I have the following code which recursively operates on each element within a List
def doMatch(list: List[Int]): Unit = list match {
case last :: Nil => println("Final element.")
case head :: tail => println("Recursing..."); doMatch(tail)
}
Now, ignoring that this functionality is available through filter() and foreach(), this works just fine. However, if I try to change it to accept any Seq[Int], I run into problems:
Seq doesn't have ::, but it does have +:, which as I understand is basically the same thing. If I try to match on head +: tail however, the compiler complains 'error: not found: value +:'
Nil is specific to List, and I'm not sure what to replace it with. I'm going to try Seq() if I ever get past the previous problem
Here is how I think the code should look, except it doesn't work:
def doMatch(seq: Seq[Int]): Unit = seq match {
case last +: Seq() => println("Final element.")
case head +: tail => println("Recursing..."); doMatch(tail)
}
Edit: So many good answers! I'm accepting agilesteel's answer as his was the first that noted that :: isn't an operator in my example, but a case class and hence the difference.
As of the ides of March 2012, this works in 2.10+:
def doMatch(seq: Seq[Int]): Unit = seq match {
case last +: Seq() => println("Final element.")
case head +: tail => println("Recursing..."); doMatch(tail)
} //> doMatch: (seq: Seq[Int])Unit
doMatch(List(1, 2)) //> Recursing...
//| Final element.
More generally, two different head/tail and init/last decomposition objects mirroring append/prepend were added for Seq in SeqExtractors:
List(1, 2) match { case init :+ last => last } //> res0: Int = 2
List(1, 2) match { case head +: tail => tail } //> res1: List[Int] = List(2)
Vector(1, 2) match { case init :+ last => last } //> res2: Int = 2
Vector(1, 2) match { case head +: tail => tail } //> res3: scala.collection.immutable.Vector[Int] = Vector(2)
Kind of cheating, but here it goes:
def doMatch(seq: Seq[Int]): Unit = seq match {
case Seq(x) => println("Final element " + x)
case Seq(x, xs#_*) => println("Recursing..." + x); doMatch(xs)
}
Don't ask me why xs* doesn't work...
There are two :: (pronounced cons) in Scala. One is an operator defined in class List and one is a class (subclass of List), which represents a non empty list characterized by a head and a tail.
head :: tail is a constructor pattern, which is syntactically modified from ::(head, tail).
:: is a case class, which means there is an extractor object defined for it.
You can actually define an object for +: to do exactly what you are looking for:
object +: {
def unapply[T](s: Seq[T]) =
if(s.nonEmpty)
Some(s.head, s.tail)
else
None
}
scala> val h +: t = Seq(1,2,3)
h: Int = 1
t: Seq[Int] = List(2, 3)
Then your code works exactly as expected.
This works because h +: t is equivalent to +:(h,t) when used for patten matching.
I don't think there is pattern matching support for arbitrary sequences in the standard library. You could do it with out pattern matching though:
def doMatch(seq: Seq[Int]) {
if (seq.size == 1) println("final element " + seq(0)) else {
println("recursing")
doMatch(seq.tail)
}
}
doMatch(1 to 10)
You can define your own extractor objects though. See http://www.scala-lang.org/node/112
object SEQ {
def unapply[A](s:Seq[A]):Option[(A, Seq[A])] = {
if (s.size == 0) None else {
Some((s.head, s.tail))
}
}
}
def doMatch(seq: Seq[Int]) {
seq match {
case SEQ(head, Seq()) => println("final")
case SEQ(head, tail) => {
println("recursing")
doMatch(tail)
}
}
}
A simple tranformation from Seq to List would do the job:
def doMatch (list: List[Int]): Unit = list match {
case last :: Nil => println ("Final element.")
case head :: tail => println ("Recursing..."); doMatch (tail)
case Nil => println ("only seen for empty lists")
}
def doMatchSeq (seq: Seq[Int]) : Unit = doMatch (seq.toList)
doMatch (List(3, 4, 5))
doMatchSeq (3 to 5)
Related
Say I want to do a simple conversion of strings to ints thus:
List("1", "3", "55", "x", "7") => List(1, 3, 55, 7)
One way to do this would be the following recursive call:
def recurse1(strs: List[String]): List[Int] = strs match {
case h :: t =>
try {
h.toInt :: recurse1(t)
}
catch {
case _ : java.lang.NumberFormatException =>
recurse1(t)
}
case _ =>
List()
}
However this cannot be compiled as tail recursive due to line 4 in the code. So to get around this I can redefine the function as follows:
def recurse2(strs: List[String], accum: List[Int] = List()): List[Int] = strs match {
case h :: t =>
try {
recurse2(t, h.toInt :: accum)
}
catch {
case _ : java.lang.NumberFormatException =>
recurse2(t, accum)
}
case _ =>
accum.reverse
}
So my question is this. Is there an idiom I can use in scala that will allow me to do this tail recursively but without having to pass a variable to accumulate the values?
Maybe your recurse method is just for illustration but for completeness I'll add to #pamu's answer how to use standard functions:
def foo(ss: List[String]): List[Int] =
ss.map(s => Try(s.toInt).toOption)
.filter(_.isDefined)
.map(_.get)
or
def foo(ss: List[String]): List[Int] =
ss.map(s => Try(s.toInt))
.collect { case Success(n) => n }
Usually, I see programmers write a helper function which takes many arguments (internally) which is specific to method/algorithm you are using. They write a minimal interface function around the ugly internal function which is tail recursive and takes only inputs required and hides the internal mechanism.
def reverse(input: List[Sting]): List[Int] = {
def helper(strs: List[String], accum: List[Int] = List()): List[Int] =
strs match {
case h :: t =>
try {
helper(t, h.toInt :: accum)
}
catch {
case _ : java.lang.NumberFormatException =>
helper(t, accum)
}
case _ =>
accum.reverse
}
helper(input, List.empty[Sting])
}
How do I rewrite the following loop (pattern) into Scala, either using built-in higher order functions or tail recursion?
This the example of an iteration pattern where you do a computation (comparison, for example) of two list elements, but only if the second one comes after first one in the original input. Note that the +1 step is used here, but in general, it could be +n.
public List<U> mapNext(List<T> list) {
List<U> results = new ArrayList();
for (i = 0; i < list.size - 1; i++) {
for (j = i + 1; j < list.size; j++) {
results.add(doSomething(list[i], list[j]))
}
}
return results;
}
So far, I've come up with this in Scala:
def mapNext[T, U](list: List[T])(f: (T, T) => U): List[U] = {
#scala.annotation.tailrec
def loop(ix: List[T], jx: List[T], res: List[U]): List[U] = (ix, jx) match {
case (_ :: _ :: is, Nil) => loop(ix, ix.tail, res)
case (i :: _ :: is, j :: Nil) => loop(ix.tail, Nil, f(i, j) :: res)
case (i :: _ :: is, j :: js) => loop(ix, js, f(i, j) :: res)
case _ => res
}
loop(list, Nil, Nil).reverse
}
Edit:
To all contributors, I only wish I could accept every answer as solution :)
Here's my stab. I think it's pretty readable. The intuition is: for each head of the list, apply the function to the head and every other member of the tail. Then recurse on the tail of the list.
def mapNext[U, T](list: List[U], fun: (U, U) => T): List[T] = list match {
case Nil => Nil
case (first :: Nil) => Nil
case (first :: rest) => rest.map(fun(first, _: U)) ++ mapNext(rest, fun)
}
Here's a sample run
scala> mapNext(List(1, 2, 3, 4), (x: Int, y: Int) => x + y)
res6: List[Int] = List(3, 4, 5, 5, 6, 7)
This one isn't explicitly tail recursive but an accumulator could be easily added to make it.
Recursion is certainly an option, but the standard library offers some alternatives that will achieve the same iteration pattern.
Here's a very simple setup for demonstration purposes.
val lst = List("a","b","c","d")
def doSomething(a:String, b:String) = a+b
And here's one way to get at what we're after.
val resA = lst.tails.toList.init.flatMap(tl=>tl.tail.map(doSomething(tl.head,_)))
// resA: List[String] = List(ab, ac, ad, bc, bd, cd)
This works but the fact that there's a map() within a flatMap() suggests that a for comprehension might be used to pretty it up.
val resB = for {
tl <- lst.tails
if tl.nonEmpty
h = tl.head
x <- tl.tail
} yield doSomething(h, x) // resB: Iterator[String] = non-empty iterator
resB.toList // List(ab, ac, ad, bc, bd, cd)
In both cases the toList cast is used to get us back to the original collection type, which might not actually be necessary depending on what further processing of the collection is required.
Comeback Attempt:
After deleting my first attempt to give an answer I put some more thought into it and came up with another, at least shorter solution.
def mapNext[T, U](list: List[T])(f: (T, T) => U): List[U] = {
#tailrec
def loop(in: List[T], out: List[U]): List[U] = in match {
case Nil => out
case head :: tail => loop(tail, out ::: tail.map { f(head, _) } )
}
loop(list, Nil)
}
I would also like to recommend the enrich my library pattern for adding the mapNext function to the List api (or with some adjustments to any other collection).
object collection {
object Implicits {
implicit class RichList[A](private val underlying: List[A]) extends AnyVal {
def mapNext[U](f: (A, A) => U): List[U] = {
#tailrec
def loop(in: List[A], out: List[U]): List[U] = in match {
case Nil => out
case head :: tail => loop(tail, out ::: tail.map { f(head, _) } )
}
loop(underlying, Nil)
}
}
}
}
Then you can use the function like:
list.mapNext(doSomething)
Again, there is a downside, as concatenating lists is relatively expensive.
However, variable assignemends inside for comprehensions can be quite inefficient, too (as this improvement task for dotty Scala Wart: Convoluted de-sugaring of for-comprehensions suggests).
UPDATE
Now that I'm into this, I simply cannot let go :(
Concerning 'Note that the +1 step is used here, but in general, it could be +n.'
I extended my proposal with some parameters to cover more situations:
object collection {
object Implicits {
implicit class RichList[A](private val underlying: List[A]) extends AnyVal {
def mapNext[U](f: (A, A) => U): List[U] = {
#tailrec
def loop(in: List[A], out: List[U]): List[U] = in match {
case Nil => out
case head :: tail => loop(tail, out ::: tail.map { f(head, _) } )
}
loop(underlying, Nil)
}
def mapEvery[U](step: Int)(f: A => U) = {
#tailrec
def loop(in: List[A], out: List[U]): List[U] = {
in match {
case Nil => out.reverse
case head :: tail => loop(tail.drop(step), f(head) :: out)
}
}
loop(underlying, Nil)
}
def mapDrop[U](drop1: Int, drop2: Int, step: Int)(f: (A, A) => U): List[U] = {
#tailrec
def loop(in: List[A], out: List[U]): List[U] = in match {
case Nil => out
case head :: tail =>
loop(tail.drop(drop1), out ::: tail.drop(drop2).mapEvery(step) { f(head, _) } )
}
loop(underlying, Nil)
}
}
}
}
list // [a, b, c, d, ...]
.indices // [0, 1, 2, 3, ...]
.flatMap { i =>
elem = list(i) // Don't redo access every iteration of the below map.
list.drop(i + 1) // Take only the inputs that come after the one we're working on
.map(doSomething(elem, _))
}
// Or with a monad-comprehension
for {
index <- list.indices
thisElem = list(index)
thatElem <- list.drop(index + 1)
} yield doSomething(thisElem, thatElem)
You start, not with the list, but with its indices. Then, you use flatMap, because each index goes to a list of elements. Use drop to take only the elements after the element we're working on, and map that list to actually run the computation. Note that this has terrible time complexity, because most operations here, indices/length, flatMap, map, are O(n) in the list size, and drop and apply are O(n) in the argument.
You can get better performance if you a) stop using a linked list (List is good for LIFO, sequential access, but Vector is better in the general case), and b) make this a tiny bit uglier
val len = vector.length
(0 until len)
.flatMap { thisIdx =>
val thisElem = vector(thisIdx)
((thisIdx + 1) until len)
.map { thatIdx =>
doSomething(thisElem, vector(thatIdx))
}
}
// Or
val len = vector.length
for {
thisIdx <- 0 until len
thisElem = vector(thisIdx)
thatIdx <- (thisIdx + 1) until len
thatElem = vector(thatIdx)
} yield doSomething(thisElem, thatElem)
If you really need to, you can generalize either version of this code to all IndexedSeqs, by using some implicit CanBuildFrom parameters, but I won't cover that.
I'm looking to match a sequence within a sequence like either in ex 1 or ex 2;
List(1, 2, 3, 4) match {
case 1 :: List(_*) :: 4 :: tail => // Ex 1
case 1 :: (seq : List[Int]) :: 4 :: tail => // Ex 2
case _ =>
}
This is a variant to the fixed length sequence pattern _*. Like the _*, I don't care about the content of the inner pattern, but it is important that the length can vary and that the pattern is surrounded by a prefix (such as the 1 above) and a suffix (like the 4).
My question is if any of you have a trick to do this by some crafty unapply magic or if you'd just iterate the list to search for the sequence manually.
Thanks in advance! :-)
This is really outside the scope of what pattern matching is supposed to be used for. Even if you could finagle a set of custom unapply methods to do what you wanted, it wouldn't be apparent whether matching was greedy or not, etc..
However, if you really want to, you can proceed as follows (for example):
import scala.collection.SeqLike
class Decon[A](a0: A, a1: A) {
def unapply[C <: SeqLike[A, C]](xs: C with SeqLike[A, C]): Option[(C, C)] = {
xs.span(_ != a1) match {
case (a0 +: pre, a1 +: post) => Some((pre,post))
case _ => None
}
}
}
val Dc = new Decon(1,4)
scala> List(1,2,3,4,5) match { case Dc(pre, post) => (pre, post); case _ => (Nil, Nil) }
res1: (List[Int], List[Int]) = (List(2, 3),List(5))
Separating the specification of the fixed elements 1 and 4 from the match is necessary; otherwise the normal algorithm would ask the unapply to return values without any knowledge of what was being sought, and then would test to make sure they were correct.
You could do something like this:
List(1,2,3,4) match {
case 1 :: tail if tail.last == 4 => println("Found it!")
}
But if you check the implementation of last in LinearSeqOptimized:
def last: A = {
if (isEmpty) throw new NoSuchElementException
var these = this
var nx = these.tail
while (!nx.isEmpty) {
these = nx
nx = nx.tail
}
these.head
}
It is iterating indeed. This is because List inherits from LinearSeq which are optimized to provide efficient head and tail operations. For what you want to do, it is better to use an IndexedSeq implementation like Vector which has an optimized length operation, used in last:
override /*TraversableLike*/ def last: A = {
if (isEmpty) throw new UnsupportedOperationException("empty.last")
apply(length-1)
}
So you could do something like this:
Vector(1,2,3,4) match {
case v if v.head == 1 && v.last == 4 => println("Found it!")
}
Given a sequence of elements and a predicate p, I would like to produce a sequence of sequences such that, in each subsequence, either all elements satisfy p or the sequence has length 1. Additionally, calling .flatten on the result should give me back my original sequence (so no re-ordering of elements).
For instance, given:
val l = List(2, 4, -6, 3, 1, 8, 7, 10, 0)
val p = (i : Int) => i % 2 == 0
I would like magic(l,p) to produce:
List(List(2, 4, -6), List(3), List(1), List(8), List(7), List(10, 0))
I know of .span, but that method stops the first time it encounters a value that doesn't satisfy p and just returns a pair.
Below is a candidate implementation. It does what I want, but, well, makes we want to cry. I would love for someone to come up with something slightly more idiomatic.
def magic[T](elems : Seq[T], p : T=>Boolean) : Seq[Seq[T]] = {
val loop = elems.foldLeft[(Boolean,Seq[Seq[T]])]((false,Seq.empty)) { (pr,e) =>
val (lastOK,s) = pr
if(lastOK && p(e)) {
(true, s.init :+ (s.last :+ e))
} else {
(p(e), s :+ Seq(e))
}
}
loop._2
}
(Note that I do not particularly care about preserving the actual type of the Seq.)
I would not use foldLeft. It's just a simple recursion of span with a special rule if the head doesn't match the predicate:
def magic[T](elems: Seq[T], p: T => Boolean): Seq[Seq[T]] =
elems match {
case Seq() => Seq()
case Seq(head, tail # _*) if !p(head) => Seq(head) +: magic(tail, p)
case xs =>
val (prefix, rest) = xs span p
prefix +: magic(rest, p)
}
You could also do it tail-recursive, but you need to remember to reverse the output if you're prepending (as is sensible):
def magic[T](elems: Seq[T], p: T => Boolean): Seq[Seq[T]] = {
def iter(elems: Seq[T], out: Seq[Seq[T]]) : Seq[Seq[T]] =
elems match {
case Seq() => out.reverse
case Seq(head, tail # _*) if !p(head) => iter(tail, Seq(head) +: out)
case xs =>
val (prefix, rest) = xs span p
iter(rest, prefix +: out)
}
iter(elems, Seq())
}
For this task you can use takeWhile and drop combined with a little pattern matching an recursion:
def magic[T](elems : Seq[T], p : T=>Boolean) : Seq[Seq[T]] = {
def magic(elems: Seq[T], result: Seq[Seq[T]]): Seq[Seq[T]] = elems.takeWhile(p) match {
// if elems is Nil, we have a result
case Nil if elems.isEmpty => result
// if it's not, but we don't get any values from takeWhile, we take a single elem
case Nil => magic(elems.tail, result :+ Seq(elems.head))
// takeWhile gave us something, so we add it to the result
// and drop as many elements from elems, as takeWhile gave us
case xs => magic(elems.drop(xs.size), result :+ xs)
}
magic(elems, Seq())
}
Another solution using a fold:
def magicFilter[T](seq: Seq[T], p: T => Boolean): Seq[Seq[T]] = {
val (filtered, current) = (seq foldLeft (Seq[Seq[T]](), Seq[T]())) {
case ((filtered, current), element) if p(element) => (filtered, current :+ element)
case ((filtered, current), element) if !current.isEmpty => (filtered :+ current :+ Seq(element), Seq())
case ((filtered, current), element) => (filtered :+ Seq(element), Seq())
}
if (!current.isEmpty) filtered :+ current else filtered
}
I am trying to understand the Scala quicksort example from Wikipedia. How could the sample be disassembled step by step and what does all the syntactic sugar involved mean?
def qsort: List[Int] => List[Int] = {
case Nil => Nil
case pivot :: tail =>
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
}
As much as I can gather at this stage qsort is a function that takes no parameters and returns a new Function1[List[Int],List[Int]] that implements quicksort through usage of pattern matching, list manipulation and recursive calls. But I can't quite figure out where the pivot comes from, and how exactly the pattern matching syntax works in this case.
UPDATE:
Thanks everyone for the great explanations!
I just wanted to share another example of quicksort implementation which I have discovered in the Scala by Example by Martin Odersky. Although based around arrays instead of lists and less of a show-off in terms of varios Scala features I personally find it much less convoluted than its Wikipedia counterpart, and just so much more clear and to the point expression of the underlying algorithm:
def sort(xs: Array[Int]): Array[Int] = {
if (xs.length <= 1) xs
else {
val pivot = xs(xs.length / 2)
Array.concat(
sort(xs filter (pivot >)),
xs filter (pivot ==),
sort(xs filter (pivot <)))
}
}
def qsort: List[Int] => List[Int] = {
case Nil => Nil
case pivot :: tail =>
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
}
let's pick apart a few bits.
Naming
Operators (such as * or +) are valid candidates for method and class names in Scala (hence you can have a class called :: (or a method called :: for that matter - and indeed both exist). Scala appears to have operator-overloading but in fact it does not: it's merely that you can declare a method with the same name.
Pattern Matching
target match {
case p1 =>
case p2 =>
}
Where p1 and p2 are patterns. There are many valid patterns (you can match against Strings, types, particular instances etc). You can also match against something called an extractor. An extractor basically extracts arguments for you in the case of a match, so:
target match {
case MyExtractor(arg1, arg2, arg3) => //I can now use arg1, arg2 etc
}
In scala, if an extractor (of which a case class is an example) exists called X, then the pattern X(a, b) is equivalent to a X b. The case class :: has a constructor taking 2 arguments and putting this together we get that:
case x :: xs =>
case ::(x, xs) =>
Are equivalent. This match says "if my List is an instance of :: extract the value head into x and tail into xs". pattern-matching is also used in variable declaration. For example, if p is a pattern, this is valid:
val p = expression
This why we can declare variables like:
val x :: xs = List(1, 2, 3)
val (a, b) = xs.partition(_ % 2 == 0 ) //returns a Tuple2 which is a pattern (t1, t2)
Anonymous Functions
Secondly we have a function "literal". tail is an instance of List which has a method called partition which takes a predicate and returns two lists; one of those entries satisfying the predicate and one of those entries which did not.
val pred = (el: Int) => e < 2
Declares a function predicate which takes an Int and returns true iff the int value is less than 2. There is a shorthand for writing functions inline
tail.partition(_ < pivot) // _ is a placeholder for the parameter
tail.partition( (e: Int) => e < pivot )
These two expressions mean the same thing.
Lists
A List is a sealed abstract class with only two implementations, Nil (the empty list) and :: (also called cons), which is a non-empty list consisting of a head and a tail (which is also a list). You can now see that the pattern match is a match on whether the list is empty or not. a List can be created by cons-ing it to other lists:
val l = 1 :: 2 :: Nil
val m = List(1, 2, 3) ::: List(4, 5, 6)
The above lines are simply method calls (:: is a valid method name in scala). The only difference between these and normal method calls is that, if a method end in a colon : and is called with spaces, the order of target and parameter is reversed:
a :: b === b.::(a)
Function Types
val f: A => B
the previous line types the reference f as a function which takes an A and returns a B, so I could then do:
val a = new A
val b: B = f(a)
Hence you can see that def qsort: List[Int] => List[Int] declares a method called qsort which returns a function taking a List[Int] and returning a List[Int]. So I could obviously do:
val l = List(2, 4, 1)
val m = qsort.apply(l) //apply is to Function what run is to Runnable
val n = qsort(l) //syntactic sugar - you don't have to define apply explicitly!
Recursion
When a method call is tail recursive, Scala will optimize this into the iterator pattern. There was a msitake in my original answer because the qsort above is not tail-recursive (the tail-call is the cons operator)
def qsort: List[Int] => List[Int] = {
case Nil => Nil
case pivot :: tail =>
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
}
Let's rewrite that. First, replace the function literal with an instance of Function1:
def qsort: List[Int] => List[Int] = new Function1[List[Int], List[Int]] {
def apply(input: List[Int]): List[Int] = input match {
case Nil => Nil
case pivot :: tail =>
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
}
}
Next, I'm going to replace the pattern match with equivalent if/else statements. Note that they are equivalent, not the same. The bytecode for pattern matches are more optimized. For instance, the second if and the exception throwing below do not exist, because the compile knows the second match will always happen if the first fails.
def qsort: List[Int] => List[Int] = new Function1[List[Int], List[Int]] {
def apply(input: List[Int]): List[Int] = if (input == Nil) {
Nil
} else if (input.isInstanceOf[::[_]] &&
scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]) != None) {
val unapplyResult = scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]).get
val pivot = unapplyResult._1
val tail = unapplyResult._2
val (smaller, rest) = tail.partition(_ < pivot)
qsort(smaller) ::: pivot :: qsort(rest)
} else {
throw new scala.MatchError(input)
}
}
Actually, val (smaller, rest) is pattern match as well, so Let's decompose it as well:
def qsort: List[Int] => List[Int] = new Function1[List[Int], List[Int]] {
def apply(input: List[Int]): List[Int] = if (input == Nil) {
Nil
} else if (input.isInstanceOf[::[_]] &&
scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]) != None) {
val unapplyResult0 = scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]).get
val pivot = unapplyResult0._1
val tail = unapplyResult0._2
val tmp0 = tail.partition(_ < pivot)
if (Tuple2.unapply(tmp0) == None)
throw new scala.MatchError(tmp0)
val unapplyResult1 = Tuple2.unapply(tmp0).get
val smaller = unapplyResult1._1
val rest = unapplyResult1._2
qsort(smaller) ::: pivot :: qsort(rest)
} else {
throw new scala.MatchError(input)
}
}
Obviously, this is highly unoptmized. Even worse, there are some function calls being done more than once, which doesn't happen in the original. Unfortunately, to fix that would require some structural changes to the code.
There's still some syntactic sugar here. There is an anonymous function being passed to partition, and there is the syntactic sugar for calling functions. Rewriting those yields the following:
def qsort: List[Int] => List[Int] = new Function1[List[Int], List[Int]] {
def apply(input: List[Int]): List[Int] = if (input == Nil) {
Nil
} else if (input.isInstanceOf[::[_]] &&
scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]) != None) {
val unapplyResult0 = scala.collection.immutable.::.unapply(input.asInstanceOf[::[Int]]).get
val pivot = unapplyResult0._1
val tail = unapplyResult0._2
val func0 = new Function1[Int, Boolean] {
def apply(input: Int): Boolean = input < pivot
}
val tmp0 = tail.partition(func0)
if (Tuple2.unapply(tmp0) == None)
throw new scala.MatchError(tmp0)
val unapplyResult1 = Tuple2.unapply(tmp0).get
val smaller = unapplyResult1._1
val rest = unapplyResult1._2
qsort.apply(smaller) ::: pivot :: qsort.apply(rest)
} else {
throw new scala.MatchError(input)
}
}
For once, the extensive explanations about each syntactic sugar and how it works are being done by others. :-) I hope this complements their answers. Just as a final note, the following two lines are equivalent:
qsort(smaller) ::: pivot :: qsort(rest)
qsort(rest).::(pivot).:::(qsort(smaller))
The pivot in this pattern matching example is the first element of the list:
scala> List(1,2,3) match {
| case x :: xs => println(x)
| case _ => println("empty")
| }
1
The pattern matching is based on extractors and the cons is not part of the language. It uses the infix syntax. You can also write
scala> List(1,2,3) match {
| case ::(x,xs) => println(x)
| case _ => println("empty")
| }
1
as well. So there is a type :: that looks like the cons operator. This type defines how it is extracted:
final case class ::[B](private var hd: B, private[scala] var tl: List[B]){ ... }
It's a case class so the extractor will be generated by the Scala compiler. Like in this example class A.
case class A(x : Int, y : Int)
A(1,2) match { case x A y => printf("%s %s", x, y)}
-> 1 2
Based on this machinary patterns matching is supported for Lists, Regexp and XML.