scala - The argument types of an anonymous function must be fully known. (SLS 8.5)

Question

I have a function literal

{case QualifiedType(preds, ty) =>
               t.ty = ty ;
               Some((emptyEqualityConstraintSet,preds)) }

Which results in an error message

missing parameter type for expanded function The argument types of an anonymous function
must be fully known. (SLS 8.5) Expected type was:
? => Option[(Typer.this.EqualityConstraintSet, Typer.this.TypeRelationSet)]

I looked in SLS 8.5, but didn't find an explanation.

If I expand the function myself to

{(qt : QualifiedType) =>
  qt match {case QualifiedType(preds, ty) =>
               t.ty = ty ;
               Some((emptyEqualityConstraintSet,preds)) }}

the error goes away.

(a) Why is this an error?

(b) What can I do to fix it?

I tried the obvious fix, which was to add : QualifiedType between the pattern and the =>, but this is a syntax error.

---

One thing I noticed is that the context makes a difference. If I use the function literal as an argument to a function declared as expecting a QualifiedType => B, there is no error. But if I use it as an argument to a function expecting an A => B, there is an error. I expect that what is going on here is that, as the pattern could conceivably be applied to an object whose type is a supertype of QualifiedType, the compiler is not willing to assign the obvious type without assurance that the function won't be applied to anything that isn't a QualifiedType. Really what I'd like is to be able to write {QualifiedType( preds, ty) => ...} and have it mean the same thing as Haskell's QualifiedType(preds,ty) -> ....

question from:https://stackoverflow.com/questions/12869251/the-argument-types-of-an-anonymous-function-must-be-fully-known-sls-8-5

Answer 1

{ case X(x) => ... } is a partial function, but the compiler still doesn't know what your input type is, except that it's a supertype of X. Normally this isn't a problem because if you're writing an anonymous function, the type is known from the context. But here is how you can provide the type:

case class Foo(x: Int)

// via annotation
val f: Foo => Int = { case Foo(x) => x }

// use pattern matching
val f = (_: Foo) match { case Foo(x) => x }

// or more normally, write as a method
def f(a: Foo) = a match { case Foo(x) => x }
def f(a: Foo) = a.x

As you've probably noticed, using function literals / pattern matching here is pretty pointless. It seems in your case you just need a regular method:

def whatever(qt: QualifiedType) = {
  t.ty = qt.ty
  Some((emptyEqualityConstraintSet, qt.preds)) 
}

although you should refactor to remove that mutable state.