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.

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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) -> ....

Answer 1

Here’s the SLS quote, for the rest of us:

The expected type of such an expression must in part be defined. It
must be either scala.Functionk[S1, . . . , Sk, R] for some k > 0, or
scala.PartialFunction[S1, R], where the argument type(s) S1, . . . ,
Sk must be fully determined, but the result type R may be
undetermined.

Otherwise, you answered your question.