copy-constructor

问题 How to write copy constructor for a template class. So that if the template parameter is another user defined class it's copy constructor is also get called. Following is my class template <typename _TyV> class Vertex { public: Vertex(_TyV in) : m_Label(in){ } ~Vertex() { } bool operator < ( const Vertex & right) const { return m_Label < right.m_Label; } bool operator == ( const Vertex & right ) const { return m_Label == right.m_Label; } friend std::ostream& operator << (std::ostream& os,

问题 Is there a better way to implement copy construcor for matlab for a handle derived class other than adding a constructor with one input and explicitly copying its properties? obj.property1 = from.property1; obj.property2 = from.property2; etc. Thanks, Dani 回答1: If you want a quick-and-dirty solution that assumes all properties can be copied, take a look at the PROPERTIES function. Here's an example of a class that automatically copies all properties: classdef Foo < handle properties a = 1;

问题 Today in university I was recommended by a professor that I'd check for (this != &copy) in the copy constructor, similarly to how you should do it when overloading operator= . However I questioned that because I can't think of any situation where this would ever be equal to the argument when constructing an object. He admitted that I made a good point. So, my question is, does it make sense to perform this checking, or is it impossible that this would screw up? Edit : I guess I was right, but

问题 Today in university I was recommended by a professor that I'd check for (this != &copy) in the copy constructor, similarly to how you should do it when overloading operator= . However I questioned that because I can't think of any situation where this would ever be equal to the argument when constructing an object. He admitted that I made a good point. So, my question is, does it make sense to perform this checking, or is it impossible that this would screw up? Edit : I guess I was right, but

问题 Today in university I was recommended by a professor that I'd check for (this != &copy) in the copy constructor, similarly to how you should do it when overloading operator= . However I questioned that because I can't think of any situation where this would ever be equal to the argument when constructing an object. He admitted that I made a good point. So, my question is, does it make sense to perform this checking, or is it impossible that this would screw up? Edit : I guess I was right, but

问题 Does a templated constructor (such as the following) override the implicit copy constructor? template <class T> struct Foo { T data; // ... template <class U> Foo(const Foo<U> &other) : data((T)doSomethingWith(other.data)) {} // ... }; If so, does it still override it if other is passed by value rather than constant reference? If so, is there any way around this without explicitly defining a copy constructor? 回答1: No, that is not a copy constructor. Section 12.8 ( [class.copy] ) of the

问题 What are pros and cons of duplication an object instance with constructor or instance function? Example A: type TMyObject = class strict private FField: integer; public constructor Create(srcObj: TMyObject); overload; //alternatively: //constructor CreateFrom(srcObj: TMyObject); property Field: integer read FField; end; constructor TMyObject.Create(srcObj: TMyObject); begin inherited Create; FField := srcObj.Field; end; Example B: type TMyObject = class strict private FField: integer; public

问题 Here we can read that no copy construct and copy assignment operator evaluable. But here we can read that qRegisterMetaType and Q_DECLARE_METATYPE have to have public default constructor, public copy constructor and public destructor. The question is: who is telling a lie? Or I did not understand it correctly? 回答1: Everything is true: 1. QObject can't be copied and all its descendants can't be copied also. 2. Q_DECLARE_METATYPE accepts objects with public constructor, copy constructor and

问题 This question already has answers here : Why is the copy constructor not called? (4 answers) Closed 5 years ago . I have an understanding that copy constructor will be called when an object is created from an already existing object and also when a function returns object by value. So why in the below code the copy constructor is not called but the default constructor? class A { public: A() { cout << "A" << endl; } A(A &) { cout << "A&" << endl; } A(const A &) { cout << "const A&" << endl; }

问题 Here is the code I use. I'd like to know if what I did is correct and safe. Normally it compiles and the tests I did are successful. But as it is the first time that I use dynamic_cast and static_cast , I'd like to be sure that I didn't miss anything. Can you please check : that I don't forget to delete any pointer (I don't really understand what my clone method does...) if I use correctly dynamic_cast and static_cast that there is not a more efficient of doing it But my main concern is that