在Typescript中声明Ramda组合/管道函数的类型

import * as R from 'ramda';

type Compose = {
  <T>(fn1: (x: T) => T): (x: T) => T;
  <T, R>(fn1: (x: T) => R, fn2: (x: R) => T): (x: T) => T;
  <T, R1, R2>(fn1: (x: T) => R1, fn2: (x: R1) => R2, fn3: (x: R2) => T): (x: T) => T;
  // 可以根据需要继续添加更多的函数重载
};

type Pipe = {
  <T>(fn1: (x: T) => T): (x: T) => T;
  <T, R>(fn1: (x: T) => R, fn2: (x: R) => T): (x: T) => T;
  <T, R1, R2>(fn1: (x: T) => R1, fn2: (x: R1) => R2, fn3: (x: R2) => T): (x: T) => T;
  // 可以根据需要继续添加更多的函数重载
};

const ramda: {
  compose: Compose;
  pipe: Pipe;
} = {
  compose: R.compose,
  pipe: R.pipe,
};

const add1 = (x: number) => x + 1;
const double = (x: number) => x * 2;
const square = (x: number) => x * x;

const composedFn = ramda.compose(add1, double, square);
const pipedFn = ramda.pipe(add1, double, square);

console.log(composedFn(2)); // 输出：11
console.log(pipedFn(2)); // 输出：18