JS 精度问题0.1+0.2=0.30000000000000004

//js 加法计算 ?//调用：accAdd(arg1,arg2) ?//返回值：arg1加arg2的精确结果 ??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).toFixed(2); ??} ??//js 减法计算 ?//调用：Subtr(arg1,arg2) ?//返回值：arg1减arg2的精确结果 ??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)); ??????//last modify by deeka ??????//动态控制精度长度 ??????n=(r1>=r2)?r1:r2; ??????return ((arg1*m-arg2*m)/m).toFixed(2); ?} ??//js 除法函数 ?//调用：accDiv(arg1,arg2) ?//返回值：arg1除以arg2的精确结果 ??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 (r1/r2)*pow(10,t2-t1); ????} ??} ????//js 乘法函数 ?//调用：accMul(arg1,arg2) ??//返回值：arg1乘以arg2的精确结果 ??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) ??}