How to inherit classes that inherits base class more than once - class

Title is not clear. Here I am explaining
I am having a package say package provide test. It is having classes. I am using Itcl. Package is having following structure
::itcl::class classA {
written something having constructor and methods
}
::itcl::class classB {
inherit ::test::classA
having its own constructor and methods
}
::itcl::class classC {
inherit ::test::classA
having its own constructor and methods
}
::itcl::class classD {
inehrit ::test::classB ::test::classC
having its own constructor and methods
}
When i am requiring package test, I am getting below error
class "::test::classD" inherits base class "::test::classA" more than once:
How can i handle the error

Diamond inheritance is not allowed due to path ambiguity.
I.e.
TopClass
/ \
LeftClass RightClass
\ /
BottomClass
As workaround you can use composition (has-a) rather than inheritance
(is-a).
::itcl::class classA {
}
::itcl::class classB {
inherit ::test::classA
}
::itcl::class classC {
inherit ::test::classA
}
::itcl::class classD {
constructor {} {
set _b [::test::classB #auto]
set _c [::test::classC #auto]
}
destructor {
::itcl::delete $_b
::itcl::delete $_c
}
private {
variable _b ""
variable _c ""
}
}
Now in classD you must be specific to classB or classC path the code gets to base.

Related

In binding a derived class must I also bind all parent classes to the root class?

True or False
To bind a derived class in a C++ class hierarchy one must also bind all the parent classes on up to the root class.
I'm looking to bind a bunch of custom data types in a project I just started on and am looking to scope out the degree of work involved. I'm looking to bind a class that's 3 levels of derivation away from a root type.
Are there any rules of thumb for when one must also bind the parent classes to successfully bind the child classes?
False
Let us consider a simple example:
#include <string>
class Parent
{
public:
Parent() { m_name = "Parent"; }
std::string foo() { return m_name + "::foo"; }
virtual std::string bar() { return m_name + "::bar"; }
protected:
std::string m_name;
};
class Derived : public Parent
{
public:
Derived() { m_name = "Derived"; }
std::string bar() override { return m_name + "::bar (override)"; }
};
This you can bind the way you describe: bind the parent, the derived class will benefit from all the derived methods (you even don't have to bind the override):
#include <pybind11/pybind11.h>
namespace py = pybind11;
PYBIND11_MODULE(example, m)
{
py::class_<Parent>(m, "Parent")
.def(py::init<>(), "Constructor description")
.def("foo", &Parent::foo, "Function description")
.def("bar", &Parent::bar, "Function description")
.def("__repr__", [](const Parent&) { return "<Parent>"; });
py::class_<Derived, Parent>(m, "Derived")
.def(py::init<>(), "Constructor description")
.def("__repr__", [](const Derived&) { return "<Derived>"; });
}
Indeed
import example
p = example.Parent()
print(p.foo())
print(p.bar())
d = example.Derived()
print(d.foo())
print(d.bar())
will print
Parent::foo
Parent::bar
Derived::foo
Derived::bar (override)
However, if you just want to bind one derived class (you don't care about the parent, nor about any of the other derived classes) you can also just bind the derived class you care about, without ever specifying the parent:
#include <pybind11/pybind11.h>
namespace py = pybind11;
PYBIND11_MODULE(example, m)
{
py::class_<Derived>(m, "Derived")
.def(py::init<>(), "Constructor description")
.def("foo", &Derived::foo, "Function description")
.def("bar", &Derived::bar, "Function description")
.def("__repr__", [](const Derived&) { return "<Derived>"; });
}
Indeed
import example
d = example.Derived()
print(d.foo())
print(d.bar())
prints
Derived::foo
Derived::bar (override)
See the docs for reference.

What's the difference between a struct with only type functionality, and a class with only type functionality?

For example, consider this class:
class Foo {
private init() {
// Private init, so no instances can be created.
// All methods and properties will by owned by the type.
// So no instances will need to be created,
// as there is no instance functionality.
}
static var count = 0
static func increment() {
Foo.count += 1
}
}
Perhaps it makes no difference whether it's a class or a struct? (assuming inheritance isn't needed).

In TypeScript, how to prevent a method from being called on derived class?

There are three classes.
// in external library, which I don't want to modify
class ComponentBase {
// I want calling this to be disallowed
forceUpdate() {}
}
class ComponentBase_MyVersion extends ComponentBase {
// I want subclasses to always call this, instead of forceUpdate()
Update() {}
}
class MyComponent extends ComponentBase_MyVersion {
DoSomething() {
// I want this to be disallowed
this.forceUpdate();
// forcing the subclass to call this instead
this.Update();
}
}
How can I accomplish this, with changes only to ComponentBase_MyVersion?
Is there a way to "hide" a base-class member?
Or perhaps a way to override the definition -- like with the "new" keyword in C# -- letting me mangle the method definition to at least make warnings appear when attempting to call it?
The OOP does not allow you to do this kind of method cancellation. You can impleement this funcion on your class with an Exception like you suggested, or use a composition: https://en.wikipedia.org/wiki/Composition_over_inheritance
Example 1:
class ComponentBase {
forceUpdate() {}
}
class ComponentBase_MyVersion extends ComponentBase {
Update() {}
forceUpdate() {
throw new Error("Do not call this. Call Update() instead.");
}
}
class MyComponent extends ComponentBase_MyVersion {
DoSomething() {
// wil raise an exception
this.forceUpdate();
this.Update();
}
}
Example 2 (composition):
class ComponentBase {
forceUpdate() {}
}
class ComponentBase_MyVersion {
private _component: ComponentBase = ...;
Update() {}
// expose _component desired members ...
}
class MyComponent extends ComponentBase_MyVersion {
DoSomething() {
// compilation error
this.forceUpdate();
this.Update();
}
}
I hope I helped.
Encapsulate implementation by replacing inheritance with composition Delegation Pattern
You can do this by adding the private access modifier on the forceUpdate method. This will result in all the subclasses being unable to access forceUpdate. However TypeScript does not support package access modifiers, but you can do this by replacing inheritance with composition.
class ComponentBase {
forceUpdate() {
}
}
class ComponentBase_MyVersion {
// Replace inheritance with composition.
private component: ComponentBase;
Update() {
this.component.forceUpdate();
}
}
class MyComponent extends ComponentBase_MyVersion {
DoSomething() {
// Now subclass can't access forceUpdate method
this.Update();
}
}
Use a symbol in order to prevent external access to the method.
If you don't want to replace inheritance with composition, you can use Symbol to define a method. If your target is es5 you must configure tsconfig.json compilerOptions.lib to include es2015.symbol. Because every symbol is unique, any external module will not be able to obtain the symbol and access the method.
// libs.ts
let forceUpdate = Symbol("forceUpdate");
export class ComponentBase {
[forceUpdate]() {
}
}
export default class ComponentBase_MyVersion extends ComponentBase {
Update() {
this[forceUpdate]();
}
}
// test.ts
import ComponentBase_MyVersion from "./libs";
class MyComponent extends ComponentBase_MyVersion {
DoSomething() {
// Now subclass can't access the forceUpdate method.
this.Update();
}
}
I found a way that seems to work -- that is, which causes warnings to appear when someone attempts to call forceUpdate() on a subclass instance.
forceUpdate(_: ()=>"Do not call this. Call Update() instead.") {
throw new Error("Do not call this. Call Update() instead.");
}
Now when I write new MyComponent().forceUpdate(), I get a compiler error, with the warning message containing a description telling me to use Update() instead.
EDIT: Apparently this only works because the base class already had this definition:
forceUpdate(callBack?: () => any): void;
If instead the base method is defined with no arguments originally (as in the OP), the above solution doesn't work.
However, if you have a case like mine (where there's an optional property like that, which you can narrow the return-type of), it works fine. (not sure if this return-type-narrowing is a bug, or intended)

angular2 / typescript class inheritance with generic types

Hope one of you angular2 / typescript wizards can help out or at least provide a pointer in the right direction, before I got crazy :-)
Here is what I'd like to
have a parent class that implements it's own defined parent Interface, however using Generic Types so I can when creating a child class provide it with the child's specific and tailored class & data Interface.
the child class should be able to extend the parent data class by
being able to overwrite default/parent set variables
overwriting parent functions() and have the child's version called instead of the parent's default
In the below pseudo code example, I would like the call to the child's (inherited) someOtherfunction() to return "2"...
Am I asking for too much?
I can't seem to find any decent examples on the web...
How do I get this right?
Thank you -
Oliver
(CODE BELOW MAY BE BROKEN, IT'S JUST FOR ILLUSTRATION)
//
// Parent Class
//
export interface ICoreData <T> {
observeItems: Observable <T[]> ;
items: Array <T>;
}
#Injectable()
export class CoreData<T> implements ICoreData<T> {
public observeItems: Observable<T[]>;
private items: Array<T>;
constructor( 'Dependency Injection...' ) {}
coreFunction(): number {
return 1;
}
someOtherfunction(){
return this.coreFunction();
}
}
//
// Child class
//
export interface IMyDataStructure {
name: string;
age: string;
}
export interface ISpecificData extends ICoreData<IMyDataStructure> {
someExtraKey: number;
}
#Injectable()
export class SpecificData extends CoreData<IMyDataStructure> implements ISpecificData {
constructor() {
super();
}
coreFunction(): number{
//
// This function should "overwrite" the parent's original function
// and be called by the parent's someOtherfunction() function
//
return 2;
}
}
You're not asking too much. However you can't use interfaces to accomplish what you're trying to accomplish. You need to extend a class, which can be generic.
An interface is simply a contract, or a blueprint if you like, for a data type. There is no functionality associated with an interface. However in your case you wanted to be able to have methods on the base class; methods you could override in the derived.
The way I usually do this is to declare an abstract base class (so that the base class can't be instantiated itself), and then extend classes from that. Here's an example:
Note, I've removed all the Angular2 cruft in order to keep the example as simple as possible.
abstract class Base<T> {
constructor(public controlled: T) { }
doIt(): string {
return `Base.doIt: ${JSON.stringify(this.controlled)}`;
}
doSomethingElse(): string {
return `Base.doSomethingElse: ${JSON.stringify(this.controlled)}`;
}
};
interface Foo {
foo: string;
bar: string;
};
class Derived extends Base<Foo> {
constructor(foo: Foo) {
super(foo);
}
doSomethingElse(): string {
return `Derived.doSomethingElse: ${JSON.stringify(this.controlled)}`;
}
};
let d: Derived = new Derived({ foo: 'foo', bar: 'bar' });
console.log(`doIt ==> ${d.doIt()}`);
console.log(`doSomethingElse ==> ${d.doSomethingElse()}`);
Output:
doIt ==> Base.doIt: {"foo":"foo","bar":"bar"}
doSomethingElse ==> Derived.doSomethingElse: {"foo":"foo","bar":"bar"}
Playground link.

final interface implementation not recognized from interface base list

How can a method from the InterfaceBaseList be implemented in the current interface ? Example:
interface bar(T)
{
void method1(T a);
void method2(T a);
}
interface baz: bar!int
{
final void method1(int a){}
}
class foo: baz
{
this(){method1(0);}
void method2(int a){}
}
void main()
{
auto Foo = new foo;
Foo.method2(0);
}
outputs:
myfile.d(xx): Error: foo interface function 'void method1(int a)'
is not implemented
It seems that the compiler doesnt get that baz.method1 is actually bar.method1.
Note that the example illustrates that in baz, for some reasons, we know that method1 will always have the same implemtation. a baz implementer mays be down-casted as a bar (so making a dummy final method1 in bar is not possible).
Interfaces can only declare virtual members without implementation, or final members with implementation. Your code is attempting to override a virtual method with a non-virtual implementation. Due to the nature of interfaces, you cannot actually override anything within them. What you want instead is an abstract class.
abstract class baz: bar!int
{
override void method1(int a){}
}
Replacing your baz interface with the above class will clear up the issue.
As an example of why this isn't allowed, consider this code: (Does not compile, of course!)
interface Root {
int foo();
}
interface BranchA : Root {
override int foo() { return 1; }
}
interface BranchB : Root {
override int foo() { return 2; }
}
class C : BranchA, BranchB { }
What would (new C()).foo() return? The result is ambiguous. It is only acceptable to override the interface methods in a class, because unlike interfaces, you can only inherit one class at a time.