Question

不确定“ discriminator”在这里是否正确。考虑以下代码：

abstract class A {
}

class A1 extends A {
}

class A2 extends A {
}

interface SomeMap {
  [key: string]: new (...args: any[]) => A;
}

class B<T extends SomeMap> {
  public test<K extends keyof T>(key: K, instance: InstanceType<T[K]>): void {}
}

const map = {
  a1: A1,
  a2: A2,
};

class C extends B<typeof map> {}

const c = new C();

c.test('a1', new A1()); // No errors from TS - good
c.test('a1', new A2()); // No errors from TS - bad, I want ts to error here!

并带有“鉴别符”：

abstract class A {
  protected abstract discriminator: undefined;
}

class A1 extends A {
  protected discriminator: undefined;
}

class A2 extends A {
  protected discriminator: undefined;
}

interface SomeMap {
  [key: string]: new (...args: any[]) => A;
}

class B<T extends SomeMap> {
  public test<K extends keyof T>(key: K, instance: InstanceType<T[K]>): void {}
}

const map = {
  a1: A1,
  a2: A2,
};

class C extends B<typeof map> {}

const c = new C();

c.test('a1', new A1()); // No errors from TS - good
c.test('a1', new A2()); // TS error - good!

在避免从A继承的类中声明“ discriminator”道具的同时，有什么方法可以使其起作用？只是想在TS中检测到A1！= A2。

有办法避免类“歧视者”吗？

0 个答案: