I understand class object. std::map<int, std::unique_ptr<MySet>> must be the owner of the indicators on MySet facilities, and therefore, when copying, it is necessary to create copies of the objects stored in the std::forward_list<std::shared_ptr<MySet>>For these purposes, either a normal cycle or standard algorithms, such as std::transform♪Below is a demonstration programme that shows how to use algorithms std::transform to copy the elements from one container to another. To illustrate the challenges of class designers and designers used as a basic element, I have included a consolation.#include <iostream> #include <memory> #include <forward_list> #include <map> #include <algorithm> #include <iterator> #include <utility> struct A { A() { std::cout << "A::A()" << std::endl; } A( const A & ) { std::cout << "A::A( const A & )" << std::endl; } ~A() { std::cout << "A::~A" << std::endl; } A & operator =( const A & ) { std::cout << "A::operator =( const A & )" << std::endl; return *this; } }; int main() { std::forward_list<std::shared_ptr<A>> lst; lst.push_front( std::shared_ptr<A>( new A ) ); std::map&lt;int, std::unique_ptr&lt;A&gt;&gt; m; std::transform( lst.begin(), lst.end(), std::inserter( m, m.end() ), []( auto p ) { return std::make_pair( 1, std::unique_ptr&lt;A&gt;( new A( *p.get() ) ) ); } ); return 0; } The withdrawal will be:A::A() A::A( const A & ) A::~A A::~A

There's a misunderstanding of the term. https://msdn.microsoft.com/ru-ru/library/x9afc042.aspx ♪ I'm afraid you should start studying the PLO framework. https://ru.wikipedia.org/wiki/%D0%9E%D0%B1%D1%8A%D0%B5%D0%BA%D1%82%D0%BD%D0%BE-%D0%BE%D1%80%D0%B8%D0%B5%D0%BD%D1%82%D0%B8%D1%80%D0%BE%D0%B2%D0%B0%D0%BD%D0%BD%D0%BE%D0%B5_%D0%BF%D1%80%D0%BE%D0%B3%D1%80%D0%B0%D0%BC%D0%BC%D0%B8%D1%80%D0%BE%D0%B2%D0%B0%D0%BD%D0%B8%D0%B5 later https://msdn.microsoft.com/ru-ru/library/67ef8sbd.aspx ♪Now let's try to deal with your question. What do you mean? If you still want to build a hierarchy of classes, you want to do it wrong. It's obvious that if the Vulcan class is a clarification of the Automobile class, it's not fair to say. I mean, the program doesn't care, but from a human perspective, it just doesn't make sense. How is the hierarchy of the comas? Just like life! For example, you have a class called Automobile, a production class from which Moscow is. Moscow inherits all of his attributes from the Automobile, but it also has its own, unique (I don't know, the method of getting gasoline AI-76 in the salon). Moscow has another class, Moskowicz-412, which is also Moscow (and consequently a vehicle), but has a number of additions, such as a fashion design and a richer market.So class hierarchy is a tree. clarifications♪ The notion of a wolf, as you hope, you know, can't clarify the concept of an automobile. Read the details of the classes on the references I made above.But I may have misunderstood you, and you need some kind of production log you see as a tree? For example, there are specific wolves and flaws that have time of production, which in turn affects the production of the engine that affects the time of the vehicle production?There are possible implementation options. For example, you might have a product class:public class Product { public float ProductionTime; public string Worker; } On the basis of the product, every detail can be described, as well as a car. However, please note that all these classes will not inherit from the car (e.g., they will all be derivative of the product.At the same time, the engine class will contain the Vulcan and Porsche fields and have its method of calculating the production vermin.Automobile class is to maintain the fields of Engineer, Carburator and Raul, etc.I don't bring the code because, frankly, I don't see much of that. Don't be insulted, but in my opinion, it's early for you to do so, see the beginning of the answer.

When the designer &apos; s body is performed, all members of the facility have already been established. That's why it's in this designer.object::object() { this->mass = 1; } in the proposal this->mass = 1; Appropriation operator, which cannot be used with constant objects, is used.If you need to determine the value of a constant data member on the basis of some calculations, use some form of initiator, such as a statistical function, a class member.object::object( some_argument ): mass( some_function( some_argument ) ) { } Arguments of the function should not necessarily be those of the designer. Statistic members of the class data may be used, not static members of the data that have already been initiated on the initialization list (the class ad language should be preceded or objects that are in the field of the design definition.The following is an example:#include <iostream> class Object { public: Object( double x ) : x ( 2 * x ), mass( init( this->x ) ) { } double get_mass() const { return mass; } private: double x; const double mass; static double init( double x ) { return x < 0 ? -x / 2 : x; } }; int main() { Object obj( 10 ); std::cout &lt;&lt; obj.get_mass() &lt;&lt; std::endl; return 0; }

There's an old-fashioned technology - x-macro. But it's very dark magic. Here's a setup that's actually doing everything, work it out as necessary.#define X_AIR_DATACLASS \ X(int, Crew) \ X(int, SeatingCapacity) \ X(int, Payload) \ X(int, Range) \ X(int, TopSpeed) \ X(int, CargoCapacity) \ X(int, FuelCapacity) \ X(int, Engines) \ X(int, Altitude) \ X(double, mach) \ X(double, Wingspan) typedef struct { #define X(type, name) type name; X_AIR_DATACLASS #undef X }Public_Airplane; void print (Public_Airplane x) { #define X(type, name) << #name": " << x.name << ", " std::cout << "(" X_AIR_DATACLASS << ")"; #undef X } bool compare(Public_Airplane x, Public_Airplane y) { #define X(type, name) x.name == y.name && return X_AIR_DATACLASS 1; #undef X }

Can be checked classand at true perform the function. It's not possible to check. class♪ optionI don't see the difference. It'll be easier. https://stackoverflow.com/questions/10213620/how-to-check-if-an-option-is-selected http://jquery-docs.ru/traversing/is/

SaveBidAndOfferHandler a = new SaveBidAndOfferHandler(...); double Bigprice = a.BigBid;

About why you're not compiling, as you've already written, your mistakes are lynchkin' errors, so it's an unconnected file. Student.cppand not found WinMain Says you're trying to create a final app, not a console.But you have the second part of the question - how to write. You don't have any mistakes, in the sense that they'll cause the program to fail. But like you said, it's not worth writing on C++. ♪I'd start with you having a structure rather than a class. In the sense that your class does not contain a specific internal condition that should be necessarily maintained in an agreed state or some other functionality other than the installation and return of fields. So, personally, I don't see a big sin just to declare your class a structure (i.e. class with all the open members) and work directly with them:struct Student{ string name; string surname; int age; int course; string group; string* marks; }; ... Student test; test.name = "Name"; If, however, some more serious work is planned or the assignment requires that this be a class, it is better for such small single-strong functions to be inline (instrued), i.e. to define them directly in the class announcement. This will allow the compiler not to call them, but simply to install the code, which is beneficial to productivity. And I would have done all the retrieval functions so that they could be called for for stable objects. I'd pass the lines as constant references to copy less.class Student{ public: Student(); void setName(const string& s) { name = s; } string getName() const { return name; } void setSurname(const string& s) { surname = s; } string getSurname() const { return surname; } ... private: string name; string surname; int age; int course; string group; string* marks; }; Also, your grades are clearly extravagant, so I wouldn't be in town with a set of lines, but I just used a line, which, in general, is a set of symbols. evaluation - i.e. joining this line of another symbol. Besides, you'd get rid of what's called magic constants-- you wouldn't have to remember that grades 6:for(char c : mark){ cout << c <<" "; } And you're all in class. copy.what's right is that the object is self-sufficient, "all of its own." And here's the body you're just handing over as an index. External in relation to the class, the essence, and if it's changed, it works, and the state of the class. I mean, your student's score isn't a setup, it's a cover with a paper that says, "and my grades are recorded over there on a class board" where any bully can fix them... I mean, even if you're working with a set of lines, it's better to bury it in, for the propulsion.And, why don't you create another designer that would create a first-named student? I don't think the name will change.

Found English SO https://stackoverflow.com/a/10526013/5381547 to a similar question. I'll get the free translation here.As in most other languages, Ruby has some basic copies of the existence that is originally written in the language. They came down from the sky, materialized from thin mothers, appeared in a magical way.Here's a list of these magical things:Object I don't have a super class, but you can't identify an object without a super class, a straight super class. Object♪ [Remark: Super class for Object You can ask, but in the end, there will always be an object that doesn't have a super class. ]Object is a copy of the class Classwhich is a subclass Object (What does that mean, Object indirectly is a copy of the most Object)Class isclass Modulewhich is a copy ClassClass is a copy ClassNone of these things can be explained to Ruby.BasicObject♪ Object♪ Module and Class - They must all be created at the same time because they have cyclical dependence on each other.The fact that such dependence cannot be explained in Ruby does not mean that Ruby's language specification cannot say it should be. This is the task of the perpetrator: to understand how to do this. After all, Ruby's the implementing officer has access to objects that the programmer simply does not have.For example, Ruby-Executive may be created for start-up BasicObject, certifying him superclass and class equals null♪Then create Object? superclass equal BasicObject and class equal null♪Then create Module? superclass equal Object and class equal null♪Finally, ClassI'm going after him. superclass equal Module and class equal null♪Now reappoint the indicators to Class for BasicObject Object,ModuleиClassравнымиclass`.It's pretty easy to do from outside, but it's hard to imagine how to do it from inside.As soon as they' started. There is, it is perfectly possible to perform most of their function in pure Ruby. All you have to do is have a naked skeleton of these classes, because thanks to the open Ruby class, we can complement them at any time.Implementation class Class doesn't create a new class by name. ClassIt's open. existing Class Classwhich has already been given us the means of implementation.So it's very simple to introduce standard behavior. Class#new on pure Ruby:class Class def new(*args, &block) obj = allocate # another magic thing that cannot be explained in Ruby obj.initialize(*args, &block) return obj end end [Remark: In fact, initialize - private method, so you need to use it. obj.send(:initialize, *args, &block)to circumvent restrictions on access. ]By the way, Class#allocate - another magic trick. He is distributing a new empty object in the area of Ruby facilities, which cannot be carried out by Ruby itself. So... Class#allocate -- Something that should also be provided with an environment of execution.I'll add that in RubyMine's reference code.rubystubs23/class.rb() There's a comment that explains everything:# Classes, modules, and objects are interrelated. In the diagram # that follows, the vertical arrows represent inheritance, and the # parentheses metaclasses. All metaclasses are instances # of the class `Class'. # +---------+ +-... # | | | # BasicObject-----|-->(BasicObject)-------|-... # ^ | ^ | # | | | | # Object---------|----->(Object)---------|-... # ^ | ^ | # | | | | # +-------+ | +--------+ | # | | | | | | # | Module-|---------|--->(Module)-|-... # | ^ | | ^ | # | | | | | | # | Class-|---------|---->(Class)-|-... # | ^ | | ^ | # | +---+ | +----+ # | | # obj--->OtherClass---------->(OtherClass)-----------... Interpretation of the comment on the chart:Classes, objects and modules are interlinked. Figure below vertical The shooters mean inheritance and the brackets are meta-class. All meth grades being copies Class Class♪

The answer to &quot; why &quot; is simple: because https://docs.oracle.com/javase/specs/jls/se8/html/jls-5.html#jls-5.1.5 ♪Each class may be inherited from other classes (and interfaces) and the left part of the assignment may be a pre-class or an interface.More interesting, why. Look, this is it. polymorial♪ It's often that a few different things have a lot in common, similar to each other, and you want to treat them the same way. For example, you have classes that are a plane and a helicopter, and you want to check for each of them if you need repair. The hold is that you announce the general parentage class of the &quot; flying machine &quot; , make a general functionality there. Now you can.To declare the listList<>) Your aircraft, and contain both aircraft and helicoptersTo declare a function that takes place at the entrance of the aircraft and conducts the same actions for the aircraft and the helicopter (e.g. inventory).Why, right after the designer's call, declare a narrower type of variable? Options may also be several:You may wish to emphasize that you treat the object as one of the groups of similar objects. For example, you can set up a plane, but you're using it as an aircraft, and you're not interested in a plane or a helicopter.You can get access to the data you've got. http://ideone.com/F2tTRl for such classesclass Base { public String name = "Base"; } class Derived extends Base { public String name = "Derived"; } access Base.name type reference Derived You won't get it.

It's Generic types, which will define the type. public class GenericsType<T> { private T t; public T get(){ return this.t; } public void set(T t1){ this.t=t1; } public static void main(String args[]){ GenericsType<String> type = new GenericsType<>(); type.set("Pankaj"); //valid GenericsType type1 = new GenericsType(); //raw type type1.set("Pankaj"); //valid type1.set(10); //valid and autoboxing support } } Another example:public interface MyInyerface<E> { void print(E e); } public class MyClass implements MyInyerface<String> { @Override public void print(String s) { } class MyClassInteger implements MyInyerface&lt;Integer&gt;{ @Override public void print(Integer integer) { } } } Now, by specifying the type of data, only this type of method can take. It's necessary, if you need only a certain type at the entrance, and if another type has been passed, you'll get a mistake not during the compilation, but during the code analysis. Details can be read http://www.journaldev.com/1663/java-generics-example-method-class-interface and https://docs.oracle.com/javase/tutorial/extra/generics/simple.html Here. https://docs.oracle.com/javase/tutorial/java/generics/types.html still.

Multiple inheritance is a very complicated and unpleasant subject. In summary, you've got a few copies of the base class, and the compiler just doesn't understand what the most: Two decisions, depending on what you need. Either contact members of this ambiguous base class with an intermediate class or declare it a virtual base class.Option 1:class A { int a; public: A(int a):a(a){} void out() { cout << "A: "<< a << endl; } }; class B: public A { public: B(int a):A(a){} }; class C: public A { public: C(int a):A(a){} }; class D: public B, public C { public: D(int a):B(a),C(a){} void out() { B::out(); } }; ♪ D::out() We call out() of volume Awho is an ancestor B♪Option 2:class A { int a; public: A(int a):a(a){} void out() { cout << "A: "<< a << endl; } }; class B: virtual public A { public: B(int a):A(a){} }; class C: virtual public A { public: C(int a):A(a){} }; class D: public B, public C { public: D(int a):A(a),B(a),C(a){} }; int main(int argc, const char * argv[]) { D d(5); d.out(); } Now. A There's only one thing you can do without ambiguity. ♪ ♪There's a lot of thinners in the second version of the designer. D I need to call the designer. ALike... In short, read the books.

Is that possible? If so, Hashkodu needs a full-time translator from the Russian-language starting language of the POPEers, just to put the problem in simple words. I was just trying to figure it out, because the class description was from /users/4772/mrdoctorwho It's not possible to open a file (previously open, but it's not possible to keep a name and a jandle open in attributs)class Config(): def __init__(self, filename): self.file = open(filename, "w") self.filename = filename def __setattr__(self, attr, val): if not (attr in ('filename', 'file')): print attr, val self.__dict__[attr] = val

Replace the fields with properties and make them virtual in the base class. And in the heirs, redefin your taste. And, of course, change access retrofits to at least protected

You're not a method of change. But even that's not enough, there's gonna be four if we call him. All because you created a copy of Class 1 multiplied it by 5, making 5 references to one object, not five new copies.Commented on the construction of five new classes1 and wrote a new one, plus adding a couple of prints to demonstrate. Try commenting on the line, and look at the id objects. I have to admit, this is a good job interviewing, and I'm not sure if I'm going to get a start.class Class1: def __init__(self): self.a = -1 self.b = -1 self.c = -1 class Class2: def init(self): #self.x = [Class1()] * 5 self.x = [Class1() for _ in range(5)] def change(self): for i in range(5): self.x[i].a = i self.x[i].b = i self.x[i].c = i print(self.x[i]) class2 = Class2() class2.change() for i in range(5): print(class2.x[i]) print(class2.x[1].a) print(class2.x[2].c) print(class2.x[3].b)

In fact, from your example, the code doesn't understand what you're trying to achieve.However, with regard to your code, the parameters of the functions are their local variables, which will cease their lives after completion of their functions. Therefore, this functionvoid set_a(int set) { a = &set; } Index a will contain the address of an object that will no longer exist, which will lead to an uncertain conduct of the programme. You need to create a copy of the local variable in dynamic memory and a copy address for a class member. a♪ In doing so, the memory that has been given dynamically should not be forgotten.The programme may look as follows.#include <iostream> class A { protected: int *a = nullptr; public: ~A() { delete a; } void set_a(int set) { if ( a == nullptr ) a = new int; *a = set; } }; class B : public A { protected: int b; public: void show_b() { b = *(A::a); std::cout << b << std::endl; } }; int main() { B b; b.set_a( 10 ); b.show_b(); return 0; } Or you can rewrite the program as follows:#include <iostream> class A { protected: int *a = nullptr; public: ~A() { delete a; } void set_a(int set) { if ( a == nullptr ) a = new int; *a = set; } }; class B : public A { protected: int b; public: void show_b() { b = *(A::a); std::cout << b << std::endl; } }; int main() { B b; A &amp;ra = b; ra.set_a( 10 ); b.show_b(); return 0; } In this example, reference is made to class object Bbut having a type A♪ Using this reference, you can only refer to the methods declared in the classroom. A♪As for your question.How to transmit the value obtained in one class in another.this shall be done by means of class methods which return the value.:)For example,#include <iostream> class A { private: int a; public: void set_a( int set ) { a = set; } int get_a() const { return a; } }; class B { private: int b; public: void show_b( const A &a ) { b = a.get_a(); std::cout << b << std::endl; } }; int main() { A a; a.set_a( 10 ); B b; b.show_b( a ); return 0; }

One. https://ru.stackoverflow.com/questions/235352/ It is that they may run some logic, but some methods may remain unrealized. Realization is required to add heirs.So if you have a class B There will be no difference between the abstract members and the heirs. The only difference will be that you won't be able to produce copies of the class. B (this is another feature of abstract classes).

or even easier:var items = document.body.querySelectorAll('.my-class-1'); for(var i = 0; i < items.length; i++){ var item = items[i]; if(item.innerHTML === "user-2"){ if( item.className === "my-class-1" ){ item.className = "my-class-2"; } } }.my-class-2 { color: red; } <div class="my-class-1">user-1</div> <div class="my-class-1">user-2</div> <div class="my-class-1">user-2</div> <div class="my-class-1">user-1</div> <div class="my-class-1">user-2</div>

Well, at least you don't initiate. pnew->firstYou have some kind of garbage in it when you try to add a list to it. Turn it off before the cycle.But it's definitely C? You have two functions with the same name.void add2list(Neighboor **nnhead, int val); void add2list(Node **pphead, int val, int neighoods[]); It works great at C++, but not C.

We programmers should help each other.Let's start with the title. <cstring>#include <cstring> Because you're using the functions declared in this heading.The length of the line is better defined as the type size_t instead int♪class String { char* s; size_t length; //... Your designers are wrong to copy the lines. They don't copy the resulting zero symbol '\0'♪ For example,String(const char* str="") { this->length = strlen(str); this->s = new char[this->length+1]; for (int i = 0; i < length; i++) // символ нуля не входи в диапазон `[0, length)` this->s[i] = str[i]; } Otherwise, it doesn't make sense why you're giving me the memory of the this->length+1 instead this->length♪ I don't know why you're using cycles to copy lines when you already use standard C functions from the heading. <cstring>It would be better to announce the designer as follows.String( const char *str = "" ) { this->length = std::strlen( str ); this->s = new char[ this->length + 1 ]; std::strcpy( this->s, str ); } The same point is true for the copy designer.The classification of the class cannot indicate the qualified name of the member functions of the class when they are declared. Although MS VC++ This allows, but this is his own expansion of a language that does not meet the standard.String& String::operator=(String& right); ^^^^^^^^ These announcementsString String::operator+(String right); friend String& operator+=(String& left, const String& right); better redesign as followsfriend String operator + ( const String &left, const String &right ); String & operator += ( const String &right ); Function show It's better to announce.std::ostream & String::show( std::ostream &os = std::cout ) const; Reclassify the functions to reflect the above observations.For the copying operator of the attachment, it may be determined as follows:String & String::operator =( const String &right ) { if ( this != &right ) { size_t n = std::strlen( right.s ); char tmp = new char[ n + 1 ]; std::strcpy( tmp, right.s ); delete [] this-&gt;s; this-&gt;s = tmp; this-&gt;length = n; } return *this; } Similarly, you need to identify the operator.String & operator += ( const String &right ); For exampleString & String::operator +=( const String &right ) { size_t n = this->length + std::strlen( right.s ); char tmp = new char[ n + 1 ]; std::strcpy( tmp, this-&gt;s ); std::strcat( tmp, right.s ); delete [] this-&gt;s; this-&gt;s = tmp; this-&gt;length = n; return *this; } Function show simplystd::ostream & String::show( std::ostream &os ) const { return os << this->s; } As for your question.Почему у меня s3 всё равно при выводе пустая? the operator.String& operator+=(String& left, const String& right) { return left+right; } doesn't change the left argument. Moreover, there is uncertainty as to how to return the reference to a temporary facility created by the expression left+right♪ EDIT: If the operator of the operator [] which I understand you must also determine, it is possible to place a symbol in the class object '\0', you should change the definitions of the copy designer copying the assignor, operator. operator +=and friendly functions operator +replacing standard C functions strXXX ♪ memXXX♪ The following is a demonstration programme which shows how this can be done in the example of some members of the class. Other members of the class will try to identify themselves.#include <iostream> #include <cstring> class String { private: char *s; size_t length; public: String( const char *str = "" ) { this->length = std::strlen( str ); this->s = new char[ this->length + 1 ]; std::strcpy( this-&gt;s, str ); } String( const String &amp;src ) { this-&gt;length = src.length; this-&gt;s = new char[ this-&gt;length + 1 ]; std::memcpy( this-&gt;s, src.s, this-&gt;length + 1 ); } String &amp; operator =( const String &amp;src ) { if ( this !=+ &amp;src ) { char *tmp = new char[ src.length + 1 ]; std::memcpy( tmp, src.s, src.length + 1 ); delete [] this-&gt;s; this-&gt;s = tmp; this-&gt;length = src.length; } return *this; } String &amp; operator +=( const String &amp;src ) { size_t n = this-&gt;length + src.length; char *tmp = new char[ n + 1 ]; std::memcpy( tmp, this-&gt;s, this-&gt;length ); std::memcpy( tmp + this-&gt;length, src.s, src.length + 1 ); delete [] this-&gt;s; this-&gt;s = tmp; this-&gt;length = n; return *this; } ~String() { delete [] this-&gt;s; } std::ostream &amp; show( std::ostream &amp;os = std::cout ) { return os &lt;&lt; this-&gt;s; } }; int main() { String s1( "Hello world!" ); String s2; s2 = s1; s2.show() &lt;&lt; std::endl; s2 += " Glad to see you!"; s2.show() &lt;&lt; std::endl; return 0; } The programme output will be as followsHello world! Hello world! Glad to see you!

In order to insulate a partially certified code, the NET contains a restriction of access in AppDomain, which is called Sandbox (sand-box launch). AppDomain.CreateDomain( string friendlyName, Evidence securityInfo, AppDomainSetup info, PermissionSet grantSet, params StrongName[] fullTrustAssemblies); Information on how to use it, as well as a number of additional considerations and nuances, is detailed in this article on MSDN: https://msdn.microsoft.com/ru-ru/library/bb763046(v=vs.110).aspx