overriding

问题 Possible Duplicate: C++ overload resolution I ran into a problem where after my class overrode a function of its base class, all of the overloaded versions of the functions were then hidden. Is this by design or am I just doing something wrong? Ex. class foo { public: foo(void); ~foo(void); virtual void a(int); virtual void a(double); }; class bar : public foo { public: bar(void); ~bar(void); void a(int); }; the following would then give a compile error saying there is no a(double) function

问题 I am using PhoneGap 1.5.0, jQuery 1.7.1 and jQuery mobile 1.0.1 and trying to override the backbutton in Android as stated here or here. document.addEventListener(\"deviceready\", onDeviceReady, false); // PhoneGap loaded function onDeviceReady() { console.log(\"PhoneGap Ready!\"); // waiting for button document.addEventListener(\"backbutton\", handleBackButton, false); } // handle the back button function handleBackButton() { console.log(\"Back Button Pressed!\"); navigator.app.exitApp(); }

问题 I was looking at this blog post and had following questions: Why do we need the new keyword, is it just to specify that a base class method is being hidden. I mean, why do we need it? If we don\'t use the override keyword, aren\'t we hiding the base class method? Why is the default in C# to hide and not override? Why have the designers implemented it this way? 回答1: Good questions. Let me re-state them. Why is it legal to hide a method with another method at all? Let me answer that question

问题 Let\'s say I have three classes A, B and C. B extends A C extends B All have a public void foo() method defined. Now from C\'s foo() method I want to invoke A\'s foo() method (NOT its parent B\'s method but the super super class A\'s method). I tried super.super.foo(); , but it\'s invalid syntax. How can I achieve this? 回答1: You can't even use reflection. Something like Class superSuperClass = this.getClass().getSuperclass().getSuperclass(); superSuperClass.getMethod("foo").invoke(this);

问题 Is it possible to call the base method from a prototype method in JavaScript if it\'s been overridden? MyClass = function(name){ this.name = name; this.do = function() { //do somthing } }; MyClass.prototype.do = function() { if (this.name === \'something\') { //do something new } else { //CALL BASE METHOD } }; 回答1: I did not understand what exactly you're trying to do, but normally implementing object-specific behaviour is done along these lines: function MyClass(name) { this.name = name; }

问题 As far as I understand, the introduction of override keyword in C++11 is nothing more than a check to make sure that the function being implemented is the override ing of a virtual function in the base class. Is that it? 回答1: That's indeed the idea. The point is that you are explicit about what you mean, so that an otherwise silent error can be diagnosed: struct Base { virtual int foo() const; }; struct Derived : Base { virtual int foo() // whoops! { // ... } }; The above code compiles, but

问题 I have two classes. public class A { public Object method() {...} } public class B extends A { @Override public Object method() {...} } I have an instance of B. How do I call A.method() from b? Basically, the same effect as calling super.method() from B. B b = new B(); Class<?> superclass = b.getClass().getSuperclass(); Method method = superclass.getMethod(\"method\", ArrayUtils.EMPTY_CLASS_ARRAY); Object value = method.invoke(obj, ArrayUtils.EMPTY_OBJECT_ARRAY); But the above code will still

问题 Is it possible to override onBackPressed() for only one activity ? On back button click I want to call a dialog on a specific Activity, but in all other activities i want it to work as it worked before (going to previous activities). EDITED Thank you everyone for your answers, I already had everything like you told me, but my problem was that when i was clicking back button on another Activity, I was going to my previous Activity (The one where i had back button Overridden) and i thought that

问题 Is there way to override return types in C#? If so how, and if not why and what is a recommended way of doing it? My case is that I have an interface with an abstract base class and descendants of that. I would like to do this (ok not really, but as an example!) : public interface Animal { Poo Excrement { get; } } public class AnimalBase { public virtual Poo Excrement { get { return new Poo(); } } } public class Dog { // No override, just return normal poo like normal animal } public class

问题 Say I have classes Foo and Bar set up like this: class Foo { public: int x; virtual void printStuff() { std::cout << x << std::endl; } }; class Bar : public Foo { public: int y; void printStuff() { // I would like to call Foo.printStuff() here... std::cout << y << std::endl; } }; As annotated in the code, I\'d like to be able to call the base class\'s function that I\'m overriding. In Java there\'s the super.funcname() syntax. Is this possible in C++? 回答1: The C++ syntax is like this: class