TypeScript 中 c# 类虚拟成员的等价物

Question

So in C# when I've been creating model classes and lazy loading things, I do something like this:

public int? User_ID { get; set; }
public int? Dept_ID { get; set; }

Then a little farther down in my class I pop in my virtuals like so:

public virtual User User {get; set;}
public virtual Department Dept {get; set;}

How would I do this in Typescript? Something like this?:

User_ID: number;
Dept_ID: number;
User: User;
Dept: Department;

I don't think interfaces is what I want/need...but then again protected doesn't seem correct either. Something tells me I'm missing an obvious answer here.

Answer 1

The following applies equally to TypeScript and JavaScript:

There is no equivalent.

Every member access in JavaScript is subject to dynamic dispatch by definition. This is because member access in JavaScript is morally a key lookup in a hash table.

An object may declare a member with the same key as one it inherits via the prototype chain and thereby shadow the inherited member but this is not the same as virtualness, a concept not present in the language.

Answer 2

Yes there are ..

//non-abstract or abstract class
class A {
    
    // The virtual method
    protected virtualStuff1?():void;

    public Stuff2(){
        //Calling overridden child method by parent if implemented
        this.virtualStuff1 && this.virtualStuff1();
        alert("Baseclass Stuff2");
    }
}

//class B implementing virtual method
class B extends A{
    
    // overriding virtual method
    public virtualStuff1()
    {
        alert("Class B virtualStuff1");
    }
}

//Class C not implementing virtual method
class C extends A{
 
}

var b1 = new B();
var c1= new C();
b1.Stuff2();
b1.virtualStuff1();
c1.Stuff2();

Answer 3

I achieved this using an arrow function as a property of the class (note this was in ts, but if you're handy at modern js it should work):

class baseClass {
    doSomething = () => {};
}

class derivedClass extends baseClass {
    constructor() {
        this.doSomeThing = () => { console.log('something'); };
    }
}

Answer 4

Javascript 和 Typescript 中的所有方法在技术上都是虚拟的，因为没有什么可以阻止您覆盖任何方法。