浮点数运算的精确

var floatObj = function () {
 
        /*
         * 判断obj是否为一个整数
         */
        function isInteger(obj) {
            return Math.floor(obj) === obj
        }
 
        /*
         * 将一个浮点数转成整数,返回整数和倍数。如 3.14 >> 314,倍数是 100
         * @param floatNum {number} 小数
         * @return {object}
         *   {times:100, num: 314}
         */
        function toInteger(floatNum) {
            var ret = {times: 1, num: 0};
            if (isInteger(floatNum)) {
                ret.num = floatNum;
                return ret
            }
            var strfi = floatNum + '';
            var dotPos = strfi.indexOf('.');
            var len = strfi.substr(dotPos + 1).length;
            var times = Math.pow(10, len);
            var intNum = parseInt(floatNum * times + 0.5, 10);
            ret.times = times;
            ret.num = intNum;
            return ret
        }
 
        /*
         * 核心方法,实现加减乘除运算,确保不丢失精度
         * 思路:把小数放大为整数(乘),进行算术运算,再缩小为小数(除)
         *
         * @param a {number} 运算数1
         * @param b {number} 运算数2
         * @param op {string} 运算类型,有加减乘除(add/subtract/multiply/divide)
         *
         */
        function operation(a, b, op) {
            var o1 = toInteger(a);
            var o2 = toInteger(b);
            var n1 = o1.num;
            var n2 = o2.num;
            var t1 = o1.times;
            var t2 = o2.times;
            var max = t1 > t2 ? t1 : t2;
            var result = null;
            switch (op) {
                case 'add':
                    if (t1 === t2) { // 两个小数位数相同
                        result = n1 + n2
                    } else if (t1 > t2) { // o1 小数位 大于 o2
                        result = n1 + n2 * (t1 / t2)
                    } else { // o1 小数位 小于 o2
                        result = n1 * (t2 / t1) + n2
                    }
                    return result / max;
                case 'subtract':
                    if (t1 === t2) {
                        result = n1 - n2
                    } else if (t1 > t2) {
                        result = n1 - n2 * (t1 / t2)
                    } else {
                        result = n1 * (t2 / t1) - n2
                    }
                    return result / max;
                case 'multiply':
                    result = (n1 * n2) / (t1 * t2);
                    return result;
                case 'divide':
                    result = (n1 / n2) * (t2 / t1);
                    return result
            }
        }
 
        // 加减乘除的四个接口
        function add(a, b) {
            return operation(a, b, 'add')
        }
 
        function subtract(a, b) {
            return operation(a, b, 'subtract')
        }
 
        function multiply(a, b) {
            return operation(a, b, 'multiply')
        }
 
        function divide(a, b) {
            return operation(a, b, 'divide')
        }
 
        // exports
        return {
            add: add,
            subtract: subtract,
            multiply: multiply,
            divide: divide
        }
    }()

  

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转载自www.cnblogs.com/LFxanla/p/11765025.html