js 计算精度问题

　　js 计算有时会因为精度问题而得不准确的值，当次遇到的是乘法问题

　　计算器的值则为：

　　贴个代码防止以后再用：

　　其他计算方法也一并加一下：

  /** 四则运算： 加 */
  public static add(arg1: number, arg2: number): number {
    const i1 = FormatHelper.toInteger(arg1),
      i2 = FormatHelper.toInteger(arg2);
    let result;
    if (i1.times > i2.times) {
      result = (i1.num + i2.num * (i1.times / i2.times)) / i1.times;
    }else {
      result = (i1.num * (i2.times / i1.times) + i2.num) / i2.times;
    }
    return result;
  }

  /** 四则运算： 减 */
  public static subtract(arg1: number, arg2: number): number {
    const i1 = FormatHelper.toInteger(arg1),
      i2 = FormatHelper.toInteger(arg2);
    let result;
    if (i1.times > i2.times) {
      result = (i1.num - i2.num * (i1.times / i2.times)) / i1.times;
    }else {
      result = (i1.num * (i2.times / i1.times) - i2.num) / i2.times;
    }
    return result;
  }

  /** 四则运算： 乘 */
  public static multiply(arg1: number, arg2: number): number {
    let m = 0;
    const 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);
  }

  /** 四则运算： 除 */
  public static divide(arg1: number, arg2: number, dec: number): number {
    let t1 = 0, t2 = 0, r1, r2;
    try {
      t1 = arg1.toString().split('.')[1].length;
    } catch (e) { }
    try {
      t2 = arg2.toString().split('.')[1].length;
    } catch (e) { }
    r1 = Number(arg1.toString().replace('.', ''));
    r2 = Number(arg2.toString().replace('.', ''));
    return round(FormatHelper.multiply((r1 / r2), Math.pow(10, t2 - t1)), dec);
  }