重构代码以不使用typeof作为参考

Question

我有下面的代码来自Enum，它为每个值创建enum +功能。

我想摆脱这2个“ typeof”是否可行？

const TEST = createEnumComparator<typeof Test, SuperCompareEnums<typeof Test>>(Test, SuperCompareEnums);

这就是我想要得到的并做同样的事情

const TEST = createEnumComparator<Test, SuperCompareEnums<Test>>(Test, SuperCompareEnums);

或者类似的东西将是最好的

const TEST = createEnumComparator<SuperCompareEnums<Test>>(Test, SuperCompareEnums);

整个代码

class CompareEnums<T> {
  constructor(protected value: keyof T) {

  }

  public is(value: keyof T) {
    return value === this.value;
  }
}

class SuperCompareEnums<T> extends CompareEnums<T> {
  sayHello() {
    console.log('hello ' + this.value);
  }
}

interface Constructor<T> {
  new(...args: any[]): T;
}

function createEnumComparator<TObj1, TObj2 extends CompareEnums<TObj1>>(obj1: TObj1, obj2: Constructor<TObj2>): EnumAsComparator<TObj1, TObj2> {
  return _.mapValues<any>(obj1, value => new obj2(value)) as any;
}

// --------------------------------------------


enum Test {
  A = 'A',
  B = 'B'
}

const TEST = createEnumComparator<typeof Test, SuperCompareEnums<typeof Test>>(Test, SuperCompareEnums);

// --------------------------------------------
console.warn(1111111111111);
console.warn(TEST.A.is(Test.A), 1);
console.warn(TEST.B.is(Test.B));
console.warn(TEST.B.sayHello());
console.warn(TEST.A.is(Test.B));

// --------------------------------------------

type EnumAsComparator<TObj1, TObj2> = {
  [P in keyof TObj1]: TObj2;
};