1. 程式人生 > >js小數相加、相乘失去精度問題解析詳解(最優方案)

js小數相加、相乘失去精度問題解析詳解(最優方案)

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;
};

/**
 ** 減
 **/
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 //動態控制精度長度
  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);
  }
};
大家可以複製下來,看看如何,歡迎大家指出錯誤