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
}
答案 0 :(得分:8)
我建议如下:
double
。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。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。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 强>
您可能希望实施的所有转换方法:
明确的MoneyAmount - &gt;十进制
隐式int - &gt; MoneyAmount
您可能想要实施的所有数学运算:
您可能想要实施的所有关系操作:
通过所有这些操作,您可以了解所有基础知识。
答案 1 :(得分:3)
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。
请考虑您可能希望对MoneyAmount
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));
}
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)