为什么在嵌套类型参数时 Scala 不能完全推断类型参数? [英] Why doesn't Scala fully infer type parameters when type parameters are nested?

问题描述

考虑以下 Scala 代码:

Consider the following Scala code:

abstract class A
abstract class B[T <: A]
class ConcreteA extends A
class ConcreteB extends B[ConcreteA]

class Example[U <: B[T], T <: A]( resolver: U )
object Test {
    new Example( new ConcreteB )
}

最后一行 new Example( new ConcreteB ) 编译失败，错误如下:

The last line new Example( new ConcreteB ) fails to compile with the following error:

错误:推断的类型参数 [ConcreteB,Nothing] 不符合类 Example 的类型参数边界 [U <: B[T],T <: A]

error: inferred type arguments [ConcreteB,Nothing] do not conform to class Example's type parameter bounds [U <: B[T],T <: A]

但是 ConcreteB 拥有解析 U 和 T 所需的所有数据.我在这里遗漏了什么?

But ConcreteB has all the necessary data to resolve both U and T. What am I missing here?

推荐答案

Kipton 接近了他的高级解决方案.不幸的是，他被 Scala 中的一个 bug 绊倒了 <2.9.1.RC1.以下在 2.9.1.RC1 和主干上按预期工作，

Kipton got close with his higher-kinded solution. Unfortunately he tripped over what appears to be a bug in Scala < 2.9.1.RC1. The following works as expected with 2.9.1.RC1 and trunk,

Welcome to Scala version 2.9.1.RC1 (Java HotSpot(TM) Server VM, Java 1.7.0).
Type in expressions to have them evaluated.
Type :help for more information.

scala> abstract class A
defined class A

scala> abstract class B[T <: A]
defined class B

scala> class ConcreteA extends A
defined class ConcreteA

scala> class ConcreteB[T <: A] extends B[T]
defined class ConcreteB

scala> class Example[T <: A, U[X <: A] <: B[X]](resolver: U[T])
defined class Example

scala> new Example(new ConcreteB[ConcreteA])
res0: Example[ConcreteA,ConcreteB] = Example@ec48e7

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