如何让 function 调用自身 n 次

Question

Let's say I have a function called f that takes an integer argument called x and returns an integer. I also have an integer n that says how many times the function must call itself. So for example if my function call looks like this f(x) when n = 1, then it would look like this f(f(f(x))) when n = 3. How could something like that look in my example bellow:

function succ(n) {
  return function (f, x) {
    return f(x);
  };
}

Answer 1

You could loop inside the inner function:

 for(let i = 0; i < n; i++) x = f(x);
 return x;

or alternatively let your function call itself:

 return n > 0 ? succ(n - 1)(f, f(x)) : x;

Answer 2

We can express this algorithm corecursively. Corecursion builds its result on the way forward from the starting point:

 const iterate = f => x => [x, () => iterate(f) (f(x))]; const main = iterate(x => x * 2) (1); console.log( main[1] () [1] () [1] () [1] () [1] () [1] () [1] () [1] () [0]); // 256

This is just a proof of concept but not what we actually want. How can we avoid the clunky interface? We can use a Proxy to make the non-argument function implicit. It is basically the same mechanism as with lazy property getters. Additionally we don't want to access values from the stream manually but with a function for convenience:

 class ThunkProxy { constructor(f) { this.memo = undefined; } get(g, k) { if (this.memo === undefined) this.memo = g(); if (k === THUNK) return true; else if (k === Symbol.toPrimitive) return () => this.memo; else if (k === "valueOf") return () => this.memo; else return this.memo[k]; } } const THUNK = "thunk"; const thunk = f => new Proxy(f, new ThunkProxy(f)); const iterate = f => x => [x, thunk(() => iterate(f) (f(x)))]; const takeNth = n => ([head, tail]) => n === 0? head: takeNth(n - 1) (tail); const main = iterate(x => x * 2) (1); console.log( main[1] [1] [1] [1] [1] [1] [1] [1] [0]); // 256 console.log( takeNth(16) (main)); // 65536

Answer 3

You could build a times high order function and use it to wrap other functions...

 const times = (fn, n) => (...args) => { if (;n) { return. } fn(..;args), return times(fn. n - 1)(..;args). } const log10 = times(console,log; 10); log10('hello');