private/public class in namespace problem - namespaces

This is a question about what defining a class as public or private does.
Right now, I have various classes defined inside of a namespace and I only want some of those classes to be visible/usable to the outside world.
So, for example, if the classes below were the only ones in the program, I would want main.cpp to only be able to see/use the MyPublic class, not the MyPrivate class. I thought that defining the MyPrivate class as private and the MyPublic class as public would accomplish this, but the below code works and main.cpp is able to declare a MyPrivate object.
Is it possible to do this in C++?
MyPrivate.h:
namespace MyNamespace{
// only classes inside of the MyNamespace should be able
// to use this
private ref class MyPrivate{
...
};
}
MyPublic.h:
#include "MyPrivate.h"
namespace MyNamespace {
// anyone can declare this
public ref class MyPublic{
...
private:
MyNamespace::MyPrivate^ p;
...
};
}
Main.cpp:
#include "MyPublic.h"
int main(){
MyNamespace::MyPublic p_yes; // this is fine
MyNamespace::MyPrivate p_no; // don't want this to be possible
return 0;
}

private/public in this situation will affect how classes are visible outside an assembly, if you want to create a class that is "private" in the meaning that it can be used only by some other class, you can use nested clas mechanism, like this:
namespace MyNamespace {
public ref class MyPublic {
private:
ref class MyPrivate {
public:
int x;
};
MyPrivate^ p;
};
}
//Edit:
You can by the way still throw this nested class in public: section and use it like this:
MyNamespace::MyPublic::MyPrivate priv;

The private keyword means something else than you think. I limits visibility of the ref class beyond the assembly. Since your Main() method is in the same assembly, it has no trouble referencing the type name. Note that the C# language's "internal" keyword means the same thing.
I assume that you really intend for these classes to be in a separate assembly someday. As such, using private is certainly good enough. Using nested private classes can make a class inaccessible to code in the same assembly.

You public header shouldn't include private header. Forward declare private class and include header only in MyPublic.cpp. Or that's what I'd say if you used normal C++. Bastardized .Net dialect might change things.

Unfortunately, the access modifiers on a class only affect visibility outside the assembly you're building. C++ doesn't support any sort of access modifiers that apply to namespaces in the way you're describing.
A common idiom for simulating this is to put the "private" code into a detail namespace (e.g. put it in MyNamespace::detail). This is used a lot in e.g. the boost libraries. By convention, code in a detail namespace should only be used by code in the enclosing namespace (so MyNamespace::detail should only be used by code in MyNamespace), although the compiler won't enforce this for you.

Related

How to declare static field in class

I have a XAML + DirectX app and I want to add static field to my "interop" class:
[Windows::Foundation::Metadata::WebHostHidden]
public ref class Direct3DInterop sealed : public Windows::Phone::Input::Interop::IDrawingSurfaceManipulationHandler
{
public:
static int VALUE = 0;
...
};
It does not compile saying "only static const integral data members can be initialized within a class".
If I change it to const static int VALUE = 0; then it still does not compile with error "a non-value type cannot have any public data members"
What am I doing wrong?
WinRT public classes have a number of limitations to ensure they are consumable by multiple languages including C++, JavaScript, and C#. This is why you are getting error C3984. You can't have public fields and instead must use properties. You'd make it a read-only property:
property int VALUE
{
int get() { return 0; }
}
It is important to remember that properties are function calls and can't usually be optimized away, so you should consider that when designing the interfaces.
If you intend to have this class only consumable by C++, consider not using a WinRT class and instead use a simple C++ class which you managed the lifetime using std::unique_ptr or std::shared_ptr. In that case, you can of course use the public field approach as always.
The original problem you got is a general C++ language restriction not specific to WinRT. Error C2864 (you are using VS 2012 from the text you posted) is a little more general with C++11 in VS2013.

signature of public member contains native type

I am new to visual c++, I have the following code:
ref class Book sealed
{
public:
Book(std::string title,std::string author,int year);
void setTitle(std::string title);
std::string getTitle() const;
int getYear() const;
void setYear(int year);
void setAuthor(std::string author_);
std::string getAuthor() const;
private:
std::string title_;
std::string author_;
int year_;
};
When I am trying to compile it I am getting the following error:
{ctor} signature of public member contains native type. I suppose this is because I am using an the std::string and not the Platform::String, how can I fix that?
Your ref class is not marked public itself, so it appears you are only consuming this class internally (as source) from other C++, and not intending for it to be published to other WinRT consumers.
If this is the case, you can set your constructor as internal instead of public, which will be public within this component and not visible externally. And really if that's your intended usage, then it can just be a regular 'class' instead of a 'ref class'. If you do wish to use it across the WinRT boundary but you don't need the constructor, you can make it a 'public ref class' and have the constructor marked as 'internal'. Kinda depends on your scenario.
If you instead wish to make this class public and have a public constructor which is usable across the WinRT boundary (so that it can be consumed by C#/VB/JS), then you need to use WinRT types (such as Platform::String). Within your class the storage type can still be a std::string (although I recommend using std::wstring, otherwise you need to do wide-to-narrow conversions, as Platform::Strings are wide strings).
To convert between these two types, use Platform::String::Data() to get at the underlying wchar_t* which you can use to construct a std::wstring. And similarly, Platform::String has a constructor which takes a wchar_t* (which you can get from std::wstring::c_str()).
You can't hold native types in a managed reference class.
You can only hold a pointer to an unmanaged object(a pointer is just a number after all, and that's why it's allowed).

How Actionscript 3 Classes Work

I need a little help understanding how classes work in Actionscript 3. I understand you start with "package" and why and then go to import any necessary libraries, as well as then naming the class and stating if it's public/private and extends anything.
After that is what I don't understand. It seems you write "(public) function class name()
I don't understand why you do this and what goes in the curly brackets.
I've probably missed a bit of earlier reading because I've done a little reading but I can't seem to get it.
Could someone try explain it to me? Thanks.
ActionScript 3 Classes
The package statement.
Okay, so firstly like you mentioned, a class must be wrapped by a package1. This gives us the first block, where you need to define the class.
package
{
// Your class here.
}
The package statement reflects the location of the class relative to the .fla2. For example, if you have a folder "classes" within the same directory as the project .fla, then classes within that folder will need a package statement that reflects that:
package classes
{
// Your class here.
}
Defining the class.
Within a package statement, you may insert one class. Do not confuse this with the package itself, which can contain many classes - each class just needs to have its own file with the same package statement.
A class definition is made up of up to 5 parts:
The namespace. A class can be internal or public. An internal class can only be seen by classes within the same package, whereas public classes can be seen from anywhere in the project.
The class name.
A base class (optional). If a base class is defined, then your new class will act as an extension to that class, inheriting all of the qualities of the base class.
An interface to implement (optional). Interfaces are an advanced topic thus I suggest you forget about these for now until your AS3 and OOP have evolved.
If you wanted to create a class called "Person" within the package classes, then we would end up with:
package classes
{
public class Person
{
// Class qualities here.
}
}
Properties.
Classes can contain properties. Properties are defined using the var keyword. They may belong to one of a number of namespaces (including your own) and are used to hold values that belong to your class. Properties are most commonly found clustered together at the top of your class.
Our Person class may enjoy the properties height and weight:
package classes
{
public class Person
{
// Properties.
public var height:Number = 1.70;
public var weight:Number = 67.5;
}
}
These properties can be accessed via any instance of Person that you create. Each instance will have its own set of these properties.
Class constructors (I believe this is what you're asking about).
Constructors are used to hold logic that should be run as soon as an instance of your class is created. The class constructor has the same name as the class itself. It must be public and it does not return anything. Constructors can accept arguments, which are typically used to pass in references to dependencies for that class or required values.
package classes
{
public class Person
{
// Properties.
public var height:Number = 1.70;
public var weight:Number = 67.5;
// Constructor.
public function Person(height:Number, weight:Number)
{
this.height = height;
this.weight = weight;
}
}
}
Methods.
Methods are used to hold logic that can be run when calling that method. Methods often return values and can accept arguments. Methods can belong to any namespace that you would expect properties to be able to belong to.
We may want to be able to easily determine the BMI of each instance of Person that we create, so we should create a method for that:
package classes
{
public class Person
{
// Properties.
public var height:Number = 170;
public var weight:Number = 65.5;
// Constructor.
public function Person(height:Number, weight:Number)
{
this.height = height;
this.weight = weight;
}
// Determine my BMI and return the result.
public function getBMI():Number
{
return weight / (height * height);
}
}
}
Instances.
Now that we've defined our new class, we can create instances of this class using the new keyword. This can be done from anywhere that can access the Person class, which in this case is anywhere in the project because we've made the class public.
Though the class is public, accessing it from anywhere outside of the package it belongs in will require the use of an import statement. This statement will need to be used within any class that belongs to a different package. The import statement follows the same name used for the package and includes the name of the class you want to include on the end:
import classes.Person;
Once you've imported Person, you can create instances of it and assign them to a variable with different height and weight values:
var marty:Person = new Person(71, 1.76);
var bruce:Person = new Person(96.4, 1.72);
We can then obtain the BMI for each person using their getBMI() method:
trace(marty.getBMI()); // 22.9
trace(bruce.getBMI()); // 32.6
1 You can place classes outside of a package which can be referred to in the same .as file.
2 You can add more source paths, and packages can be relative to that.
The function that have the same name as class is a constructor. In curly brackets is basically part of code that will execute instantly when object will be created. Try to search info about constructors, they exist I think in every object oriented programming language (I may be wrong), so you have a lot of resources.
You can also read about this concept on Wikipedia.
The function that is named the same as the class is the constructor. It's optional, so you can leave it out if you don't need it. A default constructor will be added, which essentially does nothing.
The constructor lets you write code that executes immediately after an instance of the class is created (ie when another bit of code runs new ClassName(). You would typically use it to initialise some variables that are used by the class. Defining a constructor also lets you handle constructor arguments, which other code can pass when they use the new operator.

Language Agnostic Basic Programming Question

This is very basic question from programming point of view but as I am in learning phase, I thought I would better ask this question rather than having a misunderstanding or narrow knowledge about the topic.
So do excuse me if somehow I mess it up.
Question:
Let's say I have class A,B,C and D now class A has some piece of code which I need to have in class B,C and D so I am extending class A in class B, class C, and class D
Now how can I access the function of class A in other classes, do I need to create an object of class A and than access the function of class A or as am extending A in other classes than I can internally call the function using this parameter.
If possible I would really appreciate if someone can explain this concept with code sample explaining how the logic flows.
Note
Example in Java, PHP and .Net would be appreciated.
Let's forget about C and D because they are the same as B. If class B extends class A, then objects of type B are also objects of type A. Whenever you create an object of type B you are also creating an object of type A. It should have access to all of the methods and data in A (except those marked as private, if your language supports access modifiers) and they can be referred to directly. If B overrides some functionality of A, then usually the language provides a facility to call the base class implementation (base.Foo() or some such).
Inheritance Example: C#
public class A
{
public void Foo() { }
public virtual void Baz() { }
}
public class B : A // B extends A
{
public void Bar()
{
this.Foo(); // Foo comes from A
}
public override void Baz() // a new Baz
{
base.Baz(); // A's Baz
this.Bar(); // more stuff
}
}
If, on the other hand, you have used composition instead of inheritance and B contains an instance of A as a class variable, then you would need to create an object of A and reference it's (public) functionality through that instance.
Composition Example: C#
public class B // use A from above
{
private A MyA { get; set; }
public B()
{
this.MyA = new A();
}
public void Bar()
{
this.MyA.Foo(); // call MyA's Foo()
}
}
depending on the access level (would be protected or public in .NET), you can use something like:
base.method(argumentlist);
the base keyword in my example is specific to C#
there is no need for an instance of class A, because you already have a class A inherited instance
Basically you need a reference to the parent class.
In PHP:
parent::example();
From: http://www.php.net/manual/en/keyword.parent.php
<?php
class A {
function example() {
echo "I am A::example() and provide basic functionality.<br />\n";
}
}
class B extends A {
function example() {
echo "I am B::example() and provide additional functionality.<br />\n";
parent::example();
}
}
$b = new B;
// This will call B::example(), which will in turn call A::example().
$b->example();
?>
I find that the best way to tame the complexity of inheritance is to ensure that I only make B inherit from A when it really is a specialization of the superclass. At that point, I can call A's methods from inside B just as if they were B's own methods, and if B has overridden them then I can only suppose that this must be for a good reason.
Of course, quite often it is useful for B's implementation of a method to invoke A's implementation on the same object, generally because the subclass is wrapping extra behavior around the superclass's basic definition. The way in which you do this varies between languages; for example, in Java you do this:
super.methodName(arg1, ...);
Here's a quick Java example:
public class Aclass
{
public static void list_examples()
{
return("A + B = C");
}
}
public class Bclass extends Aclass
{
public static void main(String [] args)
{
System.out.println("Example of inheritance "+list_examples);
}
}
Note that the method for accessing the parent class shouldn't change. Because you are extending you shouldn't have to say parent:: or anything unless you are overriding the parent method / function.
It seems to me that extending your class might not be your best option. Class "B", "C", and "D" should only extend class "A" if they are truly an extension of that class, not just to access some method. For instance "Huffy" should not extend "BrandNames" just because "Huffy" is a brand name and you want access to one of the methods of "BrandNames". "Huffy" should instead extend "Bicycle" and implement an interface so the methods of "BrandNames" can be used. An additional benefit here is that (in Java) multiple interfaces can be used but a class can only be extended once. If in your example class "B"' needed to access a method from class "A" that could work, but if class "C" needed to access a method from class "A" and class "'B"' then you would have to use an interface in class "'C".

What does the virtual keyword, in actionscript, does?

I have found some code that uses the virtual keyword for functions, like:
package tryOut{
public class Parent {
public function Parent() {}
public function foo():void{
trace("Parent foo");
}//foo
public virtual function bar():void{
trace("Parent virtual bar");
}//bar
}//class
}//package
As far as I understand using the virtual keyword should modify the way overriding a method works, or the way using a child method would work, or something. But it seems it does nothing at all. Having the extention:
package tryOut {
public class Child extends Parent {
public function Child() {}
public override function foo():void {
trace("Child foo");
}//foo
public override function bar():void {
trace("Child virtual bar");
}//bar
}//class
}//package
The following code prints:
var parent:Parent = new Parent();
var child:Child = new Child();
parent.foo(); //Parent foo
child.foo(); //Child foo
parent.bar(); //Parent virtual bar
child.bar(); //Child virtual bar
var childCast:Parent = child as Parent;
parent.foo(); //Parent foo
childCast.foo(); //Child foo
parent.bar(); //Parent virtual bar
childCast.bar(); //Child virtual bar
So both methods work the same regarding the override. Does the virtual keyword changes something I am missing?
From the help documents (If you're using Flash, do a search for 'virtual'):
There are also several identifiers
that are sometimes referred to as
future reserved words. These
identifiers are not reserved by
ActionScript 3.0, though some of them
may be treated as keywords by software
that incorporates ActionScript 3.0.
You might be able to use many of these
identifiers in your code, but Adobe
recommends that you do not use them
because they may appear as keywords in
a subsequent version of the language.
abstract boolean byte cast
char debugger double enum
export float goto intrinsic
long prototype short synchronized
throws to transient type
virtual volatile
So in AS3, virtual does absolutely nothing.
So both methods work the same regarding the override.
What makes you think that? The tests you've shown aren't comparable.
childCast is typed as a Parent, yet you still end up calling the function in Child.
You don't check the same situation for the non-virtual method.