js小数相加、相乘失去精度问题解析详解（最优方案）

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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);  }};

大家可以复制下来，看看如何，欢迎大家指出错误

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