对于TypeScript中的咖喱函数,我具有以下定义:
interface Curried2<A, B, Z> {
(_0: A): (_0: B) => Z;
(_0: A, _1: B): Z;
}
我有以下应该接受一个功能的功能(是否已咖喱):
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(预期):
apply<number, (_0: boolean) => string>(curried, 4);
但是TypeScript无法自行推断类型:
apply(curried, 4);
(即使只有一个重载()
采用单个值。)它抱怨:
类型'Curried2 <number,boolean,string>'的参数不能分配给类型'(_0:number)=> string'的参数...它已经正确地推断出了
T
,但可以推断出 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起作用(因为已明确指定 T
并且 U
对于我的口味来说太冗长了)?
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]
再试一次。如果您传递的curry函数不正确,那么语法是正确的,但是没有安全性:
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
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