Typescript - Get a prototype with defined values for different classes-CodePudding

So I want to create a function that takes a class and creates a prototype for it. But in this prototype, every property of the class is defined (including the optional ones). So here's what I want:

class Bar {
    public i?: number
    //...
}

class Foo {
    public a?: string;
    public b?: number;
    public c?: Array<string>;
    public l?: boolean;
    public d?: Bar;

    //...
}

const prot = getPrototype<Foo>();

console.log(prot); //My expected output: { a: "", b: 0, c: [], l: false, d: {}, ... }

const prot2 = getPrototype<Bar>();

console.log(prot2); // My expected output: {i: 0, ...}

I really can't see any way of doing this. If you think there's a way, I'd love to see one. The values given to the properties are supposedly hard-coded values for each type. Something like this:

//...
//... the logic for the getPrototype function ...
//...
    if(typeof property === "string"){ prototype[property] = ""; }
    else if(typeof property === "number"){ prototype[property] = 0; }
    //...
//...

CodePudding user response：

We can achieve this by making use of Mapped Types and Conditional Property by creating a type alias InitializeType<T> and make use of it for defining the type of getPrototype<T>

class Bar {
    public i?: number
    //...
}

class Foo {
    public a?: string;
    public b?: number;
    public c?: Array<string>;
    public l?: boolean;
    public d?: Bar;

    //...
}

type InitializeType<T> = {} & { 
  [K in keyof T]-?: 
    string extends T[K] 
      ? ""
      : number extends T[K]
      ? 0
      : boolean extends T[K]      
      ? false
      : Array<any> extends T[K]
      ? []
      : object extends T[K]
      ? {}
      : T[K]
    }


declare function getPrototype<T>(): InitializeType<T>

const prot = getPrototype<Foo>();


console.log(prot); // output: { a: "", b: 0, c: [], l: false, d: {}, ... }

const prot2 = getPrototype<Bar>();

console.log(prot2); // output: {i: 0, ...}

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