JavaScript加减乘除丢失精度问题处理

来源：互联网发布：win python 编辑：程序博客网时间：2024/05/16 10:03

 <script type="text/javascript"> window.onload=function ale(){alert(accMul(accDiv(28.24,8),5)); } function accDiv(arg1,arg2){   var t1=0,t2=0,r1,r2;   try{t1=arg1.toString().split(".")[1].length}catch(e){}   try{t2=arg2.toString().split(".")[1].length}catch(e){}   with(Math){   r1=Number(arg1.toString().replace(".",""))   r2=Number(arg2.toString().replace(".",""))  return accMul((r1/r2),pow(10,t2-t1));   }   }    //乘法   function accMul(arg1,arg2)    {    var m=0,s1=arg1.toString(),s2=arg2.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)    }  //加法  function accAdd(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  }  //减法  function Subtr(arg1,arg2){     var r1,r2,m,n;     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));     n=(r1>=r2)?r1:r2;     return ((arg1*m-arg2*m)/m).toFixed(n); } </script>

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