How to return a function in scala - function

How can I return a function side-effecting lexical closure1 in Scala?
For instance, I was looking at this code sample in Go:
...
// fib returns a function that returns
// successive Fibonacci numbers.
func fib() func() int {
a, b := 0, 1
return func() int {
a, b = b, a+b
return b
}
}
...
println(f(), f(), f(), f(), f())
prints
1 2 3 5 8
And I can't figure out how to write the same in Scala.
1. Corrected after Apocalisp comment

Slightly shorter, you don't need the return.
def fib() = {
var a = 0
var b = 1
() => {
val t = a;
a = b
b = t + b
b
}
}

Gah! Mutable variables?!
val fib: Stream[Int] =
1 #:: 1 #:: (fib zip fib.tail map Function.tupled(_+_))
You can return a literal function that gets the nth fib, for example:
val fibAt: Int => Int = fib drop _ head
EDIT: Since you asked for the functional way of "getting a different value each time you call f", here's how you would do that. This uses Scalaz's State monad:
import scalaz._
import Scalaz._
def uncons[A](s: Stream[A]) = (s.tail, s.head)
val f = state(uncons[Int])
The value f is a state transition function. Given a stream, it will return its head, and "mutate" the stream on the side by taking its tail. Note that f is totally oblivious to fib. Here's a REPL session illustrating how this works:
scala> (for { _ <- f; _ <- f; _ <- f; _ <- f; x <- f } yield x)
res29: scalaz.State[scala.collection.immutable.Stream[Int],Int] = scalaz.States$$anon$1#d53513
scala> (for { _ <- f; _ <- f; _ <- f; x <- f } yield x)
res30: scalaz.State[scala.collection.immutable.Stream[Int],Int] = scalaz.States$$anon$1#1ad0ff8
scala> res29 ! fib
res31: Int = 5
scala> res30 ! fib
res32: Int = 3
Clearly, the value you get out depends on the number of times you call f. But this is all purely functional and therefore modular and composable. For example, we can pass any nonempty Stream, not just fib.
So you see, you can have effects without side-effects.

While we're sharing cool implementations of the fibonacci function that are only tangentially related to the question, here's a memoized version:
val fib: Int => BigInt = {
def fibRec(f: Int => BigInt)(n: Int): BigInt = {
if (n == 0) 1
else if (n == 1) 1
else (f(n-1) + f(n-2))
}
Memoize.Y(fibRec)
}
It uses the memoizing fixed-point combinator implemented as an answer to this question: In Scala 2.8, what type to use to store an in-memory mutable data table?
Incidentally, the implementation of the combinator suggests a slightly more explicit technique for implementing your function side-effecting lexical closure:
def fib(): () => Int = {
var a = 0
var b = 1
def f(): Int = {
val t = a;
a = b
b = t + b
b
}
f
}

Got it!! after some trial and error:
def fib() : () => Int = {
var a = 0
var b = 1
return (()=>{
val t = a;
a = b
b = t + b
b
})
}
Testing:
val f = fib()
println(f(),f(),f(),f())
1 2 3 5 8

You don't need a temp var when using a tuple:
def fib() = {
var t = (1,-1)
() => {
t = (t._1 + t._2, t._1)
t._1
}
}
But in real life you should use Apocalisp's solution.

Related

Putting last element of list in the first index "n" times SML

I'm trying to put the last element of a list in the front of the list while keeping the rest of the elements in the same order N times. I can do it once with this function, but I want to add another parameter to the function so that the function in called N times.
Code:
fun multcshift(L, n) =
if null L then nil
else multcshift(hd(rev L)::(rev(tl(rev L))));
Thanks
To make the parameter n work, you need recursion. You need a base case at which point the function should no longer call itself, and a recursive case where it does. For this function, a good base case would be n = 0, meaning "shift the last letter in front 0 times", i.e., return L without modification.
fun multcshift(L, n) =
if n = 0
then L
else multcshift( hd(rev L)::rev(tl(rev L)) , n - 1 )
The running time of this function is terrible: For every n, reverse the list three times!
You could save at least one of those list reversals by not calling rev L twice. E.g.
fun multcshift (L, 0) = L
| multcshift (L, n) =
let val revL = rev L
in multcshift ( hd revL :: rev (tl revL) , n - 1 ) end
Those hd revL and rev (tl revL) seem like useful library functions. The process of applying a function to its own output n times seems like a good library function, too.
(* Return all the elements of a non-empty list except the last one. *)
fun init [] = raise Empty
| init ([_]) = []
| init (x::xs) = x::init xs
(* Return the last element of a non-empty list. *)
val last = List.last
(* Shift the last element of a non-empty list to the front of the list *)
fun cshift L = last L :: init L
(* Compose f with itself n times *)
fun iterate f 0 = (fn x => x)
| iterate f 1 = f
| iterate f n = f o iterate f (n-1)
fun multcshift (L, n) = iterate cshift n L
But the running time is just as terrible: For every n, call last and init once each. They're both O(|L|) just as rev.
You could overcome that complexity by carrying out multiple shifts at once. If you know you'll shift one element n times, you might as well shift n elements. Shifting n elements is equivalent to removing |L| - n elements from the front of the list and appending them at the back.
But what if you're asked to shift n elements where n > |L|? Then len - n is negative and both List.drop and List.take will fail. You could fix that by concluding that any full shift of |L| elements has no effect on the result and suffice with n (mod |L|). And what if n < 0?
fun multcshift ([], _) = raise Empty
| multcshift (L, 0) = L
| multcshift (L, n) =
let val len = List.length L
in List.drop (L, len - n mod len) #
List.take (L, len - n mod len) end
There are quite a few corner cases worth testing:
val test_zero = (multcshift ([1,2,3], 0) = [1,2,3])
val test_empty = (multcshift ([], 5); false) handle Empty => true | _ => false
val test_zero_empty = (multcshift ([], 0); false) handle Empty => true | _ => false
val test_negative = (multcshift ([1,2,3,4], ~1) = [2,3,4,1])
val test_nonempty = (multcshift ([1,2,3,4], 3) = [2,3,4,1])
val test_identity = (multcshift ([1,2,3,4], 4) = [1,2,3,4])
val test_large_n = (multcshift [1,2,3,4], 5) = [4,1,2,3])
val test_larger_n = (multcshift [1,2,3,4], 10) = [3,4,1,2])

Recursive call in if expression - ocaml

module Dfs = struct
let rec dfslsts g paths final =
let l = PrimePath.removeDuplicates (PrimePath.extendPaths g paths)
in
let f elem =
if (List.mem "%d" (List.flatten final) = false) then (dfslsts g ["%d"] (List.flatten l)::final)
else final
in
List.iter f (Graph.nodes g)
end
Error: This expression has type string but an expression was expected of type int list
This error occurred when I called dfslsts function, which is recursive, inside the if condition.
The function dfslsts returns a list of lists.
If I try to replace the complex expression in if statement to
if (List.mem "%d" (List.flatten final) = false) then "%d"
else "%d"
then I get
Error: This expression has type 'a -> string
but an expression was expected of type 'a -> unit
Type string is not compatible with type unit
at List.iter line.
How do I solve this problem and are we allowed to call a recursive function inside the if expression.
This is the definition of my graph type:
module Graph = struct
exception NodeNotFound of int
type graph = {
nodes : int list;
edges : (int * int) list;
}
let makeGraph () =
{
nodes = [];
edges = [];
}
let rec isNodeOf g n = List.mem n g.nodes
let nodes g = g.nodes
let edges g = g.edges
let addNode g n =
let nodes = n::g.nodes and edges = g.edges in
{
nodes;
edges;
}
let addEdge g (n1, n2) =
if ((isNodeOf g n1) = false) then
raise (NodeNotFound n1)
else if ((isNodeOf g n2) = false) then
raise (NodeNotFound n2)
else
let nodes = g.nodes
and edges = (n1, n2) :: g.edges in
{
nodes;
edges;
}
let nextNodes g n =
let rec findSuccessors edges n =
match edges with
[] -> []
| (n1, n2) :: t ->
if n1 = n then n2::findSuccessors t n
else findSuccessors t n
in
findSuccessors g.edges n
let rec lastNode path =
match path with
[] -> raise (NodeNotFound 0)
| n :: [] -> n
| _ :: t -> lastNode t
end
module Paths = struct
let extendPath g path =
let n = (Graph.lastNode path) in
let nextNodes = Graph.nextNodes g n in
let rec loop path nodes =
match nodes with
[] -> []
| h :: t -> (List.append path [h]) :: (loop path t)
in
loop path nextNodes
let rec extendPaths g paths =
match paths with
[] -> []
| h :: t -> List.append (extendPath g h) (extendPaths g t)
(* Given a list lst, return a new list with all duplicate entries removed *)
let rec removeDuplicates lst =
match lst with
[]
| _ :: [] -> lst
| h :: t ->
let trimmed = removeDuplicates t in
if List.mem h trimmed then trimmed
else h :: trimmed
end
Any expression can be a recursive function call. There are no limitations like that. Your problem is that some types don't match.
I don't see any ints in this code, so I'm wondering where the compiler sees the requirement for an int list. It would help to see the type definition for your graphs.
As a side comment, you almost certainly have a precedence problem with this code:
dfslsts g ["%d"] (List.flatten l)::final
The function call to dfslsts has higher precedence that the list cons operator ::, so this is parsed as:
(dfslsts g ["%d"] (List.flatten l)) :: final
You probably need to parenthesize like this:
dfslsts g ["%d"] ((List.flatten l) :: final)

How to generate a sum and carry with Chisel in one line?

Is it possible to generate sum and carry in one line in Chisel similar to this code in Verilog?
module Adder_with_carry(
input [3:0] A,
input [3:0] B,
input Carry_in,
output [3:0] Sum,
output Carry_out
);
assign {Carry_out, Sum} = A + B + Carry_in;
endmodule
I am using this code
class Adder extends Module{
val io = new Bundle{
val a = UInt(INPUT, 3)
val b = UInt(INPUT, 3)
val carry_in = UInt (INPUT, 1)
val sum = UInt (OUTPUT, 2)
val carry_out = UInt(OUTPUT, 1)
}
val SUM = io.a + io.b + io.carry_in;
io.carry_out := SUM(2)
io.sum := SUM(1,0)
}
But I think it will be more convenient if there is a one-liner for this.

Functions in scala

I'm having hard time understanding what the following means in scala:
f: Int => Int
Is the a function?
What is the difference between f: Int => Intand def f(Int => Int)?
Thanks
Assuming f: Int => Int is a typo of val f: Int => Int,
and def f(Int => Int) is a typo of def f(i: Int): Int.
val f: Int => Int means that a value f is Function1[Int, Int].
First, A => B equals to =>[A, B].
This is just a shortcut writing, for example:
trait Foo[A, B]
val foo: Int Foo String // = Foo[Int, String]
Second, =>[A, B] equals to Function1[A, B].
This is called "type alias", defined like:
type ~>[A, B] = Foo[A, B]
val foo: Int ~> String // = Foo[Int, String]
def f(i: Int): Int is a method, not a function.
But a value g is Function1[Int, Int] where val g = f _ is defined.
f: Int => Int
means that type of f is Int => Int.
Now what does that mean? It means that f is a function which gets an Int and returns an Int.
You can define such a function with
def f(i: Int): Int = i * 2
or with
def f: Int => Int = i => i * 2
or even with
def f: Int => Int = _ * 2
_ is a placeholder used for designating the argument. In this case the type of the parameter is already defined in Int => Int so compiler knows what is the type of this _.
The following is again equivalent to above definitions:
def f = (i:Int) => i * 2
In all cases type of f is Int => Int.
=>
So what is this arrow?
If you see this arrow in type position (i.e. where a type is needed) it designates a function.
for example in
val func: String => String
But if you see this arrow in an anonymous function it separates the parameters from body of the function. For example in
i => i * 2
To elaborate just slightly on Nader's answer, f: Int => Int will frequently appear in a parameter list for a high order function, so
def myHighOrderFunction(f : Int => Int) : String = {
// ...
"Hi"
}
is a dumb high order function, but shows how you say that myOrderFunction takes as a parameter, f, which is a function that maps an int to an int.
So I might legally call it like this for example:
myHighOrderFunction(_ * 2)
A much more illustrative example comes from Odersky's Programming Scala:
object FileMatcher {
private def filesHere = (new java.io.File(".")).listFiles
private def filesMatching(matcher: String => Boolean) =
for (file <- filesHere if matcher(file.getName))
yield file
def filesEnding(query: String) = filesMatching(_.endsWith(query))
def filesContaining(query: String) = filesMatching(_.contains(query))
def filesRegex(query: String) = filesMatching(_.matches(query))
}
Here filesMatching is a high order function, and we define three other functions that call it passing in different anonymous functions to do different kinds of matching.
Hope that helps.

Difference between these two method definitions

What's the difference between these two definitions?:
def sayTwords(word1: String, word2: String) = println(word1 + " " + word2)
def sayTwords2(word1: String)(word2: String) = println(word1 + " " + word2)
What is the purpose of each?
The second is curried, the first isn't. For a discussion of why you might choose to curry a method, see What's the rationale behind curried functions in Scala?
sayTwords2 allows the method to be partially applied.
val sayHelloAnd = sayTwords2("Hello")
sayHelloAnd("World!")
sayHaelloAnd("Universe!")
Note you can also use the first function in the same way.
val sayHelloAnd = sayTwords("Hello", _:String)
sayHelloAnd("World!")
sayHelloAnd("Universe!")
def sayTwords(word1: String, word2: String) = println(word1 + " " + word2)
def sayTwords2(word1: String)(word2: String) = println(word1 + " " + word2)
The first contains a single parameter list. The second contains multiple parameter lists.
They differ in following regards:
Partial application syntax. Observe:
scala> val f = sayTwords("hello", _: String)
f: String => Unit = <function1>
scala> f("world")
hello world
scala> val g = sayTwords2("hello") _
g: String => Unit = <function1>
scala> g("world")
hello world
The former has a benefit of being positional syntax. Thus you can partially apply arguments in any positions.
Type inference. The type inference in Scala works per parameter list, and goes from left to right. So given a case, one might facilitate better type inference than other. Observe:
scala> def unfold[A, B](seed: B, f: B => Option[(A, B)]): Seq[A] = {
| val s = Seq.newBuilder[A]
| var x = seed
| breakable {
| while (true) {
| f(x) match {
| case None => break
| case Some((r, x0)) => s += r; x = x0
| }
| }
| }
| s.result
| }
unfold: [A, B](seed: B, f: B => Option[(A, B)])Seq[A]
scala> unfold(11, x => if (x == 0) None else Some((x, x - 1)))
<console>:18: error: missing parameter type
unfold(11, x => if (x == 0) None else Some((x, x - 1)))
^
scala> unfold(11, (x: Int) => if (x == 0) None else Some((x, x - 1)))
res7: Seq[Int] = List(11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1)
scala> def unfold[A, B](seed: B)(f: B => Option[(A, B)]): Seq[A] = {
| val s = Seq.newBuilder[A]
| var x = seed
| breakable {
| while (true) {
| f(x) match {
| case None => break
| case Some((r, x0)) => s += r; x = x0
| }
| }
| }
| s.result
| }
unfold: [A, B](seed: B)(f: B => Option[(A, B)])Seq[A]
scala> unfold(11)(x => if (x == 0) None else Some((x, x - 1)))
res8: Seq[Int] = List(11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1)