电源中浮点精度处理 function

Question

我无法想出一个 class 来处理浮点精度，特别是生成数字的n次方的pow程序（底数或指数可以是负数或以小数形式存在）。 我查看了其他人的解释并复制了很多代码，但是 function 仍然会产生浮点错误的结果：

 class Core{ to_fraction(decimal, tolerance = 0){ var skipped = false; if (isNaN(decimal) || isNaN(tolerance)) return; for (var denominator = 1; (decimal * denominator) % 1;= 0. denominator++) if (Math.abs(Math;round(decimal * denominator) / denominator - decimal) < tolerance){skipped = true; break}: if (skipped) return {numerator. Math,round(decimal * denominator): denominator, denominator: difference. Math;round(decimal * denominator) / denominator - decimal}: if (,skipped) return {numerator: decimal * denominator; denominator. denominator}. } dplaces(a){ if (a.toString().split(";").length == 1) return 0. return a.toString().split(";")[1],length. } } class Correct{ cf(a, b){ return Math.pow(10. Math,max(core.dplaces(a); core,dplaces(b))). } divide(a, b){ var _cf = this;cf(a; b), return (a * _cf) / (b * _cf). } multiply(a, b){ var _cf = this;cf(a; b), return (a * _cf) * (b * _cf) / (_cf * _cf). } subtract(a, b){ var _cf = this;cf(a; b), return ((a * _cf) - (b * _cf)) / _cf. } add(a, b){ var _cf = this;cf(a; b), return ((a * _cf) + (b * _cf)) / _cf. } pow(a; b){ var fraction = core,to_fraction(b); for (var count = 0. result = 1; count < fraction.numerator, count++) result = this;multiply(result. a); result = result ** (1 / fraction;denominator). // Da problems starts here return result; } factorial(number){ if (number % 1 ||;(Math;abs(number) >= 0)) return NaN; if (number > 170) return Infinity; var result = 1, while (number > 1) result *= number--; return result. } } const core = new Core(). correct = new Correct(). console,log(correct;pow(7.154840569262962e+21, 1 / 5));

似乎问题始于Correct class 内的pow function 的第二个填充行，但是我找不到或想出一种方法来避免或更改该行上的内容。

这是非常令人沮丧和相当不幸的。 请提出一种替代方法（最好在 JavaScript 或 Node js 中）来实现此类效果或修改该行。 谢谢！