在多个类型的特征实例中键入方差

Question

考虑

trait Foo[+A, +B]

trait Foo1[+A] extends Foo[A, Nothing]

trait Foo2[+B] extends Foo[Nothing, B]

---

new Foo1[String] with Foo2[Nothing] 作品.

new Foo1[Nothing] with Foo2[String] 作品.

new Foo1[String] with Foo2[String] 不会:

error: illegal inheritance;
<$anon: Foo1[String] with Foo2[String]> inherits different type instances of trait Foo:
Foo[Nothing,String] and Foo[String,Nothing]
          new Foo1[String] with Foo2[String]
              ^

---

似乎在第一种情况下,实例是一个子类型Foo[String, Nothing].

同样,似乎在第二种情况下,实例应该是子类型Foo[String, String].

这两个之间有什么区别只会导致一个编译？

Answer 1

如果您看到编译器实际上解释得很好:

: Foo1[String] with Foo2[String]> inherits different type instances of trait Foo:
Foo[Nothing,String] and Foo[String,Nothing]
              new Foo1[String] with Foo2[String]

解释第一个:

new Foo1[String] with Foo2[Nothing]

看看这个例子:

scala> val x = new Foo1[String] with Foo2[Nothing]
x: Foo1[String] with Foo2[Nothing] = $anon$1@58651fd0

scala> val y:Foo[String, Nothing] = x
y: Foo[String,Nothing] = $anon$1@58651fd0

根据规范,多个with特征的实例化从左到右发生.所以首先Foo1实例化.做new Foo1[String]给你Foo[String, Nothing].with Foo2[Nothing]给你Foo[Nothing, Nothing].现在Foo是它的第一个类型参数的共变体.简而言之,这是有效的:

scala> val a = new Foo[Nothing, Nothing]{}
a: Foo[Nothing,Nothing] = $anon$1@134593bf

scala> val b:Foo[String, Nothing] = a
b: Foo[String,Nothing] = $anon$1@134593bf

因此,您可以使用Foo [String,Nothing]而不是Foo [Nothing,Nothing].这使您能够实例化y.

的情况下:

new Foo1[String] with Foo2[String]

new Foo1[String]给Foo[String, Nothing].with Foo2[String]给Foo[Nothing, String].而且它们都是矛盾的(由于它的第二个参数:

with Foo2[String] (Foo [Nothing,String])变为Foo [String,String].

scala> val p :Foo[String, String] = new Foo[Nothing, String]{}
p: Foo[String,String] = $anon$1@39529185

但是Foo [String,String] 不能成为Foo [String,Nothing].(val p :Foo[String, Nothing] = new Foo[Nothing, String]{}失败)

因此错误.

In-Variant 如果Foo的第一个参数不是in-variant,那么它不起作用:

scala> trait Foo[A, +B]
defined trait Foo

scala> trait Foo1[A] extends Foo[A, Nothing]
defined trait Foo1

scala> trait Foo2[+B] extends Foo[Nothing, B]
defined trait Foo2

scala> new Foo1[String] with Foo2[Nothing]
<console>:11: error: illegal inheritance;
 <$anon: Foo1[String] with Foo2[Nothing]> inherits different type instances of trait Foo:
Foo[Nothing,Nothing] and Foo[String,Nothing]
              new Foo1[String] with Foo2[Nothing]

Contra-Variance 如果Foo在第二种类型上具有反变量,则两种语句都有效

scala> trait Foo[+A, -B]
defined trait Foo

scala> trait Foo1[+A] extends Foo[A, Nothing]
defined trait Foo1

scala> trait Foo2[-B] extends Foo[Nothing, B]
defined trait Foo2

scala> new Foo1[String] with Foo2[String]
res0: Foo1[String] with Foo2[String] = $anon$1@77468bd9

scala> new Foo1[String] with Foo2[Nothing]
res1: Foo1[String] with Foo2[Nothing] = $anon$1@51016012

它只是起作用,因为现在:Foo [String,String] 可以成为Foo [String,Nothing]