如何表示整数无穷大？

Question

我需要一种方法来表示一个可以是无限的整数。我不想使用浮点类型（double.PositiveInfinity），因为数字永远不能是小数，这可能会使 API 混乱。最好的方法是什么？

编辑：我还没有看到的一个想法是使用 int? null 代表无穷大。有什么好的理由不这样做吗？

Answer 1

如果您不需要完整范围的整数值，可以使用int.MaxValue 和int.MinValue 常量来表示无穷大。

但是，如果需要完整的值范围，我建议要么创建一个包装类，要么直接使用双精度值。

Answer 2

一个示例部分实现沿 SLaks 和其他的 cmets 行（欢迎反馈）：

用法：

int x = 4;
iint pi = iint.PositiveInfinity;
iint ni = iint.NegativeInfinity;

Assert.IsTrue(x + pi == iint.PositiveInfinity);
Assert.IsTrue(pi + 1 == iint.PositiveInfinity);
Assert.IsTrue(pi + (-ni) == iint.PositiveInfinity);
Assert.IsTrue((int)((iint)5) == 5);

实施：

public struct iint
{
    private readonly int _int;

    public iint(int value) 
    {
        if(value  == int.MaxValue || value == int.MinValue)
            throw new InvalidOperationException("min/max value reserved in iint");
        _int = value;
    }

    public static explicit operator int(iint @this)
    {
        if(@this._int == int.MaxValue || @this._int == int.MinValue)
            throw new InvalidOperationException("cannot implicit convert infinite iint to int");

        return @this._int;
    }

    public static implicit operator iint(int other)
    {
        if(other == int.MaxValue || other == int.MinValue)
            throw new InvalidOperationException("cannot implicit convert max-value into to iint");
        return new iint(other);
    }

    public bool IsPositiveInfinity {get { return _int == int.MaxValue; } }

    public bool IsNegativeInfinity { get { return _int == int.MinValue; } }

    private iint(bool positive)
    {
        if (positive)
            _int = int.MaxValue;
        else
            _int = int.MinValue;
    }

    public static readonly iint PositiveInfinity = new iint(true);

    public static readonly iint NegativeInfinity = new iint(false);

    public static bool operator ==(iint a, iint b)
    {
        return a._int == b._int;
    }

    public static bool operator !=(iint a, iint b)
    {
        return a._int != b._int;
    }

    public static iint operator +(iint a, iint b)
    {
        if (a.IsPositiveInfinity && b.IsNegativeInfinity)
            throw new InvalidOperationException();
        if (b.IsPositiveInfinity && a.IsNegativeInfinity)
            throw new InvalidOperationException();
        if (a.IsPositiveInfinity)
            return PositiveInfinity;
        if (a.IsNegativeInfinity)
            return NegativeInfinity;
        if (b.IsPositiveInfinity)
            return PositiveInfinity;
        if (b.IsNegativeInfinity)
            return NegativeInfinity;

        return a._int + b._int;
    }

    public static iint operator -(iint a, iint b)
    {
        if (a.IsPositiveInfinity && b.IsPositiveInfinity)
            throw new InvalidOperationException();
        if (a.IsNegativeInfinity && b.IsNegativeInfinity)
            throw new InvalidOperationException();
        if (a.IsPositiveInfinity)
            return PositiveInfinity;
        if (a.IsNegativeInfinity)
            return NegativeInfinity;
        if (b.IsPositiveInfinity)
            return NegativeInfinity;
        if (b.IsNegativeInfinity)
            return PositiveInfinity;

        return a._int - b._int;
    }

    public static iint operator -(iint a)
    {
        if (a.IsNegativeInfinity)
            return PositiveInfinity;
        if (a.IsPositiveInfinity)
            return NegativeInfinity;

        return -a;
    }

    /* etc... */
    /* other operators here */
}

Answer 3

您的 API 可以使用约定，int.MaxValue 表示正无穷值，int.MinValue - 负无穷。

但是您仍然需要在某个地方记录它，并且您可能需要对无限整数进行一些操作：

 /// <summary>
/// Making int infinity
/// ...
/// </summary>
public static class IntExtension
{

    public const int PositiveInfinity = int.MaxValue;

    public const int NegativeInfinity = int.MinValue;

    public static bool IsPositiveInfinity(this int x)
    {
        return x == PositiveInfinity;
    }

    public static bool IsNegativeInfinity(this int x)
    {
        return x == NegativeInfinity;
    }

    public static int Operation(this int x, int y)
    {
        // ...

        return PositiveInfinity;
    }
}

Answer 4

另一个部分实现（我看到 Jack 更快）：

struct InfinityInt
{
  readonly int Value;

  InfinityInt(int value, bool allowInfinities)
  {
    if (!allowInfinities && (value == int.MinValue || value == int.MaxValue))
      throw new ArgumentOutOfRangeException("value");
    Value = value;
  }

  public InfinityInt(int value)
    : this(value, false)
  {
  }

  public static InfinityInt PositiveInfinity = new InfinityInt(int.MaxValue, true);

  public static InfinityInt NegativeInfinity = new InfinityInt(int.MinValue, true);

  public bool IsAnInfinity
  {
    get { return Value == int.MaxValue || Value == int.MinValue; }
  }

  public override string ToString()
  {
    if (Value == int.MinValue)
      return double.NegativeInfinity.ToString();
    if (Value == int.MaxValue)
      return double.PositiveInfinity.ToString();

    return Value.ToString();
  }

  public static explicit operator int(InfinityInt ii)
  {
    if (ii.IsAnInfinity)
      throw new OverflowException();
    return ii.Value;
  }
  public static explicit operator double(InfinityInt ii)
  {
    if (ii.Value == int.MinValue)
      return double.NegativeInfinity;
    if (ii.Value == int.MaxValue)
      return double.PositiveInfinity;

    return ii.Value;
  }
  public static explicit operator InfinityInt(int i)
  {
    return new InfinityInt(i); // can throw
  }
  public static explicit operator InfinityInt(double d)
  {
    if (double.IsNaN(d))
      throw new ArgumentException("NaN not supported", "d");
    if (d >= int.MaxValue)
      return PositiveInfinity;
    if (d <= int.MinValue)
      return NegativeInfinity;

    return new InfinityInt((int)d);
  }

  static InfinityInt FromLongSafely(long x)
  {
    if (x >= int.MaxValue)
      return PositiveInfinity;
    if (x <= int.MinValue)
      return NegativeInfinity;

    return new InfinityInt((int)x);
  }

  public static InfinityInt operator +(InfinityInt a, InfinityInt b)
  {
    if (a.IsAnInfinity || b.IsAnInfinity)
    {
      if (!b.IsAnInfinity)
        return a;
      if (!a.IsAnInfinity)
        return b;
      if (a.Value == b.Value)
        return a;

      throw new ArithmeticException("Undefined");
    }
    return FromLongSafely((long)a.Value + (long)b.Value);
  }
  public static InfinityInt operator *(InfinityInt a, InfinityInt b)
  {
    if (a.IsAnInfinity || b.IsAnInfinity)
    {
      if (a.Value == 0 || b.Value == 0)
        throw new ArithmeticException("Undefined");

      return (a.Value > 0) == (b.Value > 0) ? PositiveInfinity : NegativeInfinity;
    }
    return FromLongSafely((long)a.Value * (long)b.Value);
  }

  // and so on, and so on
}

Answer 5

C# 有一个类型，BigInteger 类是无限大小的

http://msdn.microsoft.com/en-us/library/system.numerics.biginteger.aspx

如果您希望该类具有无穷大的表示形式——然后将 BigInteger 包装在一个为其提供无穷大标志的类中。

您必须重新定义所有标准运算符和转换才能使其正常工作。

究竟如何对无穷大工作进行操作取决于您的领域。

（例如，在某些形式的数学中，您想要 2 x infinity = infinity，而在某些形式中您不想要）。

如何实现细节实际上取决于您的域问题，并且从您的问题中不清楚。