I am going through the book "Functional Programming in Scala" and have run across an example that I don't fully understand.
In the chapter on strictness/laziness the authors describe the construction of Streams and have code like this:
sealed trait Stream[+A]
case object Empty extends Stream[Nothing]
case class Cons[+A](h: () => A, t: () => Stream[A]) extends Stream[A]
object Stream {
def cons[A](hd: => A, tl: => Stream[A]) : Stream[A] = {
lazy val head = hd
lazy val tail = tl
Cons(() => head, () => tail)
}
...
}
The question I have is in the smart constructor (cons
) where it calls the constructor for the Cons
case class. The specific syntax being used to pass the head
and tail
vals doesn't make sense to me. Why not just call the constructor like this:
Cons(head, tail)
As I understand the syntax used it is forcing the creation of two Function0 objects that simply return the head
and tail
vals. How is that different from just passing head
and tail
(without the () =>
prefix) since the Cons
case class is already defined to take these parameters by-name anyway? Isn't this redundant? Or have I missed something?
In Scala when arguments pass through call-by-value function it compute the passed-in expression's or arguments value once before calling the function . But a call-by-Name function in Scala calls the expression and recompute the passed-in expression's value every time it get accessed inside the function.
In the case of Call by Value, when we pass the value of the parameter during the calling of the function, it copies them to the function's actual local argument. In the case of Call by Reference, when we pass the parameter's location reference/address, it copies and assigns them to the function's local argument.
Call by name is a a parameter passing scheme where the parameter is evaluated when it is used, not when the function is called.
For this circumstance, Scala offers call-by-name parameters. A call-by-name mechanism passes a code block to the call and each time the call accesses the parameter, the code block is executed and the value is calculated.
The difference is in => A
not being equal to () => A
.
The former is pass by name, and the latter is a function that takes no parameters and returns an A.
You can test this out in the Scala REPL.
scala> def test(x: => Int): () => Int = x
<console>:9: error: type mismatch;
found : Int
required: () => Int
def test(x: => Int): () => Int = x
^
Simply referencing x
in my sample causes the parameter to be invoked. In your sample, it's constructing a method which defers invocation of x.
First, you are assuming that => A
and () => A
are the same. However, they are not. For example, the => A
can only be used in the context of passing parameters by-name - it is impossible to declare a val
of type => A
. As case class
parameters are always val
s (unless explicitly declared var
s), it is clear why case class Cons[+A](h: => A, t: => Stream[A])
would not work.
Second, just wrapping a by-name parameter into a function with an empty parameter list is not the same as what the code above accomplishes: using lazy val
s, it is ensured that both hd
and tl
are evaluated at most once. If the code read
Cons(() => hd, () => tl)
the original hd
would be evaluated every time the h
method (field) of a Cons
object is invoked. Using a lazy val
, hd
is evaluated only the first time the h
method of this Cons
object is invoked, and the same value is returned in every subsequent invocation.
Demonstrating the difference in a stripped-down fashion in the REPL:
> def foo = { println("evaluating foo"); "foo" }
> val direct : () => String = () => foo
> direct()
evaluating foo
res6: String = foo
> direct()
evaluating foo
res7: String = foo
> val lzy : () => String = { lazy val v = foo; () => v }
> lzy()
evaluating foo
res8: String = foo
> lzy()
res9: String = foo
Note how the "evaluating foo" output in the second invocation of lzy()
is gone, as opposed to the second invocation of direct()
.
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