参数类型不是反变体吗？ [英] Isn't the argument type co- not contra-variant?

问题描述

我理解术语协方差和反方差。但有一件小事我无法理解。在课程中的Scala中的函数式编程课程中，Martin Ordersky提到：

I understand the terms co-variance and contra-variance. But there is one small thing I am unable to understand. In the course "Functional Programming in Scala" on coursera, Martin Ordersky mentions that:

函数在其参数类型和共变量中是逆变的在
中他们的返回类型

Functions are contravariant in their argument types and co-variant in their return types

所以例如在Java中，让 Dog extends Animal 。让函数成为：

So for example in Java, let Dog extends Animal. And let a function be :

void getSomething(Animal a){

我有函数调用

Dog d = new Dog();
getSomething(d)

所以基本上发生的事情是动物a = d 。根据 wiki ，协方差是将范围扩大到更窄。而且我们正在从狗转变为动物。 SO不是参数类型协变而不是逆变？

So basically what is happeneing is that Animal a = d. And according to wiki covariance is "Converting wider to narrow". And above we are converting from dog to Animal. SO isnt the argument type covariant rather than contravariant?

推荐答案

这是在Scala中定义函数的方式：

trait Function1 [-T1, +R]  extends AnyRef

英文，参数 T1 是逆变的，结果类型 R 是协变的。这是什么意思？

In English, parameter T1 is contravariant and result type R is covariant. What does it mean?

当某段代码需要 Dog =>的函数时动物类型，你可以提供 Animal =>的函数。动物类型，这要归功于参数的逆转（你可以使用更广泛的类型）。

When some piece of code requires a function of Dog => Animal type, you can supply a function of Animal => Animal type, thanks to contravariance of parameter (you can use broader type).

你还可以提供的功能狗=>狗类型，感谢结果类型的协方差（你可以使用更窄的类型）。

Also you can supply function of Dog => Dog type, thanks to covariance of result type (you can use narrower type).

这实际上是有道理的：有人想要一个函数来转换狗对任何动物。您可以提供转换任何动物（包括狗）的功能。此外，您的功能只能返回狗，但狗仍然是动物。

This actually makes sense: someone wants a function to transform dog to any animal. You can supply a function that transforms any animal (including dogs). Also your function can return only dogs, but dogs are still animals.

这篇关于参数类型不是反变体吗？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！