协方差与放大器之间的差异;孔特拉方差 [英] Difference between Covariance & Contra-variance
问题描述
我无法理解协变和逆变的区别。
I am having trouble understanding the difference between covariance and contravariance.
推荐答案
现在的问题是什么是协变和逆变的区别吗?
The question is "what is the difference between covariance and contravariance?"
协变和逆变是属性的的关联与其他一组的一个成员的映射函数。更具体地,映射可以是协变或逆变相对于关系的上组
Covariance and contravariance are properties of a mapping function that associates one member of a set with another. More specifically, a mapping can be covariant or contravariant with respect to a relation on that set.
考虑该组所有的C#类型的以下两个子集。第一:
Consider the following two subsets of the set of all C# types. First:
{ Animal,
Tiger,
Fruit,
Banana }.
第二,这显然与集:
And second, this clearly related set:
{ IEnumerable<Animal>,
IEnumerable<Tiger>,
IEnumerable<Fruit>,
IEnumerable<Banana> }
有一个的映射的从第一组到第二组操作。也就是说,每个吨的第一盘的对应的的在第二盘类型为的IEnumerable&LT; T&GT;
。或者,在短表,映射 T→IE&LT; T&GT;
There is a mapping operation from the first set to the second set. That is, for each T in the first set, the corresponding type in the second set is IEnumerable<T>
. Or, in short form, the mapping is T → IE<T>
.
随着我这么远吗?
现在让我们考虑一个的关系的。有一个在第一组的的分配兼容性关系的对类型之间。类型的值虎
可分配给类型的变量动物
,所以这些类型的被说成是分配兼容。让我们写类型的值 X
可分配给类型的变量是
在较短的形式:< code系列> X⇒是。因此,在我们的第一个子集,这里有所有的分配兼容性关系:
Now let's consider a relation. There is an assignment compatibility relationship between pairs of types in the first set. A value of type Tiger
can be assigned to a variable of type Animal
, so these types are said to be "assignment compatible". Let's write "a value of type X
can be assigned to a variable of type Y
" in a shorter form: X ⇒ Y
. So in our first subset, here are all the assignment compatibility relationships:
Tiger ⇒ Tiger
Tiger ⇒ Animal
Animal ⇒ Animal
Banana ⇒ Banana
Banana ⇒ Fruit
Fruit ⇒ Fruit
在C#4,它支持某些接口的协变分配兼容性,有一个在第二组对的类型之间的分配兼容性关系:
In C# 4, which supports covariant assignment compatibility of certain interfaces, there is an assignment compatibility relationship between pairs of types in the second set:
IE<Tiger> ⇒ IE<Tiger>
IE<Tiger> ⇒ IE<Animal>
IE<Animal> ⇒ IE<Animal>
IE<Banana> ⇒ IE<Banana>
IE<Banana> ⇒ IE<Fruit>
IE<Fruit> ⇒ IE<Fruit>
注意,映射 T→IE&LT; T&GT;
的 preserves的存在和分配兼容性的方向的。也就是说,如果 X⇒是
,那么它也确实 IE&LT; X&GT; ⇒IE&LT; Y方式&gt;
Notice that the mapping T → IE<T>
preserves the existence and direction of assignment compatibility. That is, if X ⇒ Y
, then it is also true that IE<X> ⇒ IE<Y>
.
具有这种属性相对于一特定的关系的映射被称为协变映射。这应该是有意义:可以在需要的动物的序列中使用老虎的序列,但相反的是不正确的。动物的序列可以在需要老虎的顺序不一定使用。
A mapping which has this property with respect to a particular relation is called a "covariant mapping". This should make sense: a sequence of Tigers can be used where a sequence of Animals is needed, but the opposite is not true. A sequence of animals cannot necessarily be used where a sequence of Tigers is needed.
这就是协方差。现在考虑集合所有类型的子集:
That's covariance. Now consider this subset of the set of all types:
{ IComparable<Tiger>,
IComparable<Animal>,
IComparable<Fruit>,
IComparable<Banana> }
现在我们从第一套到第三套映射 T→IC标签; T&GT;
now we have the mapping from the first set to the third set T → IC<T>
.
在C#4:
IC<Tiger> ⇒ IC<Tiger>
IC<Animal> ⇒ IC<Tiger> Backwards!
IC<Animal> ⇒ IC<Animal>
IC<Banana> ⇒ IC<Banana>
IC<Fruit> ⇒ IC<Banana> Backwards!
IC<Fruit> ⇒ IC<Fruit>
也就是说,映射 T→IC标签; T&GT;
拥有的 preserved的存在,但扭转分配兼容性的方向的。也就是说,如果 X⇒是
,那么 IC标签; X&GT; ⇐IC标签; Y方式&gt;
That is, the mapping T → IC<T>
has preserved the existence but reversed the direction of assignment compatibility. That is, if X ⇒ Y
, then IC<X> ⇐ IC<Y>
.
其中的的映射preserves但挫折的关系被称为逆变的映射。
A mapping which preserves but reverses a relation is called a contravariant mapping.
此外,这应该是显然是正确的。它可以比较两个动物的设备也可以比较两只老虎,但可以比较两个老虎的设备不一定能比较任意两个动物。
Again, this should be clearly correct. A device which can compare two Animals can also compare two Tigers, but a device which can compare two Tigers cannot necessarily compare any two Animals.
这就是在C#4方差协变和逆变的区别的 preserves 的转让的方向。逆变的逆转的吧。
So that's the difference between covariance and contravariance in C# 4. Covariance preserves the direction of assignability. Contravariance reverses it.
这篇关于协方差与放大器之间的差异;孔特拉方差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!