协方差和上溯造型的区别

问题描述

什么是协方差和上溯造型，或者，之间的差异更确切地说，他们为什么要给予不同的名字呢？

我见过被称为向上转型下面的例子：

 字符串s =你好;
对象o = S; //向上转型为'字符串'到'对象'
 

然而，下面我见过所谓的'协'：

 的String [] S =新的字符串[100];
[对象] O = S;IEnumerable的＆LT;串GT; IES =新的List＆LT;串GT;（）;
IEnumerable的＆LT;对象＆gt; IEO = IES;
 

现在，我的未经训练的眼，协方差似乎是一样向上转型，除了它是指集合的铸造。 （和类似的声明，可制成关于逆变和向下转换）。

是不是真的那么简单吗？

解决方案

Covariance isn't about upcasting, although I can see why you think it's related.

Covariance is about the following very simple idea. Let's say you have a variable derivedSequence of type IEnumerable<Derived>. Let's say you have a variable baseSequence of type IEnumerable<Base>. Here, Derived derives from Base. Then, with covariance, the following is a legal assignment, and an implicit reference conversion occurs:

baseSequence = derivedSequence;

Note that this is not upcasting. It is not the case that IEnumerable<Derived> derives from IEnumerable<Base>. Rather, it is covariance that allows you to assign the value of the variable derivedSequence to the variable baseSequence. The idea is that variables of type Base can be assigned from objects of type Derived, and since IEnumerable<T> is covariant in its parameter, objects of type IEnumerable<Derived> can be assigned to variables of type IEnumerable<Base>.

Of course, I haven't yet really explained what covariance is. In general, covariance is about the following simple idea. Let's say you have a mapping F from types to types (I'll denote this mapping by F<T>; given a type T its image under the mapping F is F<T>.) Let's say that this mapping has the following very special property:

In this case, we say that F is covariant in its parameter T. (Here, to say that "A is assignment compatible with B" where A and B are reference types means that instances of B can be stored in variables of type A.)

In our case, IEnumerable<T> in C# 4.0, an implicit reference conversion from instances of IEnumerable<Derived> to IEnumerable<Base> if Derived is derived from Base. The direction of assignment compatibility is preserved, and this is why we say that IEnumerable<T> is covariant in its type parameter.

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