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Accepted
Ram
Asked:2020-04-10 05:52:40 +0000 UTC2020-04-10 05:52:40 +0000 UTC

如何动态提取 TypeScript 返回类型

  • 772
const isNull = (v:any) : v is null => v === null;

const isString = (v:any) : v is string => typeof v === 'string';

const or = <T extends Function,K extends Function>(fn1:T, fn2:K) => (first: any): first is ReturnType<T> | first is ReturnType<K>   => (fn1(first) || fn2(first))

执行

const conditon = or(isNull, isString);

condition(v)

如何 first is ReturnType<T> | first is ReturnType<K> 获取不是返回函数的类型，而是给定的类型 v is null

这样在函数的描述中传递一个值时，它应该是first is string | null

typescript

2 个回答

  1. 我找到了这个答案 

    const or = <T extends Function,K extends Function>(fn1:T, fn2:K) =>
  <T>(first: any): first is T  => 
  (fn1(first) || fn2(first))
condition = (v:any): v is null|string => or(isNull, isString)(v)
condition(v)
    • 0
  2. Best Answer

    有一个沙箱

    这样在函数的描述中传递一个值时，它应该是first is string | null

    见t4。

    缺少功能本身的实现-超出了问题的范围。

    type $is<T> = (q: any) => q is T
// type tt = ReturnType<$is<string>> // type tt = boolean // : (

t2: {
    type $valid = $is<number | string>
    const valid: $valid = null as any

    TEST: {
        const q = null as any
        if (valid(q)) {
            const res = q //const res: string | number
        } else {
            const res = q // const res: any
        }
    }
}

t4: {
    type $or = <T extends $is<any>[]>(...f: T) => T[number]
    const or: $or = null as any

    TEST: {
        class User { }

        const isUser = (q => q instanceof User) as $is<User>
        const isString = (q => 'string' === typeof q) as $is<string>

        const conditon = or(isUser, isString)

        const q = null as any
        if (conditon(q)) {
            const res = q // const res: string | User
        }
    }
}
    • 0

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