javascript解决小数的加减乘除精度丢失的方案

原文：http://www.jb51.net/article/85463.htm

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); 
}

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functionaccDiv(arg1,arg2){

vart1=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(".",""))

returnaccMul((r1/r2),pow(10,t2-t1));

}

}

//乘法

functionaccMul(arg1,arg2)

{

varm=0,s1=arg1.toString(),s2=arg2.toString();

try{m+=s1.split(".")[1].length}catch(e){}

try{m+=s2.split(".")[1].length}catch(e){}

returnNumber(s1.replace(".",""))*Number(s2.replace(".",""))/Math.pow(10,m)

}

//加法

functionaccAdd(arg1,arg2){

varr1,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

functionSubtr(arg1,arg2){

varr1,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);

}