8

TypeScript (v3.2.2) allows me to define a union of interfaces, each with a unique string literal property which can be used as a type guard, e.g.

type Device = Laptop | Desktop | Phone;

interface Laptop {
  type: 'Laptop';
  countDriveBays: number;
  hasTouchScreen: boolean;
}

interface Desktop {
  type: 'Desktop';
  countDriveBays: number;
}

interface Phone {
  type: 'Phone';
  hasTouchScreen: boolean;
}

function printInfo(device: Device) {
  if (device.type === 'Laptop') {
    // device: Laptop
    console.log(
      `A laptop with ${device.countDriveBays} drive bays and ${
        device.hasTouchScreen ? 'a' : 'no'
      } touchscreen.`,
    );
  } else if (device.type === 'Desktop') {
    // device: Desktop
    console.log(`A desktop with ${device.countDriveBays} drive bays.`);
  } else {
    // device: Phone
    console.log(`A phone with ${device.hasTouchScreen ? 'a' : 'no'} touchscreen.`);
  }
}

I want to write a function isDeviceType in a generic way:

const isDeviceType = <T extends Device['type']>(type: T) => {
  return (device: Device): device is DeviceOf<T> => device.type === type;
}

// e.g.
const isPhone = isDeviceType('Phone');
isPhone({ type: 'Phone', hasTouchScreen: true }); // true

However, the way I have defined the DeviceOf type is pretty verbose since it lists every single type within the union:

type DeviceOf<Type extends Device['type']> =
  Type extends Laptop['type'] ? Laptop :
  Type extends Desktop['type'] ? Desktop :
  Type extends Phone['type'] ? Phone :
  never;

Is there a more concise way to define DeviceOf? I have tried these:

type DeviceOf<Type extends Device['type']> =
  (infer D)['type'] extends Type ? D : never;

// TS2536: Type '"type"' cannot be used to index type 'D'.
// TS1338: 'infer' declarations are only permitted in the 'extends' clause of a conditional type.
// TS6133: 'D' is declared but its value is never read.

type DeviceOf<Type extends Device['type']> =
  (infer D) extends Device
    ? D['type'] extends Type
    ? D
    : never
    : never;

// TS1338: 'infer' declarations are only permitted in the 'extends' clause of a conditional type.
// TS6133: 'D' is declared but its value is never read.
// TS2304: Cannot find name 'D'.

My impression is that error TS1338 is the limiting factor, and so it's impossible to define DeviceOf in a generic way in the current version of TypeScript.

4

2 回答 2

6

找到了另一种方法，只使用没有infer关键字的条件类型：

type FindByType<Union, Type> = Union extends { type: Type } ? Union : never;
type DeviceOf<Type extends Device['type']> = FindByType<Device, Type>;

type Result = DeviceOf<'Laptop'>;

基于 Ryan Cavanaugh 的评论：https ://github.com/Microsoft/TypeScript/issues/17915#issuecomment-413347828

于 2020-10-21T18:37:39.120 回答
6

知道了。您必须应用“if”两次，一次用于创建infer类型，第二次用于检查infer类型是否扩展设备。只有在分支中D extends Device，您才能使用D['type']

type DeviceOf<Type extends Device['type']> =
  Device extends (infer D) ?
  D extends Device ?
  D['type'] extends Type ? D : never : never : never;

type Result = DeviceOf<'Laptop'>;

操场

于 2019-02-08T21:25:12.920 回答