关于JavaScript0.1+0.2=？浮点数精度问题

<script>
        var a1 = 0.1;
        var a2 = 0.2;
        alert(a1+a2);
</script>

var a1 = 0.1;
        var a2 = 0.2;
        a1 = a1 *10;
        a2 = a2 *10;
        alert((a1+a2)/10);

var a1 = "0.01";
alert(a1.length);//获取a1的长度

var a1 = "0.01";
var a1_arr = a1.split(".");
var a1_arr_2 = a1_arr[1];
alert(a1_arr_2.length); //小数点后面的位数

	<script>
        var add = function (a1,a2){
            //转换为字符串
            a1 = a1.toString();
            a2 = a2.toString();
            //获取小数点的位置
            var index1 = a1.indexOf(".");
            var index2 = a2.indexOf(".");
            //如果小数点存在，那么再获取各自的小数位数
            var ws1 = 0;
            var ws2 = 0;
            if(index1 != -1){
                ws1 = a1.split(".")[1].length;
            }
            if(index2 != -1){
                ws2 = a2.split(".")[1].length;
            }

            //谁的位数大，谁的位数小，计算要补全的0
            var bigger = (ws1>ws2) ? ws1 : ws2;
            var smaller = (ws1<ws2) ? ws1 : ws2;
            var zeroCount = bigger - smaller;

            //去除小数点
            a1 = a1.replace(".","");
            a2 = a2.replace(".","");

            //比较a1和a2
            if(ws1 == smaller){
                for(var i = 0; i<zeroCount; i++){
                    a1 += "0";
                }
            }else{
                for(var i = 0; i<zeroCount; i++){
                    a2 += "0";
                }
            }

            //计算
            var sum = parseInt(a1) + parseInt(a2);

            //较大的小数位数计算倍数
            var beishu = 1;
            for(var i = 0; i<bigger; i++){
                beishu = beishu * 10;
            }
            sum = sum/beishu;
            return sum;
        }

        alert(add(1.001,2.002));
    </script>

													——本文参考《JavaScript百炼成仙》