gistfile1.scala

scala> object Foo
defined module Foo

scala> val x = Foo
x: Foo.type = Foo$@24753433

scala> val y = identity(Foo)
y: object Foo = Foo$@24753433

Singleton types aren't inferred as type arguments. Not sure of the rationale.

scala> identity[Foo.type](x)
res2: Foo.type = Foo$@20724356

scala> def i[X <: Singleton](x: X) = x
i: [X <: Singleton](x: X)X

scala> i(x)
<console>:9: error: inferred type arguments [object Foo] do not conform to method i's type parameter bounds [X <: Singleton]
       i(x)
       ^
scala> i[Foo.type](x)
res4: Foo.type = Foo$@20724356

Hmm... yeah, I think I remember Martin saying that singleton types aren't inferred because then a singleton type would pretty much always be inferred.

Also, James Iry found this gem:

scala> def test(x: Foo.type) = x
test: (x: Foo.type)object Foo

scala> test(x)
res6: object Foo = Foo$@600ab1bf

From @collinmbulluck: Somehting to do with type inference and stable vs unstable identifiers?

scala> val x = Foo
x: Foo.type = Foo$@600ab1bf

scala> var y = Foo
y: object Foo = Foo$@600ab1bf

a) If singleton types were inferred, we wouldn't have the martin classic "I invite everyone to change this rule, and observe what breaks!"

https://lampsvn.epfl.ch/trac/scala/ticket/1208

b) See also comments in:

https://lampsvn.epfl.ch/trac/scala/ticket/3494 .