无精度损失整数运算

Question

如果操作导致整数（否则抛出），我将如何创建一个类型，它可以处理整数，至少支持加法减法乘法和保证以及整数。

例如，我希望能够执行以下操作：

Precise A = 10;
A.Divide(3);
A.GetNumber(); // This would throw an exception as 10/3 isn't an int.
A.Multiply(6);
int result = A.GetNumber; // I want result to be = to 20, not to a floating point type that would round to 2 or be very close like 1.9999999999999999999999998992

我意识到这是一个奇怪的用例，但我确实有这个需求（执行一系列操作，在浮点数中可能会被错误表示，但保证最终会成为一个有效的 int）。

Answer 1

然后使用 Math.Round() 舍入最终答案。

Answer 2

如果你不允许任意精度的有理数，看起来你在没有更多限制的情况下问不可能。

将1除以65537两次，再乘以65537两次得到1：32位整数放不下。

Answer 3

因为我们不知道 10 / 3 最终会得到一个精确的整数答案，直到 * 6 之后，我们必须将它推迟到那时，并承诺：

public sealed class Precise
{
  private interface IOperation
  {
    int Calculate(int value);
    IOperation Combine(IOperation next);
  }
  private sealed class NoOp : IOperation
  {
    public static NoOp Instance = new NoOp();
    public int Calculate(int value)
    {
      return value;
    }
    public IOperation Combine(IOperation next)
    {
      return next;
    }
  }
  private sealed class Combo : IOperation
  {
    private readonly IOperation _first;
    private readonly IOperation _second;
    public Combo(IOperation first, IOperation second)
    {
      _first = first;
      _second = second;
    }
    public int Calculate(int value)
    {
      return _second.Calculate(_first.Calculate(value));
    }
    public IOperation Combine(IOperation next)
    {
      return new Combo(_first, _second.Combine(next));
    }
  }
  private sealed class Mult : IOperation
  {
    private readonly int _multiplicand;
    public Mult(int multiplicand)
    {
      _multiplicand = multiplicand;
    }
    public int Calculate(int value)
    {
      return value * _multiplicand;
    }
    public int Multiplicand
    {
      get { return _multiplicand; }
    }
    public IOperation Combine(IOperation next)
    {
      var nextMult = next as Mult;
      if(nextMult != null)
        return new Mult(_multiplicand * nextMult._multiplicand);
      var nextDiv = next as Div;
      if(nextDiv != null)
      {
        int divisor = nextDiv.Divisor;
        if(divisor == _multiplicand)
          return NoOp.Instance;//multiplcation by 1
        if(divisor > _multiplicand)
        {
          if(divisor % _multiplicand == 0)
            return new Div(divisor / _multiplicand);
        }
        if(_multiplicand % divisor == 0)
          return new Mult(_multiplicand / divisor);
      }
      return new Combo(this, next);
    }
  }
  private sealed class Div : IOperation
  {
    private readonly int _divisor;
    public Div(int divisor)
    {
      _divisor = divisor;
    }
    public int Divisor
    {
      get { return _divisor; }
    }
    public int Calculate(int value)
    {
      int ret = value / _divisor;
      if(value != ret * _divisor)
        throw new InvalidOperationException("Imprecise division");
      return ret;
    }
    public IOperation Combine(IOperation next)
    {
      var nextDiv = next as Div;
      if(nextDiv != null)
        return new Div(_divisor * nextDiv._divisor);
      var nextMult = next as Mult;
      if(nextMult != null)
      {
        var multiplicand = nextMult.Multiplicand;
        if(multiplicand == _divisor)
          return NoOp.Instance;
        if(multiplicand > _divisor)
        {
          if(multiplicand % _divisor == 0)
            return new Mult(multiplicand / _divisor);
        }
        else if(_divisor % multiplicand == 0)
          return new Div(multiplicand / _divisor);
      }
      return new Combo(this, next);
    }
  }
  private sealed class Plus : IOperation
  {
    private readonly int _addend;
    public Plus(int addend)
    {
      _addend = addend;
    }
    public int Calculate(int value)
    {
      return value + _addend;
    }
    public IOperation Combine(IOperation next)
    {
      var nextPlus = next as Plus;
      if(nextPlus != null)
      {
        int newAdd = _addend + nextPlus._addend;
        return newAdd == 0 ? (IOperation)NoOp.Instance : new Plus(newAdd);
      }
      return new Combo(this, next);
    }
  }
  private readonly int _value;
  private readonly IOperation _operation;
  public static readonly Precise Zero = new Precise(0);
  private Precise(int value, IOperation operation)
  {
    _value = value;
    _operation = operation;
  }
  public Precise(int value)
    : this(value, NoOp.Instance)
  {
  }
  public int GetNumber()
  {
    return _operation.Calculate(_value);
  }
  public static explicit operator int(Precise value)
  {
    return value.GetNumber();
  }
  public static implicit operator Precise(int value)
  {
    return new Precise(value);
  }
  public override string ToString()
  {
    return GetNumber().ToString();
  }
  public Precise Multiply(int multiplicand)
  {
    if(multiplicand == 0)
      return Zero;
    return new Precise(_value, _operation.Combine(new Mult(multiplicand)));
  }
  public static Precise operator * (Precise precise, int value)
  {
    return precise.Multiply(value);
  }
  public Precise Divide(int divisor)
  {
    return new Precise(_value, _operation.Combine(new Div(divisor)));
  }
  public static Precise operator / (Precise precise, int value)
  {
    return precise.Divide(value);
  }
  public Precise Add(int addend)
  {
    return new Precise(_value, _operation.Combine(new Plus(addend)));
  }
  public Precise Subtract(int minuend)
  {
    return Add(-minuend);
  }
  public static Precise operator + (Precise precise, int value)
  {
    return precise.Add(value);
  }
  public static Precise operator - (Precise precise, int value)
  {
    return precise.Subtract(value);
  }
}

这里每个 Precise 都有一个整数值和将对其执行的操作。进一步的操作产生一个新的 Precise （做这种事情作为一个可变的是疯狂的）与一个新的操作，但在可能的情况下，这些操作被组合成一个更简单的操作。因此 "divide by three then multiply by six" 变成 "multiply by two".

我们可以这样测试：

public static void Main(string[] args)
{
  Precise A = 10;
  A /= 3;
  try
  {
    var test = (int)A;
  }
  catch(InvalidOperationException)
  {
    Console.Error.WriteLine("Invalid operation attempted");
  }
  A *= 6;
  int result = (int)A;
  Console.WriteLine(result);
  // Let's do 10 / 5 * 2 = 4 because it works but can't be pre-combined:
  Console.WriteLine(new Precise(10) / 5 * 2);
  // Let's do 10 / 5 * 2 - 6 + 4 == 2 to mix in addition and subtraction:
  Console.WriteLine(new Precise(10) / 5 * 2 - 6 + 4);
  Console.Read();
}

一个好的解决方案还可以很好地处理左轴为整数、右轴为 Precise 以及两者均为 Precise 的操作；留作 reader 的练习 ;)

遗憾的是，我们必须变得更加复杂才能处理 (10 / 3 + 1) * 3，必须在 Combine 实现中进行改进。

编辑：进一步思考上述问题是否做得足够好以至少捕获大多数边缘情况，我认为它应该从只处理两个 Precise 对象之间的操作开始，因为int -> Precise 是微不足道的，可以很容易地放在最上面，但是 Precise -> int 需要调用计算，也许还为时过早。我还将操作作为关键操作（让操作存储一个或两个对象，这些对象又包含一个操作或一个值）。然后，如果您从总和 (10 / 3) + 5 的表示开始并将其乘以 6，则将其转换为 (10 * (6 / 3)) + (5 * 6) 更容易，最终计算可以给出精确的结果 50 而不是失败，因为它命中不精确 10 / 3.

Answer 4

我会为操作结果使用小数，如果有“.”，我会在 .ToString 的 GetNumber 检查中使用小数。如果是，我抛出异常，如果不是，我将其转换为 int。

无精度损失整数运算

No precision loss integer arithmetic

c#

math

floating-point