JS数字计算精度问题解决

       add(a, b) {//相加
           var c, d, e;
           try {
             c = a.toString().split(".")[1].length;
           } catch (f) {
             c = 0;
           }
           try {
             d = b.toString().split(".")[1].length;
           } catch (f) {
             d = 0;
           }
           return e = Math.pow(10, Math.max(c, d)), (this.mul(a, e) + this.mul(b, e)) / e;
       },
       sub(a, b) {//相减
         var c, d, e;
         try {
           c = a.toString().split(".")[1].length;
         } catch (f) {
           c = 0;
         }
         try {
           d = b.toString().split(".")[1].length;
         } catch (f) {
           d = 0;
         }
         return e = Math.pow(10, Math.max(c, d)), (this.mul(a, e) - this.mul(b, e)) / e;
       },
       mul(a, b) {//主体
         var c = 0,
           d = a.toString(),
           e = b.toString();
         try {
           c += d.split(".")[1].length;
         } catch (f) { }
         try {
           c += e.split(".")[1].length;
         } catch (f) { }
         return Number(d.replace(".", "")) * Number(e.replace(".", "")) / Math.pow(10, c);
       },
       div(a, b) {//除
         var c, d, e = 0,
           f = 0;
         try {
           e = a.toString().split(".")[1].length;
         } catch (g) { }
         try {
           f = b.toString().split(".")[1].length;
         } catch (g) { }
         return c = Number(a.toString().replace(".", "")), d = Number(b.toString().replace(".", "")), this.mul(c / d, Math.pow(10, f - e));
       }

js在数字计算时，因为IEEE 754会有精度丢失，完善一下，需要用到哪个，只要mul和你用到的函数就OK