overriding

Overriding a method contract in an extended interface that uses generics (Java)?

余生长醉 提交于 2019-12-06 01:38:55
I am attempting to override a method declaration within an interface that extends another interface. Both of these interfaces use generics. According to the Java tutorials, this should be possible, but the example does not use generics. When I try to implement it, the compiler shows the following error (I've replaced names because some of the code is not my own.): myMethod(T) in InterfaceExtended clashes with myMethod(T) in Interface; both methods have the same erasure, but neither overrides the other. Code looks like this: public interface Interface<T> { public void myMethod(T x); } public

ExtJS 'datefield' validation override

我只是一个虾纸丫 提交于 2019-12-06 01:32:00
I needed a datecolumn with some custom behaviour, specifically I needed to be able to enter in a date or and age in the same field (with an age staying rendered as an age and a date staying rendered as a date) Eg. Entering "23" will leave "23" in the field as a valid value, or entering "22/1/88" will leave "22/1/88" as valid value. So I tried having a datecolumn with the editor defined like this (note the validate override): editor: { xtype: 'datefield', format: 'd/m/Y', validate: function(){ if(!this.value.match(SOME_REGEX){ if(!this.value.match(SOME_REGEX){ return false; } } return true; } }

Can a base class determine if a derived class has overridden a virtual member?

99封情书 提交于 2019-12-06 01:10:14
问题 Here's a simplified version of my class: public abstract class Task { private static object LockObject = new object(); protected virtual void UpdateSharedData() { } protected virtual void UpdateNonSharedData() { } public void Method() { lock(LockObject) { UpdateSharedData(); } UpdateNonSharedData(); } } I'm trying to hide the locking code from derived classes. But I only want to obtain the lock if the derived class overrides UpdateSharedData; if it doesn't, I don't want the method to block

Inherit from jQuery UI dialog and call overridden method

青春壹個敷衍的年華 提交于 2019-12-06 00:01:57
问题 The simple code below describes my question (at least I hopse so): $.widget("ui.mydialog", $.ui.dialog, { _create: function() { // How to call _create method of dialog? } }); I tried to call $.ui.dialog.prototype._create() from within the above create method, but get the below error in Firebug: this.element is undefined this.originalTitle = this.element.attr('title'); jquery...5667348 (line 5864) How else can I call that "super" method? jQuery UI version 1.8.8 回答1: I guess I just found a

C++ Overriding Methods

谁都会走 提交于 2019-12-05 23:37:52
问题 I can't figure out what is up with this. I have a Scene class that has a vector of Entities and allows you to add and get Entities from the scene: class Scene { private: // -- PRIVATE DATA ------ vector<Entity> entityList; public: // -- STRUCTORS --------- Scene(); // -- PUBLIC METHODS ---- void addEntity(Entity); // Add entity to list Entity getEntity(int); // Get entity from list int entityCount(); }; My Entity class is as follows (output is for testing): class Entity { public: virtual void

Why if static method don't involve in polymorphism(late binding) I see error that static method cannot be overridden

余生颓废 提交于 2019-12-05 23:03:32
问题 please consider following code: class A{ public static void m(Number n){ System.out.println("Number A"); }; } class B extends A{ public static int m(Number n){ System.out.println("Number B"); return 1; }; } output: java: m(java.lang.Number) in inheritanceTest.B cannot override m(java.lang.Number) in inheritanceTest.A return type int is not compatible with void I know that static methods doen't involve in polymorphism hence I infer that overriding is impossible for my code. This compiler

What is the reason behind Dynamic Method Resolution in a staticlly typed language like Java

半世苍凉 提交于 2019-12-05 21:48:37
I'm a bit confused to the, I guess concept, of the Dynamic/Static types of a reference variable and Dynamic Method Resolution in Java. Consider: public class Types { @Override public boolean equals(Object obj){ System.out.println("in class Types equals()"); return false;//Shut-up compiler! } public static void main(String[] args){ Object typ = new Types(); typ.equals("Hi");//can do this as String is a subclass of Object } } First: the reference variable typ is of type Types, isn't it?! So what is the reason behind typ having a Static Type Object and a Dynamic Type Types for method overriding?

Why should we override a method? [duplicate]

只愿长相守 提交于 2019-12-05 21:44:08
This question already has answers here : Why is method overloading and overriding needed in java? [duplicate] (3 answers) Closed 5 years ago . Recently I was asked this question "why should one override a method? " I replied, if I have a class with 10 methods and I want to use all of its functionality except one method, then I will override that method to have my own functionality. Then the interviewer replied in that case why cant we write a new method with a different name and use that method instead. Yes this is also right. Now I am confused. What is the real objective in overriding a

java override method invocation

我只是一个虾纸丫 提交于 2019-12-05 21:12:46
I have a super class: public class SuperClass { public void dosomething() { firstMethod(); secondMethod(); } public void firstMethod() { System.out.println("Super first method"); } public void secondMethod() { System.out.println("Super second method"); } } A sub class: public class SubClass extends SuperClass { public void dosomething() { super.dosomething(); } public void firstMethod() { System.out.println("Sub first method"); } public void secondMethod() { System.out.println("Sub second method"); } } A test class: public static void main(String[] args) { SubClass sub = new SubClass(); sub

Overriding a portion of a google.com anonymous function

。_饼干妹妹 提交于 2019-12-05 20:51:55
问题 If a javascript function is declared anonymously is there any way to override it or portions of it? I am attempting to stop google.com's instant search from hijacking the up and down arrow keys to move through your search rankings. I have identified what I believe is the problematic section of code. Keycodes 38 and 40 are for the down and up keys. if (b == 40) aa(f); else if (b == 38) aa(j); else if (b == 37 || b == 39) if (!ca(b == 39)) return f; a.preventDefault && a.preventDefault();