javascript小数精度丢失的完美解决方法

Posted 2020-12-10 vwvwvwgwg

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);
}

 

 来源：锌闻网