c#,class,protected,derived-class
When you are calling base on the constructor, those variables don't even exist yet (the base class hasn't been constructed). Thus, you can't pass the class member to it in the base constructor call. Instead, you needed: public Plant(int itemNumber, string description, decimal price, decimal size) : base(itemNumber, description, price)...
Yes, you should call change() using a getter like getPermissions(): temp->getPermissions().change(); ...
“direct access” to an object's protected field is allowed in this particular example case, because the protected field declaration was made in a class that resides inside the same package as the "calling code". I guess Java doesn't look to see in which package the field is actually being used...
You must specify an absolute path for the .htpasswd file. Try using an absolute path from the / folder from your machine. Example: AuthType Basic AuthName "My Protected Area" AuthUserFile /var/yourpath/.htpasswd Require valid-user TIP: I created a Gist on Github to help developers creating a basic .htaccess file. You can...
c++,class,inheritance,protected
Derived can access protected members of Base on all instances of Derived only. But obj isn't an instance of Derived, it's an instance of Base - so access is forbidden. The following would compile fine, since now obj is a Derived class Derived : public Base { public: bool operator==(const...
javascript,google-apps-script,google-spreadsheet,protected,named-ranges
Change: var protections = ss.getProtections(SpreadsheetApp.ProtectionType.RANGE); To: var protections = ss.getProtections(SpreadsheetApp.ProtectionType.SHEET); That way, you'll be assured that you are getting a reference to everything that is protected, and future conflicts don't happen....
This has nothing to do with the member being protected; you'd get the exact same error with a public member. The real reason is that templates are involed. More specifically, that your class template has a base class which depends on a template parameter. This means that when parsing the...
Just have them friend any manager: class A { template <typename T> friend struct Manager; }; template<class BASE> class B: public BASE { template <typename T> friend struct Manager; }; // etc. ...
jquery,jquery-ui,widget,factory,protected
I've found the problem. It seems that shouldn't define properties in your plugin root like so: ;(function ( $, window, document, undefined ) { $.widget( "namespace.mywidgetname" , { myPrivateVar = {}, _create: function(){ this.myPrivateVar.test = 2; .... This keeps a reference for all the instances. But instead you should declare...
c++,inline,private,member,protected
The Compiler can Access everything. The restrictions are only valid for the programmer. This means there are no restrictions for the Compiler to Access any variables! At the end every variable is just translated to an address which can be accessed. So for the Compiler it is no Problem to...
c++,language-lawyer,c++14,protected,derived-class
This is simply a compiler bug. The normative text of the standard supports the example. The fact that multiple compilers have the same bug means this is part of the standard is tricky to get right. There are open bugs about this for GCC and for clang. Note that a...
java,oop,object,inheritance,protected
If class C2 extends C1, and C1 contains a public method, then the method in C2 (if overridden) must also be public; Java makes it illegal to put additional restrictions on a method's access when overriding. If C1 contains a protected method, then an overriding method in C2 may be...
c++,inheritance,protected,member-function-pointers
B is allowed to access protected members of A as long as the access is performed through an object of type B. In your example you're trying to access foo through A, and in that context it is irrelevant whether B derives from A or not. From N3337, §11.4/1 [class.protected]...
nhibernate,mapping,nhibernate-mapping,protected,mapping-by-code
As we can see the overloaded method here: PropertyContainerCustomizer.cs public void Bag<TElement>(string notVisiblePropertyOrFieldName , Action<IBagPropertiesMapper<TEntity, TElement>> collectionMapping , Action<ICollectionElementRelation<TElement>> mapping) { ... } What we have to pass as the generic template is the TElement, the one used as ICollection<TElement>. And because the defintion is: // TElement is Items protected...
As far as I could understand from the docs, you have to call something like $receivedObject->getTaskSeries()->getName() Or there is a suggestion to apply toArray or toJson to returned object - that should work....
java,android,static,protected,non-static
you are trying to change non-static member from static function. you need to make that varible static as well or need to create object of that class. lets suppose, class Test { int node = 0; static int node1 =10; } class changeNode { public static void changeNode(){ new Test().node...
A protected member can be accessed in a subclass outside the package only through inheritance. Try this instead : public class C extends A { void foo() { int b = i; } } ...
Mostly the problem is that if C++ allowed you to access the non-public members of the referent of a base class pointer directly, then you could gain easy access to the data of an object simply by deriving from a common base. Still this is a known loophole in the...
c++,class,inheritance,member,protected
The error has nothing to do with the access level of the members of class rsa. Even if you declare those members public you will still get the error. The problem is that a derived class's initialization list is run solely in the context of itself, you do not have...
c++,pointers,inheritance,protected,circular-dependency
You have a forward declaration of SceneGame, likely in entity.hpp in the form of class SceneGame;. This is enough to use it as a pointer. In player.cpp, you actually use the class and you need to know its details ( that you don't need in the header). Likely, your player.cpp...
scala,class,generics,covariance,protected
In Scala, writing class Stack[+A](elems: List[A]) also implements a default constructor. If you know Java, then in Java this would be something like class Stack<A> { private List<A> elems; public Stack<A>(List<A> elems){this.elems = elems;} } Now, you have two protected keywords in your example: protected val elems: List[A] protected (/*...*/)...
c++,arrays,class,private,protected
You can indeed make the main function a friend of a class like so: int main(int, char**); namespace N { struct C { friend int ::main(int, char**); private: int privateer = 42; }; } int main(int, char**) { ::std::cout << N::C().privateer << "\n"; } However, why not just make doSomething...
c++,templates,inheritance,protected
Since your create() function won't be able to deal with further inherited classes you can take advantage of this and not create a Button but, instead, a generic derived, protected, derived class which would have access to your protected constructor: class Component { uint32_t id; template <typename T> struct Concrete:...
c++,constructor,protected,base,derived
§11.4/1: As described earlier, access to a protected member is granted because the reference occurs in a friend or member of some class C. If the access is to form a pointer to member (5.3.1), […]. All other accesses involve a (possibly implicit) object expression (5.2.5). In this case, the...
Note some things about protected access -They are available in the package using object of class -They are available outside the package only through inheritance -You cannot create object of a class and call the `protected` method outside package on it they are only called through inheritance outside package. You...
No, it's not. You are casting B to _A which can change along the way. The fact that currently _A is identical to A is just a coincidence on which you cannot depend on. If your goal is to access a protected function, you can use a pImpl approach :...
JavaScript only provides you with references to objects (including arrays). When you return _occurrences, you return a reference to the array, so you can manipulate it. If you want to prevent that, return a copy of the array instead. return _occurrences.concat(); ...
You have your friendship backwards - a class can't declare itself to be friend of somebody else. Unlike in Java's "inner classes", a class defined within a class does not automatically have access to an instance of the class the defines it - the "inner" class is completely independent, and...
java,private,public,protected,access-specifier
Try calling finalize: new Object().finalize(); It won't compile: subclasses only have access to protected methods on other objects of the same type. If equals, toString, wait and notify were protected, we could not access them freely. In JLS terms... Let C be the class in which a protected member is...
Declare a class for accessing protected members, typecast the Parent to this class, and do not use the OnClick event, instead use Click. type TControlAccess = class(TControl); procedure TSubCom.Click; begin inherited Click; if ParentControl <> nil then TControlAccess(ParentControl).Click; end; ...
You can't access a protected superclass field in a different instance of the class. There's a good reason: you don't know whether it has the same subclass as yourself, or a completely different subclass. If it were allowed to access the protected field, you would be allowed to access the...