protected

问题 Simple as that, can we emulate the "protected" visibility in Javascript somehow? 回答1: Do this: /* Note: Do not break/touch this object */ ...code... Or a bit of google found this on the first page: http://blog.blanquera.com/2009/03/javascript-protected-methods-and.html 回答2: Sure you can. Here's another example. 回答3: What could that possibly mean? You don't have classes . I suppose you could analyze caller to determine whether it meets some set of criteria for being permitted to call a method.

问题 I've got some doubts regarding protected identifier. In the first chapter of Sun Certified Java Programmer Study Guide by K.Sierra I found the following information: "Once the subclass-outside-the-package inherits the protected member, that member (as inherited by the subclass) becomes private to any code outside the subclass, with the exception of subclasses of the subclass." I provided sample code which reflects the above statement and it is absolutely clear to me. // Parent class package

问题 This question already has answers here : Understanding java's protected modifier (6 answers) Closed 4 years ago . I am in process of learning the Java access modifiers. For that, I have created a class Machine : package udemy.beginner.interfaces; public class Machine { public String name; private int id; protected String description; String serialNumber; public static int count; public Machine(){ name = "Machine"; count++; description = "Hello"; } } Then, in another package , I have created a

问题 I have a protected function that is defined within a certain class. I want to be able to call this protected function outside of the class within another function. Is this possible and if so how may I achieve it class cExample{ protected function funExample(){ //functional code goes here return $someVar }//end of function }//end of class function outsideFunction(){ //Calls funExample(); } 回答1: That's the point of OOP - encapsulation: Private Only can be used inside the class. Not inherited by

问题 I would like to pass the userid and password in the itms-services link so that the protected plist can be accessed. To clarify, in the following link, the plist cannot be accessed directly as the access requires the userid and password to be entered so that plist is accessible. <a href="itms-services://?action=download-manifest&url=http://example.com/app.plist"> Currently the above link gives an error cannot connect to example.com 回答1: I Was installing the IPA and PLIST on a Windows IIS

问题 In this answer to the question " Why can't my object access protected members of another object defined in common base class? ", one can read: You can only access protected members from your own base class instance. Either I don't get it correctly or the following MCVE (live on coliru) proves it wrong: struct Base { void f(); protected: int prot; }; struct Derived : Base { void g(); private: int priv; }; void Base::f() { Base b; b.prot = prot; (void) b; } void Derived::g() { { Derived d;

问题 I have a class like this : class Foo { public: Foo() { for(int i = 0; i < 10; ++i) v.push_back(i); }; const vector<double>& V() const {return v;}; protected: vector<double>& V() {return v;}; private: vector<double> v; }; And then a piece of code like this : Foo foo; for(int i = 0; i < (int) foo.V().size(); ++i) cout << foo.V().at(i) << endl; However, the latter raises a compilation error saying the V() call is a protected method while i am just trying to read from it, not modify it. I have

问题 #include <iostream> class A { protected: void foo() {} }; class B : public A { public: void bar() { std::cout << (&A::foo) << std::endl; } }; int main() { B b; b.bar(); } Here I am trying to get address of protected member function of base class. I am getting this error. main.cpp: In member function ‘void B::bar()’: main.cpp:5: error: ‘void A::foo()’ is protected main.cpp:13: error: within this context make: *** [all] Error 1 Changing foo to public works. Also printing &B::foo works. Can you

问题 #include <iostream> class A { protected: void foo() {} }; class B : public A { public: void bar() { std::cout << (&A::foo) << std::endl; } }; int main() { B b; b.bar(); } Here I am trying to get address of protected member function of base class. I am getting this error. main.cpp: In member function ‘void B::bar()’: main.cpp:5: error: ‘void A::foo()’ is protected main.cpp:13: error: within this context make: *** [all] Error 1 Changing foo to public works. Also printing &B::foo works. Can you

问题 From [class.access]/7 we have the following sentence: Similarly, the use of A::B as a base-specifier is well-formed because D is derived from A , so checking of base-specifier s must be deferred until the entire base-specifier-list has been seen. class A { protected: struct B { }; }; struct D: A::B, A { }; See live example with clang. As a matter of fact, clang also complains about this snippet, where no deferment is necessary. class A { protected: struct B { }; }; struct D: A, A::B { }; Why