When is a scala partial function not a partial function?

Question

While creating a map of String to partial functions I ran into unexpected behavior. When I create a partial function as a map element it works fine. When I allocate to a val it invokes instead. Trying to invoke the check generates an error. Is this expected? Am I doing something dumb? Comment out the check() to see the invocation. I am using scala 2.7.7

def PartialFunctionProblem() = {
    def dream()() = {
        println("~Dream~");
        new Exception().printStackTrace()
    }
    val map = scala.collection.mutable.HashMap[String,()=>Unit]()
    map("dream") = dream()      // partial function
    map("dream")()              // invokes as expected
    val check = dream()         // unexpected invocation
    check()                     // error: check of type Unit does not take parameters 
}

Answer 1

For convenience, Scala lets you omit empty parens when calling a method, but it's clever enough to see that the expected type in the first case is ()=>Unit, so it doesn't remove all the parens for you; instead, it converts the method into a function for you.

In the val check case, however, it looks just like a function call result getting assigned to a variable. In fact, all three of these do the exact same thing:

val check = dream
val check = dream()
val check = dream()()

If you want to turn the method into a function, you place _ after the method in place of the argument list(s). Thus,

val check = dream() _

will do what you want.

Answer 2

Well, the problem is that you got it all wrong. :-)

Here are some conceptual mistakes:

def dream()() = {
    println("~Dream~");
    new Exception().printStackTrace()
}

This is not a partial function. This is a curried method with two empty parameter lists which returns Unit.

val map = scala.collection.mutable.HashMap[String,()=>Unit]()

The type of the values in this map is not partial function, but function. Specifically, Function0[Unit]. A partial function would have type PartialFunction[T, R].

map("dream") = dream()      // partial function

What happens here is that Scala converts the partially applied method into a function. This is not a simple assignment. Scala does the conversion because the type inferencer can guess the correct type.

val check = dream()         // unexpected invocation

Here there's no expected type to help the type inferencer. However, empty parameter lists can be ommitted, so this is just a method call.