TypeScript 中用于 curried 函数的类型推断

Question

我对 TypeScript 中的柯里化函数有以下定义：

interface Curried2<A, B, Z> {
    (_0: A): (_0: B) => Z;
    (_0: A, _1: B): Z;
}

我有以下函数应该接受一个函数（curried 与否）：

function apply<T, U>(f: (_0: T) => U, x: T): U {
    return f(x);
}

现在，给定

let curried: Curried2<number, boolean, string> = ...

以下作品（如预期）：

apply<number, (_0: boolean) => string>(curried, 4);

但 TypeScript 无法自行推断类型：

apply(curried, 4);

（即使() 只有一个重载，它采用单个值。）它抱怨：

'Curried2' 类型的参数不可分配给'(_0: number) => string' 类型的参数...

它已经正确推断出T，但推断出U 是 string。为什么是这样？在这种情况下，我该怎么做才能使类型推断对我有用（因为明确指定 T 和 U 对于我的口味来说太冗长了）？

提前致谢！

Answer 1

以下现在有效：

interface F1<A, Z> {
    (_0: A): Z;
}

interface F2<A, B, Z> extends F1<A, F1<B, Z>> {
    (_0: A, _1: B): Z;
}

function apply<T, U>(f: F1<T, U>, x: T): U {
    return f(x);
}

function apply2<T, U, V>(f: F2<T, U, V>, x: T, y: U): V {
    return f(x, y);
}

let curried: F2<number, boolean, string> = null;

const x: F1<boolean, string> = apply(curried, 4);
const y: string = apply2(curried, 4, false);

Answer 2

我可以试试这个方法：

type Curried<T1, T2, T3> = (x: T1) => (y: T2) => T3;

function apply<T, U>(f: (value: T) => U, x: T): U {
    return f(x);
}

//simple demo
var someFunc: Curried<string, number, [string, number]> = x => y => [x, y];
var curriedSomeFunc = apply(someFunc, "string here");
var result1 = curriedSomeFunc(0); //will be ["string here", 0]
var result2 = curriedSomeFunc(1); //will be ["string here", 1]

再试一次。如果您传递的不是正确的柯里化函数，语法是正确的，但不安全：

interface Curried<T1, T2, T3> {
    (x: T1, y: T2): T3;
    (x: T1): (y: T2) => T3
}

let f1 = <Curried<string, number, [string, number]>>
    ((x: string, y: number) => [x, 1]);
let f2 = <Curried<string, number, [string, number]>>
    ((x: string) => (y: number) => [x, y]);

function apply<T, V>(f: (x: T) => V, x: T): V {
    return f(x);
}

let curriedF1 = apply(f1, "42"); //Won't work (no function to curry) but type inference is OK
let curriedF2 = apply(f2, "11"); //Will work and type inference is also OK