3-16 javascript精度问题

javascript 历史的浮点计算精度问题

加法 plus

function plus(arg1, arg2) {
  let 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 minus

function minus(arg1, arg2) {
  let 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 Number(((arg1 * m - arg2 * m) / m).toFixed(n));
}

除法 div

function div(num1, num2) {
  let t1, t2, r1, r2;
  try {
    t1 = num1.toString().split(".")[1].length;
  } catch (e) {
    t1 = 0;
  }
  try {
    t2 = num2.toString().split(".")[1].length;
  } catch (e) {
    t2 = 0;
  }
  r1 = Number(num1.toString().replace(".", ""));
  r2 = Number(num2.toString().replace(".", ""));
  return (r1 / r2) * Math.pow(10, t2 - t1);
}

乘法 times

function times(num1, num2) {
  let m = 0,
    s1 = num1.toString(),
    s2 = num2.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)
  );
}