有人能告诉我是否有一种方法可以用泛型将泛型类型参数T限制为:
Int16 Int32 Int64 UInt16 UInt32 UInt64
我知道哪里关键字,但不能找到只有这些类型的接口,
喜欢的东西:
static bool IntegerFunction<T>(T value) where T : INumeric
有人能告诉我是否有一种方法可以用泛型将泛型类型参数T限制为:
Int16 Int32 Int64 UInt16 UInt32 UInt64
我知道哪里关键字,但不能找到只有这些类型的接口,
喜欢的东西:
static bool IntegerFunction<T>(T value) where T : INumeric
当前回答
.NET 6有一个预览功能:
https://devblogs.microsoft.com/dotnet/preview-features-in-net-6-generic-math/#generic-math
下面是文章中的一个例子:
static T Add<T>(T left, T right)
where T : INumber<T>
{
return left + right;
}
INumber是一个实现其他接口的接口,比如IAdditionOperators,它允许通用的+用法。现在这是可能的,因为另一个预览特性是接口中的静态抽象,因为+操作符重载是一个静态方法:
/// <summary>Defines a mechanism for computing the sum of two values.</summary>
/// <typeparam name="TSelf">The type that implements this interface.</typeparam>
/// <typeparam name="TOther">The type that will be added to <typeparamref name="TSelf" />.</typeparam>
/// <typeparam name="TResult">The type that contains the sum of <typeparamref name="TSelf" /> and <typeparamref name="TOther" />.</typeparam>
[RequiresPreviewFeatures(Number.PreviewFeatureMessage, Url = Number.PreviewFeatureUrl)]
public interface IAdditionOperators<TSelf, TOther, TResult>
where TSelf : IAdditionOperators<TSelf, TOther, TResult>
{
/// <summary>Adds two values together to compute their sum.</summary>
/// <param name="left">The value to which <paramref name="right" /> is added.</param>
/// <param name="right">The value which is added to <paramref name="left" />.</param>
/// <returns>The sum of <paramref name="left" /> and <paramref name="right" />.</returns>
static abstract TResult operator +(TSelf left, TOther right);
}
其他回答
不幸的是,在这种情况下,只能在where子句中指定struct。不能具体指定Int16、Int32等,这看起来确实很奇怪,但我相信,在决定不允许在where子句中使用值类型的基础上,有一些深层的实现原因。
我想唯一的解决方案是执行运行时检查,这不幸地阻止了在编译时拾取问题。大概是这样的:-
static bool IntegerFunction<T>(T value) where T : struct {
if (typeof(T) != typeof(Int16) &&
typeof(T) != typeof(Int32) &&
typeof(T) != typeof(Int64) &&
typeof(T) != typeof(UInt16) &&
typeof(T) != typeof(UInt32) &&
typeof(T) != typeof(UInt64)) {
throw new ArgumentException(
string.Format("Type '{0}' is not valid.", typeof(T).ToString()));
}
// Rest of code...
}
我知道这有点难看,但至少提供了所需的约束条件。
我还将研究此实现可能的性能影响,也许有更快的方法。
我会使用一个通用的,你可以处理外部…
/// <summary>
/// Generic object copy of the same type
/// </summary>
/// <typeparam name="T">The type of object to copy</typeparam>
/// <param name="ObjectSource">The source object to copy</param>
public T CopyObject<T>(T ObjectSource)
{
T NewObject = System.Activator.CreateInstance<T>();
foreach (PropertyInfo p in ObjectSource.GetType().GetProperties())
NewObject.GetType().GetProperty(p.Name).SetValue(NewObject, p.GetValue(ObjectSource, null), null);
return NewObject;
}
.NET 6有一个预览功能:
https://devblogs.microsoft.com/dotnet/preview-features-in-net-6-generic-math/#generic-math
下面是文章中的一个例子:
static T Add<T>(T left, T right)
where T : INumber<T>
{
return left + right;
}
INumber是一个实现其他接口的接口,比如IAdditionOperators,它允许通用的+用法。现在这是可能的,因为另一个预览特性是接口中的静态抽象,因为+操作符重载是一个静态方法:
/// <summary>Defines a mechanism for computing the sum of two values.</summary>
/// <typeparam name="TSelf">The type that implements this interface.</typeparam>
/// <typeparam name="TOther">The type that will be added to <typeparamref name="TSelf" />.</typeparam>
/// <typeparam name="TResult">The type that contains the sum of <typeparamref name="TSelf" /> and <typeparamref name="TOther" />.</typeparam>
[RequiresPreviewFeatures(Number.PreviewFeatureMessage, Url = Number.PreviewFeatureUrl)]
public interface IAdditionOperators<TSelf, TOther, TResult>
where TSelf : IAdditionOperators<TSelf, TOther, TResult>
{
/// <summary>Adds two values together to compute their sum.</summary>
/// <param name="left">The value to which <paramref name="right" /> is added.</param>
/// <param name="right">The value which is added to <paramref name="left" />.</param>
/// <returns>The sum of <paramref name="left" /> and <paramref name="right" />.</returns>
static abstract TResult operator +(TSelf left, TOther right);
}
我想知道和samjudson一样,为什么只对整数有效?如果是这样的话,你可能想要创建一个helper类或者类似的东西来保存你想要的所有类型。
如果你想要的只是整数,不要使用泛型,那不是泛型;或者更好的是,通过检查其类型来拒绝任何其他类型。
使用策略的解决方案:
interface INumericPolicy<T>
{
T Zero();
T Add(T a, T b);
// add more functions here, such as multiplication etc.
}
struct NumericPolicies:
INumericPolicy<int>,
INumericPolicy<long>
// add more INumericPolicy<> for different numeric types.
{
int INumericPolicy<int>.Zero() { return 0; }
long INumericPolicy<long>.Zero() { return 0; }
int INumericPolicy<int>.Add(int a, int b) { return a + b; }
long INumericPolicy<long>.Add(long a, long b) { return a + b; }
// implement all functions from INumericPolicy<> interfaces.
public static NumericPolicies Instance = new NumericPolicies();
}
算法:
static class Algorithms
{
public static T Sum<P, T>(this P p, params T[] a)
where P: INumericPolicy<T>
{
var r = p.Zero();
foreach(var i in a)
{
r = p.Add(r, i);
}
return r;
}
}
用法:
int i = NumericPolicies.Instance.Sum(1, 2, 3, 4, 5);
long l = NumericPolicies.Instance.Sum(1L, 2, 3, 4, 5);
NumericPolicies.Instance.Sum("www", "") // compile-time error.
该解决方案是编译时安全的。CityLizard框架提供了。net 4.0的编译版本。文件名为“lib/NETFramework4.0/ citylizad . policy .dll”。
它也可以在Nuget: https://www.nuget.org/packages/CityLizard/。参见CityLizard.Policy.I结构。