我有一個可變參數提升函數，它允許沒有深度嵌套的函數組合的扁平 monadic 鏈：const varArgs = f => {  const go = args =>    Object.defineProperties(      arg => go(args.concat(arg)), {        "runVarArgs": {get: function() {return f(args)}, enumerable: true},        [TYPE]: {value: "VarArgs", enumerable: true}      });  return go([]);};const varLiftM = (chain, of) => f => { // TODO: replace recursion with a fold  const go = (ms, g, i) =>    i === ms.length      ? of(g)      : chain(ms[i]) (x => go(ms, g(x), i + 1));  return varArgs(ms => go(ms, f, 0));};它有效，但我想通過折疊從遞歸中抽象出來。正常的折疊似乎不起作用，至少不能與Task類型一起使用，const varLiftM = (chain, of) => f =>  varArgs(ms => of(arrFold(g => mx => chain(mx) (g)) (f) (ms))); // A因為行中的代數A會Task為每次迭代返回 a ，而不是部分應用的函數。如何用折疊替換非尾遞歸？這是當前遞歸實現的一個工作示例：const TYPE = Symbol.toStringTag;const struct = type => cons => {  const f = x => ({    ["run" + type]: x,    [TYPE]: type,  });  return cons(f);};// variadic argument transformerconst varArgs = f => {  const go = args =>    Object.defineProperties(      arg => go(args.concat(arg)), {        "runVarArgs": {get: function() {return f(args)}, enumerable: true},        [TYPE]: {value: "VarArgs", enumerable: true}      });  return go([]);};// variadic monadic lifting functionconst varLiftM = (chain, of) => f => { // TODO: replace recursion with a fold  const go = (ms, g, i) =>    i === ms.length      ? of(g)      : chain(ms[i]) (x => go(ms, g(x), i + 1));  return varArgs(ms => go(ms, f, 0));};// asynchronous Taskconst Task = struct("Task") (Task => k => Task((res, rej) => k(res, rej)));const tOf = x => Task((res, rej) => res(x));const tMap = f => tx =>  Task((res, rej) => tx.runTask(x => res(f(x)), rej));const tChain = mx => fm =>  Task((res, rej) => mx.runTask(x => fm(x).runTask(res, rej), rej));// mock functionconst delay = (ms, x) =>  Task(r => setTimeout(r, ms, x));// test dataconst tw = delay(100, 1),  tx = delay(200, 2),  ty = delay(300, 3),  tz = delay(400, 4);