基于区分联合的缩小typescript泛型类型

Question

我有一些类型定义，如下所示。

type PayloadType = 'A' | 'B';

interface Payload<T extends PayloadType> {
  type: T;
}

interface PayloadA extends Payload<'A'> {
  state: string
}

interface PayloadB extends Payload<'B'> {
  serialNumber: string;
}

type TPayload = PayloadA | PayloadB;


type PayloadInterpretation<T extends TPayload> = {
  payload: T;
  entries: T[]; // This property is only for demonstration purpose
};

type TPayloadInterpretation = PayloadInterpretation<PayloadA> | PayloadInterpretation<PayloadB>;

function f(interpretation: TPayloadInterpretation) {
  if (interpretation.payload.type === 'B') {
    const payload = interpretation.payload; // payload is of type PayloadB
    const entries = interpretation.entries; // entries is of type PayloadA[] | PayloadB[]
  }
}

注释表明，即使是有效负载的类型也可以根据区分后的联合正确地缩小到PayloadB，但是entries的T[]类型仍然是PayloadA[] | PayloadB[]。

我在想，如果typescript知道有效负载的类型T是PayloadA，那么它也应该能够将entries: T[]缩小到entries: PayloadB[]。我知道我可以像这样进行类型转换：

function f(interpretation: TPayloadInterpretation) {
  if (interpretation.payload.type === 'B') {
    const payloadBInterpretation = interpretation as PayloadInterpretation<PayloadB>;
    ...
  }
}

但我的问题是，有没有其他方法可以做到这一点？

代码是typescript playground中的here。

谢谢!

Answer 1

当您检查interpretation.payload.type时，您只是缩小了interpretation.payload对象的范围。您实际上并没有做任何事情来缩小interpretation.entries的范围。

换句话说，当你缩小interpretation.payload的范围时，typescript并不知道interpretation.entries也可以缩小。

如果希望同时缩小它们的范围，则需要在PayloadInterpretation类型中使用另一个鉴别器：

// ...

type PayloadInterpretation<T extends TPayload> = {
  type: T['type']; // the new discriminator for the whole PayloadInterpretation
  payload: T;
  entries: T[];
};

// ...

function f(interpretation: TPayloadInterpretation) {
  if (interpretation.type === 'B') { // narrowing the whole interpretation instead of only interpretation.payload
    const payload = interpretation.payload; // payload is of type PayloadB
    const entries = interpretation.entries; // entries is of type PayloadB[]
  }
}