protected

Inheritance of private and protected methods in Python

由 寵の児 提交于 2019-11-26 14:23:52
I know, there are no 'real' private/protected methods in Python. This approach isn't meant to hide anything; I just want to understand what Python does. class Parent(object): def _protected(self): pass def __private(self): pass class Child(Parent): def foo(self): self._protected() # This works def bar(self): self.__private() # This doesn't work, I get a AttributeError: # 'Child' object has no attribute '_Child__private' So, does this behaviour mean, that 'protected' methods will be inherited but 'private' won't at all? Or did I miss anything? Python has no privacy model, there are no access

Objective-C: How to change the class of an object at runtime?

由 左心房为你撑大大i 提交于 2019-11-26 14:01:38
问题 I tried to answer Using a UITableView subclass with a UITableViewController with ISA Switching like so: self.tableView->isa = [MyTableView class]; But, I get the compile error: Instance variable 'isa' is protected. Is there a way to get around this? And, if so, is it safe to do? I'm asking because @AmberStar's answer to that question seems slightly flawed. (See my comment.) 回答1: If your tableview class provides ANY storage this will break. I would not recommend the path you're going down. But

Protected member access from different packages in java - a curiosity [duplicate]

由 牧云@^-^@ 提交于 2019-11-26 13:54:20
问题 This question already has an answer here: Understanding java's protected modifier 6 answers package packageOne; public class Base { protected void display(){ System.out.println("in Base"); } } package packageTwo; public class Derived extends packageOne.Base { public void show(){ new Base().display(); //this is not working throws compilation error that display() from the type Base is not visible new Derived().display(); //is working display(); //is working } } The two packages are in two

Isn't “package private” member access synonymous with the default (no-modifier) access?

由 二次信任 提交于 2019-11-26 12:24:00
问题 I am a little confused over the term \"package private\" that some of the documentation uses, along with the usage of \"default access.\" Aren\'t package-private and default access both synonymous with protected? 回答1: Yes, it's almost the same. The protected modifier specifies that the member can only be accessed within its own package (as with package-private) and, in addition , by a subclass of its class in another package. 回答2: The "default" access modifier (the one where none of them are

Why do we actually need Private or Protected inheritance in C++?

由 被刻印的时光 ゝ 提交于 2019-11-26 12:18:06
问题 In C++, I can\'t think of a case in which I would like to inherit private/protected from a base class: class Base; class Derived1 : private Base; class Derived2 : protected Base; Is it really useful? 回答1: It is useful when you want to have access to some members of the base class, but without exposing them in your class interface. Private inheritance can also be seen as some kind of composition: the C++ faq-lite gives the following example to illustrate this statement class Engine { public:

Protected access modifier in Java

由 我的未来我决定 提交于 2019-11-26 11:16:23
问题 I am having a little trouble understanding the protected access modifier in java (or the design behind it). I thought it meant package access and access through objects that inherit the class containing an abstract member. I wrote the following sample code. I see that the commented out line produces a compilation error if uncommented. Why can I access pro through a Second object in Second but not through a First object in Second? package first; public class First { protected void pro(){

Why can't I have protected interface members?

由 自作多情 提交于 2019-11-26 10:25:42
问题 What is the argument against declaring protected-access members on interfaces? This, for example, is invalid: public interface IOrange { public OrangePeel Peel { get; } protected OrangePips Seeds { get; } } In this example, the interface IOrange would guarantee that implementors at least provide an OrangePips instance to their inheritors. If the implementor wanted to, they could expand the scope to full public : public class NavelOrange : IOrange { public OrangePeel Peel { get { return new

Why we should not use protected static in java

由 不打扰是莪最后的温柔 提交于 2019-11-26 10:12:48
问题 I was going through this question Is there a way to override class variables in Java? The first comment with 36 upvotes was: If you ever see a protected static , run. Can anyone explain why is a protected static frowned upon? 回答1: It's more a stylistic thing than a direct problem. It suggests that you haven't properly thought through what is going on with the class. Think about what static means: This variable exists at class level, it does not exist separately for each instance and it does

Why can't my subclass access a protected variable of its superclass, when it's in a different package?

由 孤人 提交于 2019-11-26 09:39:16
问题 I have an abstract class, relation in package database.relation and a subclass of it, Join , in package database.operations . relation has a protected member named mStructure . In Join : public Join(final Relation relLeft, final Relation relRight) { super(); mRelLeft = relLeft; mRelRight = relRight; mStructure = new LinkedList<Header>(); this.copyStructure(mRelLeft.mStructure); for (final Header header :mRelRight.mStructure) { if (!mStructure.contains(header)) { mStructure.add(header); } } }

Why can&#39;t I access C# protected members except like this?

由 半城伤御伤魂 提交于 2019-11-26 06:39:50
问题 This code: abstract class C { protected abstract void F(D d); } class D : C { protected override void F(D d) { } void G(C c) { c.F(this); } } Generates this error: Cannot access protected member \'C.F(D)\' via a qualifier of type \'C\'; the qualifier must be of type \'D\' (or derived from it) What in the world were they thinking? (Would altering that rule break something?) And is there a way around that aside from making F public? Edit: I now get the reason for why this is (Thanks Greg) but I
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