此代码应在Scala中编译:
trait Pipe {
type Input
type Output
def apply(input: Input): Output
}
object Pipe {
trait Start extends Pipe {
override type Input = Seq[String]
}
abstract class Connect(val prev: Pipe) extends Pipe {
override type Input = prev.Output
}
}
object Pipe1 extends Pipe.Start {
override type Output = Int
override def apply(input: Input): Output =
input.length
}
object Pipe2 extends Pipe.Connect(prev = Pipe1) {
override type Output = Boolean
override def apply(input: Input): Output =
input%2 == 0
}
Pipe1
编译好,但Pipe2
无法编译:
value % is not a member of Pipe2.this.Input
input%2 == 0
^
我知道我可以用泛型而不是依赖类型解决这个问题,但这应该起作用Pipe2.Input
应该从Int
得到Pipe1.Output
在构造函数的调用中prev = Pipe
的东西不是一个正确的路径,编译器不能将任何类型信息绑定到那个,所以你最终得到一个相当无用的prev.Output =:= Input
,用于某些不确定的prev: Pipe
,它已被设置为构造函数中的某些东西。
通过最小的更改,它按预期工作:
trait Pipe {
type Input
type Output
def apply(input: Input): Output
}
object Pipe {
trait Start extends Pipe {
override type Input = Seq[String]
}
abstract class Connect extends Pipe {
val prev: Pipe
override type Input = prev.Output
}
}
object Pipe1 extends Pipe.Start {
override type Output = Int
override def apply(input: Input): Output =
input.length
}
object Pipe2 extends Pipe.Connect {
val prev = Pipe1
override type Output = Boolean
override def apply(input: Input): Output = input % 2 == 0
}
这就是为什么它被称为路径依赖(不依赖于成员,而不依赖于值等)。
@ Andrey-Tyukin的回答如上所述。我也找到了这个工作:
trait Pipe {
type Input
type Output
def apply(input: Input): Output
}
object Pipe {
trait Start extends Pipe {
override type Input = Seq[String]
}
abstract class Connect[O](val prev: Pipe.Emitting[O]) extends Pipe {
override type Input = O
}
type Emitting[O] = Pipe {type Output = O}
}
object Pipe1 extends Pipe.Start {
override type Output = Int
override def apply(input: Input): Output =
input.length
}
object Pipe2 extends Pipe.Connect(prev = Pipe1) {
override type Output = Boolean
override def apply(input: Input): Output =
input%2 == 0
}