Typescript function 生成实现接口的 class

Question

My question is on a point I don't understand in a mechanism of typescript: why in the following code, ComposerA is correct and ComposerB is not?

type Ctor<T = {}> = new (...args: any[]) => T;

function ComposerA<T extends Ctor>(Target: T) {
  return class extends Target {
    ...
  }
}

function ComposerB<T extends Ctor>(Target: T) {
  return class implements Target {
    ...
  }
}

In ComposerB I obtain the error -> 'Target' refers to a value, but is being used as a type here.

I think there is a concept I haven't understood. Is it possible to realize a function which take an object and implements it. I need implements and not extends, this example is a simplification of my original code and I've already an extend of another class.

Answer 1

You cannot implement anything at run time, but you can add methods to the prototype of the class and change the constructor signature with generics:

type Class<T={}> = new (...args: any) => T;

interface Fooable {
    foo(n: number):string
}

function withFoo<T>(k: Class<T>): Class<T & Fooable> {

    k.prototype.foo = function(n: number): string {
        return "foo";
    }

    return k as Class<T & Fooable>;
}

class Blah {
    blah(s: string) {}    
}

let BlahFooable = withFoo(Blah);

let b = new BlahFooable;

b.blah("hey") // ok
b.foo(42) // ok

Play

In a more generic way:

type Class<T={}> = new (...args: any) => T;

function Mixin<T, M extends object>(cls: Class<T>, mixin: M): Class<T & M> {
    Object.assign(cls.prototype, mixin);
    return cls as Class<T & M>;
} 

class Blah {
    blah(s: string) {}
}

let BlahWithFoo = Mixin(Blah, {
    foo(n: number): string {
        return "foo";
    }
});

let b = new BlahWithFoo();

b.blah("hey") // ok
b.foo(42) // ok