Typescript 将超类型断言为类型脚本中的泛型参数

问题是

在TypeScrip中，我们可以使用extends来断言一个泛型参数作为另一种类型的‘子类型’.例如.

// class Base {}
// class Child extends Base {}
// edited:
class Base { a = 1 }
class Child extends Base { b = 2 } // Child is a sub class of Base

class Test {
  // assert T is a subtype of Base
  hello<T extends Base>() { ... }
}

new Test().hello<Child>();

但是如何将泛型参数断言为另一种类型的"超类型"呢？例如.

class Test2 {
  // HERE, how to ensure T is a super type of Child
  hello<T ??? Child>() { ... }
}

new Test2().hello<Base>();

用例

// class Actions<T = unknown> {
// edited:
class Actions<T = never> {
  execs: (() => T)[]

  append(exec: () => T) {
    this.execs.push(exec)
    return this;
  }

  // HERE, how to ensure NewT is a super type of T?
  setT<NewT ??? T>() {
    return this as Actions<NewT>;
  }
}

// to make the following work
new Actions()
  .setT<Child>()
  .append(() => new Child())
  // before this line, all exec should return Child
  // after this line, all exec should return Base
  .setT<Base>()
  .append(() => new Base())
  .execs.map(e => e()) // => Base[]

如果我们仍然像这样使用extends:

setT<NewT extends T>() {
  return this as Actions<NewT>;
}

收到错误:

Conversion of type 'this' to type 'Actions<NewT>' may be a mistake because neither type sufficiently overlaps with the other. If this was intentional, convert the expression to 'unknown' first.
  Type 'Actions<T>' is not comparable to type 'Actions<NewT>'.
    Type 'T' is not comparable to type 'NewT'.
      'NewT' could be instantiated with an arbitrary type which could be unrelated to 'T'.ts(2352)

是的，我可以使用unknown来转换返回类型，但是函数setT的签名不能表达我们想要的类型约束.