Question

例如，我希望以下函数返回一个可以接受任何字符串或任何数字的函数，这取决于调用初始函数的方式，但它仅接受等于a的字符串或数字：

function foo<A extends string | number>(a: A) : (b: A) => boolean {
  return (b) => a === b;
}

foo("test") // creates a function that only accepts "test", want one that accepts any string.
foo(3) // creates a function that only accepts 3, want one that accepts any number.

如何键入此函数，使其按需工作？

进行编辑以阐明问题所在

Answer 1

如果您只删除extends string，它似乎可以满足您的要求

function foo<A>(a: A) : (b: A) => boolean {
  return (b) => a === b;
}

Typescript Playground Link

如果您想限制输入类型并且仍然让它们通过，则可以执行以下操作

function foo(a:string): (b:string) => boolean
function foo(a:number): (b:number) => boolean
function foo(a:boolean): (b:boolean) => boolean

function foo (a: string|number|boolean) : (b: string|number|boolean) => boolean {
  return (b) => a === b;
}

foo("test")("a")  //expects string
foo(1)(2)  //expects number

Playground Link

Answer 2

似乎这样做的方法是使用重载：

function foo(a1: number): (a2: number) => boolean; 
function foo(a1: string): (a1: string) => boolean;
function foo(a1: number | string): (a2: number|string) => boolean {
  return (a2) => a1 === a2;
}

具有泛型函数返回具有相同泛型的函数

2 个答案: