Question

TypeScript（v3.2.2）允许我定义接口的并集，每个接口都有一个唯一的字符串文字属性，可以用作类型保护，例如

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.`);
  }
}

我想以通用方式编写函数isDeviceType：

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

但是，我定义DeviceOf类型的方式非常冗长，因为它列出了并集内的每种类型：

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

是否有更简洁的方法来定义DeviceOf？我尝试了以下方法：

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'.

我的印象是错误TS1338是限制因素，因此不可能在当前版本的TypeScript中以通用方式定义DeviceOf。

Answer 1

知道了。您必须应用两次“ if”，一次创建class Solution { public boolean checkSubarraySum(int[] nums, int k) { int curSum=nums[0]; int start=0; int k1=Math.abs(k); for(int i=1;i<nums.length;i++){ while (curSum>k1 && start<i-1){ curSum = curSum - nums[start]; start++; } if(k1>0 && (curSum%k1)==0){ return true; } if(curSum==0 && k1==0){ return true; } curSum=curSum+nums[i]; } return false; }类型，第二次检查infer类型是否扩展了设备。只有在分支infer中，您才能使用D extends Device

D['type']

Playground

Answer 2

找到了另一种方法，仅使用不带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在这里的评论：https://github.com/Microsoft/TypeScript/issues/17915#issuecomment-413347828

TypeScript：通常基于字符串文字属性推断联合类型成员

2 个答案: