为什么这些类型参数不符合类型细化? [英] Why do these type arguments not conform to a type refinement?

问题描述

为什么这段 Scala 代码无法进行类型检查?

Why does this Scala code fail to typecheck?

trait T { type A }
trait GenFoo[A0, S <: T { type A = A0 }]
trait Foo[S <: T] extends GenFoo[S#A, S]

我不明白为什么类型参数 [S#A,S] 不符合 trait GenFoo 的类型参数边界 [A0,S <: T{type A = A0}]".有解决办法吗?

I don't understand why "type arguments [S#A,S] do not conform to trait GenFoo's type parameter bounds [A0,S <: T{type A = A0}]". Is there a work-around?

编辑:正如已经指出的，一致性错误源于未能验证S <: T{type A = S#A}.Daniel Sobral 指向 -explaintypes，它告诉我们:

Edit: As has been pointed out, the conformance error stems from the failure to verify S <: T{type A = S#A}. Daniel Sobral pointed to -explaintypes, which tells us:

S <: T{type A = S#A}?
  S <: T?
  true
  S specializes type A?
    this.A = this.A?
      S = this.type?
      false
    false
  false
false

我不知道如何解释这个.

I'm not sure how to interpret this.

注意，如果我们尝试定义，我们会得到一个非法的循环引用，

Note that we get an illegal cyclic reference if we try to define,

trait Foo[S <: T { type A = S#A } ] extends GenFoo[S#A, S]

虽然这里的类型细化似乎没有添加任何新信息.(另见 为什么这个循环引用带有类型投影非法吗?)

although the type refinement here doesn't seem to add any new information. (See also Why is this cyclic reference with a type projection illegal?)

我的动机是创建一个特性 Foo[S <: T] 专门用于 S#A，如:如何专注于 Scala 中的类型投影? 为了让它起作用，我我试图将 S#A 作为显式参数 A0 显示在实现特征 GenFoo 中，它可以直接特化.我希望将 Miles Sabin 的回答中的类型细化思想应用到 为什么这个带有类型投影的循环引用是非法的? 但我遇到了这个一致性错误.

My motivation is to create a trait Foo[S <: T] that specializes on S#A, as in: How to specialize on a type projection in Scala? To get this to work, I'm trying to surface S#A as an explicit parameter A0 in the implementation trait GenFoo, which can be specialized directly. I was hoping to apply the type refinement idea from Miles Sabin's answer to Why is this cyclic reference with a type projection illegal? but I run into this conformance error.

推荐答案

这似乎就是答案:

S 专攻 A 型?

关于specialize 的问题来自这里:T { type A = A0 }.这是带有type A specialized 的T 类型——意思是，它比原来的T 更受限制.

The question about specialize comes from here: T { type A = A0 }. This is the type T with type A specialized -- meaning, it is more restricted than the original T.

这个问题的答案是否定的——S 没有限制它是专门的.

The answer to that question is no -- there's no constrains on S that it be specialized.

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