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The existential type inferred from function literal is useless. #8515

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scabug opened this issue Apr 18, 2014 · 5 comments
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The existential type inferred from function literal is useless. #8515

scabug opened this issue Apr 18, 2014 · 5 comments
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@retronym
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duplicate existential infer typer

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@scabug
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scabug commented Apr 18, 2014 

Welcome to Scala version 2.10.3 (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_45).
Type in expressions to have them evaluated.
Type :help for more information.

scala> class C { trait T }
defined class C

scala> val c = new C
c: C = C@1130ed80

scala> val t1: c.T = (new { def apply(c: C) = new c.T {} })(c)
t1: c.T = $anon$2$$anon$1@23309add

scala> val t2: c.T = ({c: C => new c.T {} })(c)
<console>:9: error: type mismatch;
 found   : $anon where type $anon <: c.T
 required: c.T
       val t2: c.T = ({c: C => new c.T {} })(c)
                                            ^

scala> val f1: { def apply(c: C): c.T } = {c: C => new c.T {} }
<console>:9: error: type mismatch;
 found   : C => $anon forSome { type $anon <: c.T; val c: C }
 required: AnyRef{def apply(c: C): c.T}
       val f1: { def apply(c: C): c.T } = {c: C => new c.T {} }
                                                ^

The function literal is inferred to existential type {code} C => $anon forSome { type $anon <: c.T; val c: C } {code}, which looks complex, but it is no more special than simple C => C#T.

Could you infer the function literal to {code} (C => C#T) with { def apply(c: C): c.T } {code}
@scabug
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scabug commented Apr 18, 2014

Imported From: https://issues.scala-lang.org/browse/SI-8515?orig=1
Reporter: @Atry
Affected Versions: 2.10.3, 2.10.4
Duplicates #4751

@scabug
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scabug commented Apr 18, 2014

@retronym said:
The type of a a function literal is always of the form: FunctionN[A, B]

In your example B is an existential type.

Scala doesn't support dependent function types. There is a feature request for this in #4751. I will merge this ticket with that one, by adding your example to the comments of that ticket.

@scabug scabug closed this as completed Apr 18, 2014
@scabug
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scabug commented Apr 18, 2014

@Atry said (edited on Apr 18, 2014 1:01:21 PM UTC):
No, dependent function types are not required to pass this example.

I assume the function literal is a shortcut of {code}new Function1[C, ] { { def apply(c: C): = new c.T {} } }{code}. I feel surprise when I found the different behavior between function literals and anonymous classes.

I guess referring the function literal to a compound type would suppress the compiler error.

@scabug
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scabug commented Apr 18, 2014

@retronym said:
Section 6.3 of the spec explains how anonymous functions are type checked:

The type of the anonymous function is scala.Functionn[S1, ..., Sn, T ], where T is the packed type (§6.1) of e. T must be equivalent to a type which does not refer to any of the formal parameters xi .

The expansion to an anonymous subclass of FunctionN is only mentioned as the evaluation model:

The anonymous function is evaluated as the instance creation expression
new scala.Functionn[T1, ..., Tn, T] { def apply(x1: T1,...,xn: Tn): T = e
}

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scabug commented Apr 18, 2014

@Atry said:
You are right

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