javascript(js)小数精度丢失的解决方案，0.2+0.1≠0.3

 

原因:js按照2进制来处理小数的加减乘除,在arg1的基础上 将arg2的精度进行扩展或逆扩展匹配,所以会出现如下情况.

javascript(js)的小数点加减乘除问题，是一个js的bug如0.3*1 = 0.2999999999，0.2+0.1=0.30000000000000004等，下面列出可以完美求出相应精度的四种js算法

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);
}