使用 Typescript，如何键入功能性 True 函数？

Question

For background, I am going through the "Flock of Functions" and doing my best to convert these Javascript examples to typed Typescript. See https://github.com/glebec/lambda-talk/blob/master/src/index.js#L152 for reference. The True function returns the first curried argument, and ignores the second.

Consider the following Typescript code:

interface ElsFn<T> {
  (els: unknown): T;
}

interface True extends Function {
  <T>(thn: T): ElsFn<T>;
}

// eslint-disable-next-line @typescript-eslint/explicit-function-return-type
const T: True = (thn) => (_els) => thn;

console.log(T('true')('false'));

Assuming I want to keep the "explicit-function-return-type" rule, how do I get rid of the ESLint disable comment? In other words, I want to properly type the True function.

My editor tells me the problem is with the (_els) => thn portion of the code. It needs to be typeed somehow.

]

What can I do to set the return type or otherwise get this thing properly typed so that I don't need to disable the ESLint rule?

Answer 1

您仍然需要指定泛型参数和返回类型：

const T: True = <T_>(thn) => <T_>(_els):T_ => thn;

Answer 2

(_els): boolean => thn;

它可能适用于您的情况吗？