c＃创建自定义“双”类型

Question

在我的应用程序中，我希望所有存储金额的属性都舍入到n位小数。

为了清楚代码，我宁愿拥有一个自定义类型MoneyAmount ，它可以包含所有相应字段，而不必在所有属性getter / setter中放置一个`Math.Round（value，n）'。

有没有一种巧妙的方法来实现这一目标？

我看到这篇关于重载赋值运算符的帖子 - 这是建议的方法吗？

编辑：鉴于多个视图，我发布了我在此处派生的完整代码：

public struct MoneyAmount {
const int N = 4;
private readonly double _value;

public MoneyAmount(double value) {
  _value = Math.Round(value, N);
}

#region mathematical operators
public static MoneyAmount operator +(MoneyAmount d1, MoneyAmount d2) {
  return new MoneyAmount(d1._value + d2._value);
}

public static MoneyAmount operator -(MoneyAmount d1, MoneyAmount d2) {
  return new MoneyAmount(d1._value - d2._value);
}

public static MoneyAmount operator *(MoneyAmount d1, MoneyAmount d2) {
  return new MoneyAmount(d1._value * d2._value);
}

public static MoneyAmount operator /(MoneyAmount d1, MoneyAmount d2) {
  return new MoneyAmount(d1._value / d2._value);
}
#endregion

#region logical operators
public static bool operator ==(MoneyAmount d1, MoneyAmount d2) {
  return d1._value == d2._value;
}
public static bool operator !=(MoneyAmount d1, MoneyAmount d2) {
  return d1._value != d2._value;
}
public static bool operator >(MoneyAmount d1, MoneyAmount d2) {
  return d1._value > d2._value;
}
public static bool operator >=(MoneyAmount d1, MoneyAmount d2) {
  return d1._value >= d2._value;
}
public static bool operator <(MoneyAmount d1, MoneyAmount d2) {
  return d1._value < d2._value;
}
public static bool operator <=(MoneyAmount d1, MoneyAmount d2) {
  return d1._value <= d2._value;
}
#endregion

#region Implicit conversions
/// <summary>
/// Implicit conversion from int to MoneyAmount. 
/// Implicit: No cast operator is required.
/// </summary>
public static implicit operator MoneyAmount(int value) {
  return new MoneyAmount(value);
}

/// <summary>
/// Implicit conversion from float to MoneyAmount. 
/// Implicit: No cast operator is required.
/// </summary>
public static implicit operator MoneyAmount(float value) {
  return new MoneyAmount(value);
}

/// <summary>
/// Implicit conversion from double to MoneyAmount. 
/// Implicit: No cast operator is required.
/// </summary>
public static implicit operator MoneyAmount(double value) {
  return new MoneyAmount(value);
}

/// <summary>
/// Implicit conversion from decimal to MoneyAmount. 
/// Implicit: No cast operator is required.
/// </summary>
public static implicit operator MoneyAmount(decimal value) {
  return new MoneyAmount(Convert.ToDouble(value));
}
#endregion

#region Explicit conversions
/// <summary>
/// Explicit conversion from MoneyAmount to int. 
/// Explicit: A cast operator is required.
/// </summary>
public static explicit operator int(MoneyAmount value) {
  return (int)value._value;
}

/// <summary>
/// Explicit conversion from MoneyAmount to float. 
/// Explicit: A cast operator is required.
/// </summary>
public static explicit operator float(MoneyAmount value) {
  return (float)value._value;
}

/// <summary>
/// Explicit conversion from MoneyAmount to double. 
/// Explicit: A cast operator is required.
/// </summary>
public static explicit operator double(MoneyAmount value) {
  return (double)value._value;
}

/// <summary>
/// Explicit conversion from MoneyAmount to decimal. 
/// Explicit: A cast operator is required.
/// </summary>
public static explicit operator decimal(MoneyAmount value) {
  return Convert.ToDecimal(value._value);
}
#endregion
}

Answer 1

我建议如下：

1. 创建一个名为MoneyAmount的新结构。
2. 它包含一个字段： double 。
3. 具有一个double参数的构造函数，此构造函数对值进行舍入并将其分配给内部字段。
4. 将您可能需要的成员/运算符添加到结构中，使其具有与double相同的操作，如+， - 等。还可以从/向其他类型转换/转换。 每个操作都会生成一个带有舍入值的MoneyAmount新实例。
5. 还要考虑实现IFormattable ， IComparable和IConvertible接口。

简短的例子：

public struct MoneyAmount
{
    const int N = 4;
    private readonly double _value;

    public MoneyAmount(double value)
    {
        _value = Math.Round(value, N);
    }

    // Example of one member of double:
    public static MoneyAmount operator *(MoneyAmount d1, MoneyAmount d2) 
    {
        return new MoneyAmount(d1._value * d2._value);
    }

    /// <summary>
    /// Implicit conversion from double to MoneyAmount. 
    /// Implicit: No cast operator is required.
    /// </summary>
    public static implicit operator MoneyAmount(double value)
    {
        return new MoneyAmount(value);
    }

    /// <summary>
    /// Explicit conversion from MoneyAmount to double. 
    /// Explicit: A cast operator is required.
    /// </summary>
    public static explicit operator double(MoneyAmount value)
    {
        return value._value;
    }

    /// <summary>
    /// Explicit conversion from MoneyAmount to int. 
    /// Explicit: A cast operator is required.
    /// </summary>
    public static explicit operator MoneyAmount(int value)
    {
        return new MoneyAmount(value);
    }

    /// <summary>
    /// Explicit conversion from MoneyAmount to int. 
    /// Explicit: A cast operator is required.
    /// </summary>
    public static explicit operator int(MoneyAmount value)
    {
        return (int)value._value;
    }

    // All other members here...
}

我意识到： double有很多成员......

使用这些运算符，可以使用以下代码：

MoneyAmount m = 1.50; // Assignment from a double.
MoneyAmount n = 10; // Assignment from an integer.
m += n; // Mathematical operation with another MoneyAmount .
m *= 10; // Mathematical operation with an integer.
m -= 12.50; // Mathematical operation with a double.

编辑

您可能想要实现的所有转换方法：

• 显式MoneyAmount - > int
• 显式MoneyAmount - > float
• 显式MoneyAmount - > double

• 显式MoneyAmount - >十进制

• 隐式int - > MoneyAmount

• 隐式浮点数 - > MoneyAmount
• 隐式双 - > MoneyAmount
• 隐式小数 - > MoneyAmount

您可能想要实现的所有数学运算：

• MoneyAmount + MoneyAmount
• MoneyAmount - MoneyAmount
• MoneyAmount * MoneyAmount
• MoneyAmount / MoneyAmount

您可能想要实现的所有关系操作：

• MoneyAmount == MoneyAmount
• MoneyAmount！= MoneyAmount
• MoneyAmount> MoneyAmount
• MoneyAmount> = MoneyAmount
• MoneyAmount <MoneyAmount
• MoneyAmount <= MoneyAmount

通过所有这些操作，您可以了解所有基础知识。

Answer 2

这很快变大了。 编写结构很容易，如@ MartinMulder的回答所示，但考虑到你需要重载多个运算符组合，以及包括一些隐式/显式转换。

数学和逻辑运算

考虑您可能想要在MoneyAmount上进行数学运算

• MoneyAmount + MoneyAmount
• MoneyAmount + double
• MoneyAmount + int
• MoneyAmount + decimal

这是+运算符的4次重载。 冲洗并重复- ， / ， * （可能为% ）。 您还需要重载< ， <= ， ==和> ， >= 。 这就像30个运算符重载。 唷！ 这是很多静态方法。

public static MoneyAmount operator +(MoneyAmount d1, double d2) 
{
    return new MoneyAmount((decimal)(d1._value + d2));
}

显式/隐式演员

现在考虑代替这个代码

MoneyAmount m = new MoneyAmount(1.234);

你想这样做：

MoneyAmount m = 1.234;

这可以通过隐式转换运算符来实现。

public static implicit operator MoneyAmount(double d)
{
    return new MoneyAmount((decimal)d);
}

（你想要允许隐式转换的每种类型都需要一个）

另一个：

int i = 4;
MoneyAmount m = (MoneyAmount)i;

这是通过显式转换运算符重载完成的。

public static explicit operator MoneyAmount(double d)
{
    return new MoneyAmount((decimal)d);
}

（同样，对于您希望允许显式转换的每种类型都为1）