参数必须是输入安全错误 [英] Parameter must be input-safe error

问题描述

这是我的一部分代码：

public interface IA<in TInput>
{
    void Method(IB<TInput> entities);
}

public interface IB<in T> { }

我不知道为什么会出现以下编译错误：
Parameter无效的差异：类型参数| TInput |必须在 IB< IB <

I can't figure out why I get following compile error: "Parameter must be input-safe. Invalid variance: The type parameter |TInput| must be contravariantly valid on "IB< in T>".

任何帮助将不胜感激。

推荐答案

在C＃中使用逆变的指示符（即在$code>中的）在您使用接受泛型类型的参数的方法时才直观。但是，自变量表示关系的反演（问题与解答附有解释），因此使用 IA 中的 in 使其与 IB 不兼容。

The designator of contravariance in C# (i.e. in) is intuitive only at the immediate level, when you make a method that "takes in" a parameter of generic type. Internally, however, contravariance means an inversion of a relation (Q&A with an explanation) so using in inside IA makes it incompatible with IB.

问题最好用一个例子来说明。考虑类 Animal 及其派生类 Tiger 。我们还假设 IB< T> 具有方法 void MethodB（T input） ，这是从 IA 的方法：

The problem is best illustrated with an example. Consider class Animal and its derived class Tiger. Let's also assume that IB<T> has a method void MethodB(T input), which is called from IA's Method:

class A_Impl<T> : IA<T> {
    T data;
    public void Method(IB<TInput> entities) {
        entities.MethodB(data);
    }
}

在TInput中声明 IA< ; 和 IB< TInput> 表示您可以做到

Declaring IA<in TInput> and IB<in TInput> means that you can do

IA<Animal> aForAnimals = new A_Impl<Animal>();
IA<Tiger> aForTigers = aForAnimals;

IA< TInput> 有一个方法它需要 IB< TInput> ，我们可以这样称呼它：

IA<in TInput> has a method that takes IB<TInput>, which we can call like this:

aForTigers.Method(new B_Impl<Tiger>());

这是一个问题，因为现在 A_Impl< Animal> 将 Animal 传递给需要 Tiger MethodB c $ c>。

This is a problem, because now A_Impl<Animal> passes an Animal to MethodB of an interface that expects a Tiger.

尽管 IB< out T> 您不会有问题-两者均具有协方差和协变：

You would have no problem with IB<out T>, though - both with covariance and contravariance:

public interface IB<out T> {
//                  ^^^
}
// This works
public interface IA<in TInput> {
    void Method(IB<TInput> x);
}
// This works too
public interface IC<out TInput> {
    void Method(IB<TInput> x);
}

这篇关于参数必须是输入安全错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！