[英]c# creating a custom “double” type
In my application I want all my properties storing money amounts to be rounded to n
decimal places. 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。
For code clarity, I'd rather have a custom type MoneyAmount
which all my corresponding fields would have, instead of having to put a `Math.Round(value, n)' in all the property getters/setters. 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)'。
Is there a neat way to achieve this? 有没有一种巧妙的方法来实现这一目标?
I saw this post about overloading assignment operators - is this the suggested approach? 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?
EDIT: Given the multiple views, I post the full code I derived here: 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
}
I'd suggest the following: 我建议如下:
double
. double
。 double
parameter, this constructor rounds the value and assigns it to the internal field. 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。 double
, like +, -, etc. But also casts/conversions from/to other types. double
相同的操作,如+, - 等。还可以从/向其他类型转换/转换。 Every operation produces a new instance of MoneyAmount with a rounded value. IFormattable
, IComparable
and IConvertible
. IFormattable
, IComparable
和IConvertible
接口。 Short example: 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...
}
I realize: The double
has a lot of members... 我意识到:
double
有很多成员......
With these operators, the following code is possible: 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.
EDIT 编辑
All conversion methods you may want to implement: 您可能想要实现的所有转换方法:
Explicit MoneyAmount --> decimal 显式MoneyAmount - >十进制
Implicit int--> MoneyAmount 隐式int - > MoneyAmount
All mathematical operations you may want to implement: 您可能想要实现的所有数学运算:
All relational operations you may want to implement: 您可能想要实现的所有关系操作:
With all these operations your have all basics covered. 通过所有这些操作,您可以了解所有基础知识。
This gets big very quickly. 这很快变大了。 Writing a struct is easy, as demonstrated in @MartinMulder's answer, but consider that you will want to overload a number of combinations of operators, as well as including a few implicit/explicit casts as well.
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。
Consider that you may want to do mathematical operations on MoneyAmount
考虑您可能想要在
MoneyAmount
上进行数学运算
MoneyAmount
+ MoneyAmount
MoneyAmount
+ MoneyAmount
MoneyAmount
+ double
MoneyAmount
+ double
MoneyAmount
+ int
MoneyAmount
+ int
MoneyAmount
+ decimal
MoneyAmount
+ decimal
That is 4 overloads of the +
operator. 这是
+
运算符的4次重载。 Rinse and repeat for -
, /
, *
(and possibly %
). 冲洗并重复
-
, /
, *
(可能为%
)。 You'll also want to overload <
, <=
, ==
and >
, >=
. 您还需要重载
<
, <=
, ==
和>
, >=
。 Thats something like 30 operator overloads. 这就像30个运算符重载。 Phew!
唷! Thats a lot of static methods.
这是很多静态方法。
public static MoneyAmount operator +(MoneyAmount d1, double d2)
{
return new MoneyAmount((decimal)(d1._value + d2));
}
Now consider that instead of this code 现在考虑代替这个代码
MoneyAmount m = new MoneyAmount(1.234);
You wanted to do this: 你想这样做:
MoneyAmount m = 1.234;
That can be achieved with an implicit cast operator. 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);
}
(You'll need one for every type you want to allow implicit casts) (你想要允许隐式转换的每种类型都需要一个)
Another one: 另一个:
int i = 4;
MoneyAmount m = (MoneyAmount)i;
This is done with an explicit cast operator overload. 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);
}
(Again, 1 for every type you want to allow explicit casts) (同样,对于您希望允许显式转换的每种类型都为1)
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.