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

 

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20210104加更

 

  function adder(me) {
    var slice = Array.prototype.slice,
      __args = slice.call(arguments);
    return function () {
      var __inargs = slice.call(arguments);
      if( arguments.length == 0 ){
        var me = 0 ;
        for(var i in __args){
          me = Add(me, __args[i]);//这里为了处理精度问题
        }
        return me ;
      }
      else
        return adder.apply(this, __args.concat(__inargs));
    };
  };
  function Add(arg1,arg2){
    var r1,r2,m;
    try{r1=arg1.toString().split(".")[1].length}catch(e){r1=0}
    try{r2=arg2.toString().split(".")[1].length}catch(e){r2=0}
    m=Math.pow(10,Math.max(r1,r2))
    return (arg1*m+arg2*m)/m
  }

多个数字相加，柯里化实现adder(x)(c)(a)....()