Implicit conversion to Runnable?

Question

As an exercise I tried to create an implicit conversion that would accept a function and produce a Runnable. That way you could call java methods that accept Runnables and use them like closures.

the implicit conversion is easy enough:

    implicit def funToRunnable(fun : Unit) = new Runnable() { def run = fun }

however I don't know how to call it. How do you pass in a no-arg function that returns Unit, without having it be evaluated at once? For example, I'd like the following to print "12" but instead it prints "21" because 'print("2")' is evaluated at once.

    var savedFun : Runnable = null
    def save(r : Runnable) = { savedFun = r }

    save(print("2"))
    print("1");
    savedFun.run()

how do I tell the compiler to treat 'print("2")' as the body of a function, not something to be evaluated at once? Some possibilies I tried, such as

    save(() => print("2"))

or

    save(=> print("2"))

are not legal syntax.

Answer 1

arg, just answered my own question. I implemented the implicit conversion incorrectly. The correct implementation is

implicit def funToRunnable(fun: () => Unit) = new Runnable() { def run() = fun() }

and you call it like this:

save(() => print("2"))

This yields "12"

Answer 2

Interesting, this way you can execute code that receives a Runnable and pass it a closure.

See:

scala> new Thread( ()  => print( "Hello" ) ).start()
<console>:5: error: overloaded method constructor Thread with alternatives (java.lang.ThreadGroup,java.lang.Runnable,java.lang.String,Long)java.lang.Thread <and> (java.lang.ThreadGroup,java.lang.Runnable,java.lang.String)java.lang.Thread <and> (java.lang.Runnable,java.lang.String)java.lang.Thread <and> (java.lang.ThreadGroup,java.lang.String)java.lang.Thread <and> (java.lang.String)ja...
       new Thread( ()  => print( "Hello" ) ).start()


scala> implicit def funcToRunnable( func : () => Unit ) = new Runnable(){ def run() = func() }
funcToRunnable: (() => Unit)java.lang.Object with java.lang.Runnable

scala> def doRun( runnable: Runnable ) = runnable.run
doRun: (Runnable)Unit

scala> doRun( () => print("Hola"))
Hola

scala> new Thread(()=>print("Hello")).start()

scala> Hello

Answer 3

If you wanted to live dangerously, you could convert anything to a runnable:

implicit def whateverToRunnable[F](f: => F) = new Runnable() { def run() { f } }

scala> val t = new Thread(println("Hello"))
t: java.lang.Thread = Thread[Thread-2,5,main]

scala> t.start()
Hello

Or you could create your own thread-creator-and-starter:

def thread[F](f: => F) = (new Thread( new Runnable() { def run() { f } } )).start

scala> thread { println("Hi"); Thread.sleep(1000); println("Still here!") }
Hi

scala> Still here!

If you wanted to return the thread, then

def thread[F](f: => F) = {
  val t = new Thread( new Runnable() { def run() { f } } )
  t.start()
  t
}

But all of this, while useful, is perhaps even less useful than scala.actors.Futures (tested only on 2.8):

scala> import scala.actors.Futures

scala> val x = Futures.future { Thread.sleep(10000); "Done!" }
x: scala.actors.Future[java.lang.String] = <function0>

scala> x.isSet
res0: Boolean = false

scala> x.isSet
res1: Boolean = false

scala> x()   // Waits until the result is ready....
res2: java.lang.String = Done!