C＃比较两个通用值 [英] c# compare two generic values

问题描述

可能重复：结果
  <一href=\"http://stackoverflow.com/questions/390900/cant-operator-be-applied-to-generic-types-in-c\">Can’t ==操作符在C＃中被应用到泛型类型？

我有codeD是这样的：

I've coded something like this:

public bool IsDataChanged()
{           
    T value1 = GetValue2;
    T value2 = GetValue1();

    return (valueInDB != valueFromView);
}

现在的功能不会错误编译的运算符'！='不能应用于类型为'T'和'T'的操作数。我有什么做的，使这项功能工作的？

Right now the function doesn't compile with the error "Operator '!=' cannot be applied to operands of type 'T' and 'T'". What do I have to do to make this function work ?

推荐答案

您不能使用泛型类型的运营商（除富== null，其特殊的套管），除非你加入其中T：类以表明它是引用类型（那么Foo就==酒吧是合法的）

You cannot use operators on generic types (except for foo == null which is special cased) unless you add where T : class to indicate it is a reference type (then foo == bar is legal)

使用 EqualityComparer＆LT; T＆GT; .DEFAULT来为你做它。这将不会在工种，只为==提供一个操作符重载也没有任何：

Use EqualityComparer<T>.Default to do it for you. This will not work on types which only supply an operator overload for == without also either:

• 执行 IEquatable＆LT; T＆GT;


• 重写Object.Equals（）

在总体实施==操作符，而不是也做的，它们中的至少一个将是一个非常不好的想法，所以这是不太可能是一个问题。

In general implementing the == operator and not also doing at least one of these would be a very bad idea anyway so this is not likely to be an issue.

public bool IsDataChanged<T>()
{           
    T value1 = GetValue2;
    T value2 = GetValue1();

    return !EqualityComparer<T>.Default.Equals(value1 , value2);
}

如果你不限制 IEquatable＆LT; T＆GT; 然后用值类型使用，如果他们不执行，当EqualityComparer默认故障可能会导致拳击 IEquatable＆LT; T＆GT; （如果你控制正在使用，这可能不是问题的类型）。我假设你正在使用=！性能虽然如此限制的泛型类型将避免偶然通过的Object.Equals拳（对象）的路线。

If you do not restrict to IEquatable<T> then the EqualityComparer default fallback may cause boxing when used with value types if they do not implement IEquatable<T> (if you control the types which are being used this may not matter). I am assuming you were using =! for performance though so restricting to the Generic type will avoid accidental boxing via the Object.Equals(object) route.

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