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小数位。

为了清晰代码，我宁愿拥有一个自定义类型MoneyAmount，我所有相应的字段都有，而不必在所有属性getter /中放入一个`Math.Round（value，n）' setter方法。

有没有一种巧妙的方法来实现这一目标？

我看到this post关于重载赋值运算符 - 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：

创建一个名为MoneyAmount的新结构。
它包含一个字段：A double。
具有一个double参数的构造函数，此构造函数对该值进行舍入并将其分配给内部字段。
将您可能需要的成员/运算符添加到结构中，使其具有与double完全相同的操作，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的新实例。
还考虑实施接口IFormattable，IComparable和IConvertible。

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.

<强> 修改

您可能希望实施的所有转换方法：

明确的MoneyAmount - ＆gt; int
明确的MoneyAmount - ＆gt;浮
明确的MoneyAmount - ＆gt;双
明确的MoneyAmount - ＆gt;十进制
隐式int - ＆gt; MoneyAmount
隐式浮动 - ＆gt; MoneyAmount
隐含双重 - ＆gt; MoneyAmount
隐式小数 - ＆gt; MoneyAmount

您可能想要实施的所有数学运算：

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

您可能想要实施的所有关系操作：

MoneyAmount == MoneyAmount
MoneyAmount！= MoneyAmount
MoneyAmount＆gt; MoneyAmount
MoneyAmount＆gt; = MoneyAmount
MoneyAmount＆lt; MoneyAmount
MoneyAmount＆lt; = 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的答案所示，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。

数学与数学逻辑操作

请考虑您可能希望对MoneyAmount

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-，/，*（以及可能%）。您还要重载<，<=，==和>，>=。这就像30个运算符重载。唷！那就是很多静态方法。

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

明确/隐含的演员阵容

现在考虑代替此代码

MoneyAmount m = new MoneyAmount(1.234);

你想这样做：

MoneyAmount m = 1.234;

可以使用隐式强制转换运算符来实现。

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

（对于你想允许隐式演员表的每种类型，你都需要一个）

另一个：

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

这是通过显式强制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）

c＃创建自定义“双”类型

2 个答案:

数学与数学逻辑操作

明确/隐含的演员阵容