在TypeScript中输入curry函数的推理
对于TypeScript中的curried函数，我有以下定义：
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（咖啡或不）：
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);
但是TypeScript无法自行推断出类型：
apply(curried, 4);
（即使只有一个超载()会采用一个值）。它抱怨说：
Argument of type 'Curried2<number, boolean, string>' is not assignable to parameter of type '(_0: number) => string' ...它正确地推断了
T ，但推断U是string 。 为什么是这样？ 在这种情况下，我能做些什么来为我进行类型推理（因为明确指定T和U对我来说过于冗长）？ 提前致谢！
我可以尝试这种方法：
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]
另一个尝试。 语法正确但不安全，如果您传递的curry函数不正确：
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
以下现在工作：
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);