函数式编程 (更新中)

高阶函数

1. 函数可以作为参数被传递
2. 函数可以作为返回值输出

柯里化(curry)

只传递给函数一部分参数来调用它，让它返回一个函数去处理剩下的参数。

代码组合(compose)

通过将一个个功能单一的纯函数组合起来实现一个复杂的功能，就像乐高拼积木一样，这种称为函数组合（代码组合）;

把层级嵌套的那种函数调用(一个函数的运行结果当作实参传给下一个函数的这种操作)扁平化，这就是compose函数

export default function compose(...funcs) {
  if (funcs.length === 0) {
    return arg => arg
  }

  if (funcs.length === 1) {
    return funcs[0]
  }

  return funcs.reduce((a, b) => (...args) => a(b(...args)))
}
//2
var compose = (...args) => (initValue) => args.reduceRight((a, c) => c(a), initValue)
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github.com/reduxjs/red…

type Func<T extends any[], R> = (...a: T) => R

/**
 * Composes single-argument functions from right to left. The rightmost
 * function can take multiple arguments as it provides the signature for the
 * resulting composite function.
 *
 * @param funcs The functions to compose.
 * @returns A function obtained by composing the argument functions from right  #后面的函数先执行(洋葱芯)
 *   to left. For example, `compose(f, g, h)` is identical to doing
 *   `(...args) => f(g(h(...args)))`.
 */
export default function compose(): <R>(a: R) => R

export default function compose<F extends Function>(f: F): F

/* two functions */
export default function compose<A, T extends any[], R>(
  f1: (a: A) => R,
  f2: Func<T, A>
): Func<T, R>

/* three functions */
export default function compose<A, B, T extends any[], R>(
  f1: (b: B) => R,
  f2: (a: A) => B,
  f3: Func<T, A>
): Func<T, R>

/* four functions */
export default function compose<A, B, C, T extends any[], R>(
  f1: (c: C) => R,
  f2: (b: B) => C,
  f3: (a: A) => B,
  f4: Func<T, A>
): Func<T, R>

/* rest */
export default function compose<R>(
  f1: (a: any) => R,
  ...funcs: Function[]
): (...args: any[]) => R

export default function compose<R>(...funcs: Function[]): (...args: any[]) => R

export default function compose(...funcs: Function[]) {
  if (funcs.length === 0) {
    // infer the argument type so it is usable in inference down the line
    return <T>(arg: T) => arg
  }

  if (funcs.length === 1) {
    return funcs[0]
  }

  return funcs.reduce(
    (a, b) =>
      (...args: any) =>
        a(b(...args))
  )
}
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