Typescript 如何键入类装饰器工厂？

我发现了无数的线程，其中没有一个与我的用例和TypeScript版本（2.8.1）完全匹配。我尝试过一些极为基础的装饰工厂，但运气不佳

type Constructable<T> = { new (): T }

function decoratorFactory<T> (): (Base: Constructable<T>) => Constructable<T> {
  return function decorator (Base: Constructable<T>): Constructable<T> {
    return class extends Base { // line 5
      constructor () {
        super()
        console.log('in constructor')
      }
    }
  }
}

@decoratorFactory()
class Example {}

我已经看到一些示例使用了

T extends

，但这对我来说并没有解决任何问题：

interface ClassOrInterface {}
type Constructable<T> = { new (): T }

function decoratorFactory<T extends ClassOrInterface> (): (Base: Constructable<T>) => Constructable<T> {
  return function decorator (Base: Constructable<T>): Constructable<T> {
    return class extends Base {
      constructor () {
        super()
        console.log('in constructor')
      }
    } as Constructable<T>
  }
}

@decoratorFactory()
class Example {}

我看到的答案表明你可以。我看到的答案是交替使用

new（）=>T

和

{new（）：T}

。我见过。我看到的答案只是，而不是一个新的构造函数/类

我怎么能做我想做的

type Constructable<T> = { new (): T }

function decoratorFactory<T> (): (Base: Constructable<T>) => Constructable<T> {
  return function decorator (Base: Constructable<T>): Constructable<T> {
    return class extends Base {
      constructor () {
        super()
        console.log('in constructor')
      }
    } as Constructable<T>
  }
}

@decoratorFactory()
class Example {}

index.ts(5,26): error TS2509: Base constructor return type 'T' is not a class or interface type.

interface ClassOrInterface {}
type Constructable<T> = { new (): T }

function decoratorFactory<T extends ClassOrInterface> (): (Base: Constructable<T>) => Constructable<T> {
  return function decorator (Base: Constructable<T>): Constructable<T> {
    return class extends Base {
      constructor () {
        super()
        console.log('in constructor')
      }
    } as Constructable<T>
  }
}

@decoratorFactory()
class Example {}

index.ts(6,26): error TS2509: Base constructor return type 'T' is not a class or interface type.