您可以將蹦床視為程序中優(yōu)化的編譯器優(yōu)化。因此，是什么使我們無法以完全相同的方式來適應(yīng)更通用的優(yōu)化技術(shù)。這是尾遞歸模態(tài)缺點(diǎn)的草圖：const loop = f => {  let step = f(); while (step && step[step.length - 1] && step[step.length - 1].type === recur) {    let step_ = step.pop();    step.push(...f(...step_.args));  }  return step;};const recur = (...args) =>  ({type: recur, args});const push = (xs, x) => (xs.push(x), xs);const map = f => xs =>  loop((i = 0) =>    i === xs.length      ? []      : push([f(xs[i])], recur(i + 1)));const xs =   map(x => x * 2) (Array(1e6).fill(0).map((x, i) => i))  .slice(0,5);console.log(xs); // [0, 2, 4, 6, 8]這種優(yōu)化取決于表達(dá)式的關(guān)聯(lián)性。乘法也具有關(guān)聯(lián)性，因此存在尾遞歸模乘。但是，我必須作弊才能在Javascript中實(shí)現(xiàn)它：const loop = f => {  let step = f();  const acc = [];  while (step && step[1] && step[1].type === recur) {    acc.push(step[0]);    step = f(...step[1].args);  }  return acc.reduce((acc, f) => f(acc), step);};const recur = (...args) =>  ({type: recur, args});const mul = x => step => [y => x * y, step];const pow = (x_, n_) =>  loop((x = x_, n = n_) =>    n === 0 ? 1    : n === 1 ? x    : mul(x) (recur(x, n - 1)));    console.log(  pow(2, 1e6)); // Infinity, no stack overflow如您所見，我不能使用mul不是特別令人滿意的Regular 。這與Javascript蜜蜂語言緊密相關(guān)嗎？有沒有更好的方法來實(shí)現(xiàn)JS中的尾遞歸模乘而不必引入笨拙的二進(jìn)制運(yùn)算符？