How to define scala type for no argument function?

Question

This works

def func(f: => Int) = f

This dosn't (inside class for example)

type EmptyFunct = => Int

or

 type EmptyFunct = (=> Int)

Scala version 2.9 Two questions:

1. Why dosn't syntax sugar work in second case?
2. How to define this function without syntax sugar?

Answer 1

=> Int is not exactly a function without parameter, it is an Int argument with call by name passing convention. (Of course, that is rather fine a point, as it is implemented by passing a function without parameter ).

The function without argument is written () => Int. You can do type EmptyFunct = () => Int.

It is not a type. Inside func, f will be typed as an Int. An argument of type () => Int will not.

def func(f: => Int) = f *2

func (: => Int) Int

But

def func(f: () => Int) : Int = f*2

error: value * is not a member of () => Int

Answer 2

You should use Function0

In the first case it does not work because you declare a non argument Function but because you indicates that the parameter is called by name.

Answer 3

I don't see much sense in a method, returning an int, invoked with no parameter. Either it returns a constant, so you could use the constant, or it would use a var?

So let's go with a var:

var k = 10 
val fi = List (() => k * 2, () => k - 2)
val n = fi(0)
n.apply 
k = 11
n.apply

result is 20, then 22.