Question

class ClosureClass {
  def printResult[T](f: => T) = {
    println("choice 1")
    println(f)
  }

  def printResult[T](f: String => T) = {
    println("choice 2")
    println(f("HI THERE"))
  }
}

object demo {
  def main(args: Array[String]) {
    val cc = new ClosureClass
    cc.printResult()   // call 1
    cc.printResult("Hi")  // call 2
  }
}

I play with the above code, and the result showed me. I have two questions

1) why both call 1 and call 2 go into choice 1?

2) How can I pass a parameter so that I can get into choice 2.

Thanks,

choice 1
()
choice 1
Hi

Answer 1

The type => T means that f is a call-by-name parameter. This means that f is of type T, and the expression passed into the function will be evaluated when it is used (not when the function is called.

The type String => T means that f is a Function[String,T], that is, a function from a String to a T.

When you call with "Hi", a String as the argument, Scala sees that there are two choices for printResult:

1) => T: In this case case T would bind to String, which is fine.

2) String => T: This does not work since String is not a function from String to anything… it’s not a function at all.

How you fix this depends on what you are trying to do. If you just want to fix how printResult is being called, then you call call it with:

val g: (String => String) = (s: String) => s + "***"
cc.printResult(g)

which prints:

choice 2
HI THERE***

since you are now are now correctly passing a function, g, from String to some T. That T here is String, since the function is just adding *** onto the end of whatever it is passed and returning the modified string.