JavaScript加减乘除丢失精度问题处理

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转载最新推荐文章于 2026-04-28 17:30:10 发布 419 阅读 AI 写同款· GEO 优化 ›

浮点数运算精度丢失和toFixed精度丢失一、浮点数计算的几种特殊情况 0.1+0.2 = 0.30000000000000004 10.22*100 = 1022.0000000000001 2.4/0.8 = 2.9999999999999996 32.2*100 = 3220.0000000000005 (32.2*100 + 3.14*100) / 100 = 35.3... 阅读详情

 <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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