JS浮点类型运算精度丢失问题解决办法

//加法  
Number.prototype.add = function(arg){  
    var r1,r2,m;  
    try{r1=this.toString().split(".")[1].length}catch(e){r1=0}  
    try{r2=arg.toString().split(".")[1].length}catch(e){r2=0}  
    m=Math.pow(10,Math.max(r1,r2))  
    return (this*m+arg*m)/m  
}  
 
//减法  
Number.prototype.sub = function (arg){  
    return this.add(-arg);  
}  
 
//乘法  
Number.prototype.mul = function (arg)  
{  
    var m=0,s1=this.toString(),s2=arg.toString();  
    try{m+=s1.split(".")[1].length}catch(e){}  
    try{m+=s2.split(".")[1].length}catch(e){}  
    return Number(s1.replace(".",""))*Number(s2.replace(".",""))/Math.pow(10,m)  
}  
 
//除法  
Number.prototype.div = function (arg){  
    var t1=0,t2=0,r1,r2;  
    try{t1=this.toString().split(".")[1].length}catch(e){}  
    try{t2=arg.toString().split(".")[1].length}catch(e){}  
    with(Math){  
        r1=Number(this.toString().replace(".",""))  
        r2=Number(arg.toString().replace(".",""))  
        return (r1/r2)*pow(10,t2-t1);  
    }  
}

    alert(0.1+0.2)                  //  0.30000000000000004      

alert(Number(0.1).add(0.2));         //  0.3

 

原文地址:

http://zm6.sm-img5.com/?src=http%3A%2F%2Fblog.csdn.net%2Fqq282030166%2Farticle%2Fdetails%2F8364343&uid=786557245b09f79404ff7abb851dd3a0&hid=a5df8b9ce3035b5c528b414288832958&pos=4&cid=9&time=1440477098910&from=click&restype=1&pagetype=0040000002000408&bu=web&query=js%E7%B2%BE%E5%BA%A6%E4%B8%A2%E5%A4%B1&uc_param_str=dnntnwvepffrgibijbprsvpi

 

 
 
posted @ 2017-06-01 16:25  苏州coder  阅读(830)  评论(0编辑  收藏  举报