typescript - 请教在Typescript中如何使用递归函数返回正确的数据类型？

Question

TS演示：https://tsplay.dev/mLO04N ， 问题：当我将reduce中返回对象的children注释去掉，返回的类型就变成any了，请问正确姿势应该是？

interface LocationItem {    code: string;    name: string;    cities?: Record<string, LocationItem>;    districts?: Record<string, string>;}interface Option {    value: string | number;    label: string;    children?: Option[];}const objectEntries = <T extends object, K = keyof T>(obj: T) => Object.entries(obj) as Array<[K, T[keyof T]]>;const transform = (data: Record<string, LocationItem | string>|undefined = {}) =>    objectEntries(data).reduce((buf, [key, value]) => {        if (typeof value === "string") {            return [...buf, { label: value, value: key }];        }        const { code, name, cities, districts } = value;        const _cities = transform(cities);        const _districts = transform(districts);        const children = _cities.length ? _cities : _districts.length ? _districts : undefined;        return [            ...buf,            {                label: name,                value: code,                // children,            },        ];    }, [] as Option[]);

我把提取的值像这样复制给一个对象，获取的类型是对的，但是如果把这个对象作为返回数据类型就不对了

请不要试图类似用下面断言的方式来解决，我试图通过正确的姿势获取数据类型：

const _cities = transform(cities) as Options[];const _districts = transform(districts) as Options[];

Answer 1

可以考虑把return的数据项单独声明并赋予类型:

// ...省略const item: Option = {  label: name,  value: code,  children: _cities.length    ? _cities    : _districts.length    ? _districts    : undefined,};return [...buf, item];

完整示例

本质原因是因为 Option[] | undefined 导致的类型匹配错误，或者在各个中间变量的定义中都加上正确的类型声明也能解决