JS浮点数计算精度

JS的浮点数，计算都是会有问题的。下面给出加减乘除的解决办法

• 131.20*100 = 13119.999999
• 0.1+0.2 = 0.3000000000004
• 1.2/12 = 0.099999999999
• 0.4-0.1= 0.30000000000004

export const accAdd = (arg1, arg2) => {
  // 增加浮点数计算方法
  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 (accMul(arg1, m) + accMul(arg2, m)) / (m * 1.0)
}
export const accMul = (arg1, arg2) => {
  // 增加浮点数的相乘方法
  var m = 0,
    s1 = arg1.toString(),
    s2 = arg2.toString();
  try {
    m += s1.split('.')[1].length;
  } catch (e) {
    // ee
  }
  try {
    m += s2.split('.')[1].length;
  } catch (e) {
    //e
  }
  return (Number(s1.replace('.', '')) * Number(s2.replace('.', ''))) / Math.pow(10, m);
}
export const accDiv = (arg1, arg2) => {
  var t1 = 0,
    t2 = 0,
    r1,
    r2;
  try {
    t1 = arg1.toString().split('.')[1].length;
  } catch (e) {
    // e
  }
  try {
    t2 = arg2.toString().split('.')[1].length;
  } catch (e) {
    // e
  }
  r1 = Number(arg1.toString().replace('.', ''));
  r2 = Number(arg2.toString().replace('.', ''));
  return (r1 / r2) * Math.pow(10, t2 - t1);
}

export const accSub = (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 ((accMul(arg1, m) - accMul(arg2, m)) / (m * 1.0)).toFixed(n);
}