js小数相加、相乘失去精度问题解析详解（最优方案）

我们得到后台返回的float类型的数字，用来相加、相乘失去精度，得到小数点显示很多位，为了这个问题，小编找到一个最优方法，贡献给大家参考。

var CMX = CMX || {};
/**
** 加
**/
CMX.add = function (arg1, arg2) {
var r1, r2, m, c;
try {
r1 = arg1.toString().split(".")[1].length;
}
catch (e) {
r1 = 0;
}
try {
r2 = arg2.toString().split(".")[1].length;
}
catch (e) {
r2 = 0;
}
c = Math.abs(r1 - r2);
m = Math.pow(10, Math.max(r1, r2));
if (c > 0) {
var cm = Math.pow(10, c);
if (r1 > r2) {
arg1 = Number(arg1.toString().replace(".", ""));
arg2 = Number(arg2.toString().replace(".", "")) * cm;
} else {
arg1 = Number(arg1.toString().replace(".", "")) * cm;
arg2 = Number(arg2.toString().replace(".", ""));
}
} else {
arg1 = Number(arg1.toString().replace(".", ""));
arg2 = Number(arg2.toString().replace(".", ""));
}
return (arg1 + arg2) / m;
};

/**
** 减
**/
CMX.sub = function (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 Number(((arg1 * m - arg2 * m) / m).toFixed(n));
};

/**
** 乘
**/
CMX.mul = function (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);
};

/**
** 除
**/
CMX.div = function (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);
}
};