorld.o Myworld.o RECTANGLE.o CIRCLE.o -o solvePlanningProblem Undefined symbols: "vtable for Obstacle", referenced from: Obstacle::Obstacle()in Myworld.o "typeinfo for Obstacle", referenced from: typeinfo for RECTANGLEin RECTANGLE.o typeinfo for CIRCLEin CIRCLE.o ld: symbol(s) not found collect2: ld

问题 Suppose there is this interface: class A{ public: virtual foo()=0; }; And a class B which implements this interface: class B:public A{ public: virtual foo(){} //Foo implemented by B } Finally, a class C which has classes A and B as base classes: Class C : public A, public B { }; My question is, there is a way to tell the compiler that the implementation for foo is the one from class B without doing an explicit call to B::foo() ? 回答1: As @BenVoigt pointed out in the comments, the below answer

问题 Consider an example where a method is pure virtual, takes a parameter of a templated type (injected from an outer type), and that templated type is a local type (defined in a function body). This scenario causes a compile-time error under g++. Admittedly, this is quite a corner case, but it does originate from real code. Here's a compilable, reproducible example: #include <cstdio> template<typename T> struct Outer { struct InnerBase { virtual void foo(T const&) = 0; virtual void bar(T const&)

问题 Regarding this post: For implementations that use vtable, the answer is: Yes, usually. You might think that vtable isn't required for abstract classes because the derived class will have its own vtable, but it is needed during construction: While the base class is being constructed, it sets the vtable pointer to its own vtable. Later when the derived class constructor is entered, it will use its own vtable instead. I'm assuming the answer is correct, but I don't quite get it. Why is the

问题 I recently switched back from Java and Ruby to C++, and much to my surprise I have to recompile files that use the public interface when I change the method signature of a private method, because also the private parts are in the .h file. I quickly came up with a solution that is, I guess, typical for a Java programmer: interfaces (= pure virtual base classes). For example: BananaTree.h: class Banana; class BananaTree { public: virtual Banana* getBanana(std::string const& name) = 0; static

问题 First off I apologize if there is another post out there that answers this, all the similar posts I found dealt with diamond inheritance schemes or defined functions, which this does not. In short, I'm wondering if it is possible to have one class inherit from two other classes where both child classes has a function with the same name and arguments but it is defined in one child class, and pure-virtual in another. Furthermore if I can do this, would invoking the function on the pure-virtual

问题 I have a purely virtual class defined as such: class BaseClass { protected: const int var; public: void somefun() = 0; // what I mean by a purely virtual class // stuff... }; If I don't add a constructor defined as such: BaseClass(const int & VAR) : var(VAR) {}; that I would have to subsequently use in ever derived class, my derived class can't initialize the const variable var to whichever value it wants to. Now I actually understand what's going on here. Before constructing a derived class,

问题 I have found some code that I am working on, and was wondering what the best design implementation is. If a base class defines a method as virtual, but implements an empty body as well, thus not requiring the derived classes to implement a body, should it not be made pure instead? virtual void AMethod1() {} // 1 virtual void AMethod2() {assert(false);} // 2 virtual void AMethod3() = 0; // 3 Current code. Idea1: Alerts user that this derived object has not implemented this method body. Idea2:

问题 Answer can be found here: An Excerpt from Effective C++, Third Edition, by Scott Meyers url posted by : hmjd Please read that page so you understand why it is happening. Also you know why substituting virtual void OnRelease() = 0; by: virtual void OnRelease(){}; will work but isn't the correct way to resolve. Original question R6025: pure virtual function call #include <Windows.h> // static lib //file.h class cBaseApplication { public: virtual ~cBaseApplication(){ Release(); } virtual void