通过方法类型参数分配给类型成员的值打破类型等效

由 Anne发布于 2020-05-18 05:28:56

为什么以下类型对等成立

trait Foo { type T }
val fa = new Foo { type T = Int }

implicitly[fa.T =:= Int] // OK

but when type member T is assigned via method parameter A then type equivalence does not hold

def makeFoo[A]: Foo = new Foo { type T = A }
val fb = makeFoo[Int]

implicitly[fb.T =:= Int] // ERROR

Intuitively I would expect if T = A and A = Int then T = Int?

尝试放纵

2020-05-18 05:28:56

This is because the return type of makeFoo is only Foo and not Foo { type T = A }. If you don't explicitly declare it or if you refine it, the error will go away.

def makeFoo[A] = new Foo { type T = A }
val fb = makeFoo[Int]

implicitly[fb.T =:= Int] // No more error

or

def makeFoo2[A]: Foo { type T = A } = new Foo { type T = A }
val fc: Foo { type T = Int } = makeFoo[Int]

implicitly[fc.T =:= Int] // No more error

Scastie:

<script src="https://scastie.scala-lang.org/knIfPcXqSQKXXCXJ2teHkg.js"></script>

Anne

这家伙很懒，什么都没留下

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