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Question

let Obj = function Obj<T>(gen: T) {
    return {
        data: <{[Key in keyof T]: T[Key]}>{},// need generic, result T[Key]
        gen,
    }
}({
    foo(min: number, max: number, container: { data: any }) {
        let res = ((max - min) / 2)
        container.data.foo = res
        return res//number
    }
});
Obj.gen.foo(20,50,Obj)
let aaaaaaa: number = Obj.data.foo;//Obj.data.foo typeof function. must be number

Obj.data.foo具有类型的功能。但必须是数字

请查看screen shot - 突出显示

我需要通用T [Key]的结果

。
。
。

更新

type qqw<R> = () => R
    ;
function ZcddF<Q>(foo: qqw<Q>): Q {
    return ;
}
function foo(): number { return ; }
let aaaa = ZcddF(foo)// 4a is number
    ;
type a = <Z>(foo: qqw<Z>) => Z
type b = () => number
    ;
let ff:a;
let bbbb = ff(foo)// 4b is number too
    ;
// type cccc = a(b) // <------------  QN how get this type/generic

Answer 1

let fooN = () => 78
let fooS = () => ""
let fooArr = (): [string, number] => ["", 78]

interface iGen<T, R, K extends keyof T> {
    [key: string]: () => R
}

function Obj<T, R, K extends keyof T, S extends {}>(gen: T & iGen<T, R, K>, Ssw?: S) {
    let res: S & {
        data: {[K in keyof T]: R}
        gen: T
    } = <any>Object.assign(Ssw, {
        data: {},
        gen,
    })
    return res
}
let obj = (
    Obj({ fooS },
        Obj({ fooN },
            Obj({ fooArr },
            )))
    /*obj*/);

let gS = obj.gen.fooS         //() => string
let S = obj.data.fooS         //string

let gN = obj.gen.fooN         //() => number
let N = obj.data.fooN         //number

let gArr = obj.gen.fooArr     //() => [string, number]
let Arr = obj.data.fooArr     //[string, number]
let Arr0 = obj.data.fooArr[0] //string
let Arr1 = obj.data.fooArr[1] //number

等待发布并查看提案：获取typeof #6606和6606#issuecomment-178362393

的任何表达式的类型

获取＆＃34;函数结果类型＆＃34; （数）

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1 个答案: