Expansion of wildcard to existential type ignores bound #2385

scabug · 2009-09-24T01:56:06Z

Given the following:

abstract class A[T <: A[T]] {
  def foo: X[T]
}

class B extends A[B] {
  def foo = new X[B]
} 

class X[T <: A[T]]

object Test {
  def test: X[_] = {
    val a: A[_] = new B

    def f[T <: A[T]](x: X[T]) = x

    f(a.foo)
  }
}

compilation fails as:

Test.scala:18: error: inferred type arguments [_$$2] do not conform to method f's type parameter bounds [T <: A[T]]
    f(a.foo)
    ^

even though it is not possible for any instance of A to fail to satisfy this bound. Explicitly specifying the existential type allows correct compilation:

object Test {
  def test: X[_] = {
    val a: A[T] forSome {type T <: A[T]} = new B

    def f[T <: A[T]](x: X[T]) = x

    f(a.foo)
  }
}

It seems like A[_] is naively being expanded to A[T] forSome {type T} instead of respecting the bound declared in A. Minor, but unexpected.

scabug · 2009-09-24T01:56:06Z

Imported From: https://issues.scala-lang.org/browse/SI-2385?orig=1
Reporter: Kris Nuttycombe (nuttycom)

scabug · 2009-09-28T12:40:42Z

@cunei said:
What you write corresponds to what is stated in the Scala specification (section "Placeholder Syntax for Existential Types").

The syntax A[_] is equivalent to A[T] forSome { type T >: scala.Nothing <: scala.Any }. When you explicitly specify val a:A[_] you are telling the compiler to assume that that one is the type of a.

You can just write val a = new B instead, and type inference will give you the specific type you are looking for.

scabug · 2009-09-28T12:58:32Z

Kris Nuttycombe (nuttycom) said:

The syntax A[_] is equivalent to A[T] forsome { type T >: scala.Nothing <: scala.Any }.

In this case, shouldn't the declaration

val a: A[_]

be a type error?

scabug · 2009-09-28T14:28:55Z

@cunei said:
No, in this case A[T] forsome { type T >: scala.Nothing <: scala.Any } only means that there is one such T, you don't know what it is. The fact that in reality it can only be within a subset of the whole set of types doesn't make the declaration any less valid.

To make an analogy, if I tell you "I will arrive on Monday, because I can only arrive on Mondays", and in turn you tell to someone else "He will arrive on some weekday", your assertion is still valid.

scabug closed this as completed May 18, 2011

scabug mentioned this issue Apr 7, 2017

propagate bounds from class type params to existential quantifiers #6169

Closed

Provide feedback

Saved searches

Use saved searches to filter your results more quickly

Expansion of wildcard to existential type ignores bound #2385

Expansion of wildcard to existential type ignores bound #2385

scabug commented Sep 24, 2009

scabug commented Sep 24, 2009

scabug commented Sep 28, 2009 •

edited by retronym

scabug commented Sep 28, 2009

scabug commented Sep 28, 2009 •

edited by retronym

Expansion of wildcard to existential type ignores bound #2385

Expansion of wildcard to existential type ignores bound #2385

Comments

scabug commented Sep 24, 2009

scabug commented Sep 24, 2009

scabug commented Sep 28, 2009 • edited by retronym

scabug commented Sep 28, 2009

scabug commented Sep 28, 2009 • edited by retronym

scabug commented Sep 28, 2009 •

edited by retronym

scabug commented Sep 28, 2009 •

edited by retronym