.NET泛型中的重载运算符约束的解决方案 [英] Solution for overloaded operator constraint in .NET generics

问题描述

如果我想有一个通用方法，只接受类型已经重载了一个运算符，例如减法运算符，我该怎么办。我尝试使用接口作为约束，但接口不能有操作符重载。

What would I do if I want to have a generic method that only accepts types that have overloaded an operator, for instance the subtraction operator. I tried using an interface as a constraint but interfaces can't have operator overloading.

这是最好的方法是什么？

What is the best way to achieve this?

推荐答案

没有立即回答;运算符是静态的，并且不能在约束中表示 - 并且现有的原语不实现任何特定的接口（与可以用于模拟大于/小于的模拟的[13]的对比）。

There is no immediate answer; operators are static, and cannot be expressed in constraints - and the existing primatives don't implement any specific interface (contrast to IComparable[<T>] which can be used to emulate greater-than / less-than).

但是;如果你只是想要它的工作，那么在.NET 3.5有一些选项...

However; if you just want it to work, then in .NET 3.5 there are some options...

我把一个库这里允许高效和简单地访问带有泛型的操作符，例如：

I have put together a library here that allows efficient and simple access to operators with generics - such as:

T result = Operator.Add(first, second); // implicit <T>; here

它可以作为 MiscUtil

此外，在C＃4.0中，通过 dynamic ：

Additionally, in C# 4.0, this becomes possible via dynamic:

static T Add<T>(T x, T y) {
    dynamic dx = x, dy = y;
    return dx + dy;
}

我还有一个.NET 2.0版本较少测试。另一个选项是创建一个接口，例如

I also had (at one point) a .NET 2.0 version, but that is less tested. The other option is to create an interface such as

interface ICalc<T>
{
    T Add(T,T)() 
    T Subtract(T,T)()
} 

等等，但是你需要通过所有的方法传递一个 ICalc ; 。

etc, but then you need to pass an ICalc<T>; through all the methods, which gets messy.

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