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I've been diving into scala, and am absolutely loving it so far. I'm getting around to pattern matching and case classes, and the following has me somewhat stumped. I know what it does, but I want to understand exactly what is happening

var list = List(2,3,4)

If I am correct in my understanding so far. Then the :: represents a case object. If so my question is:

how am I "left applying" it to the 1? instead of :: being a method of 1. Basically can someone pull this statement 1::list apart somewhat, showing what is really happening (ie. what methods are being called on what object)


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5 Answers

up vote 16 down vote accepted

It's annoying that some of the most visible and awesome features of Scala have so much complexity right beneath the surface. So, consider this simple line:

val (head :: tail): ::[Int] = 1 :: Nil

Each of the three places where :: appears refers to a different ::, and a different mechanism in Scala. Let's go through each of them, in order.

head :: tail

What is happening here is pattern matching, just like one sees with case statements. Pattern matching can appear on val assignments, on the left side of <- in for comprehensions, and on case statements.

So, how does this particular pattern matching happens? Well, whenever the pattern is in the format a b c, Scala translates this into b(a, c), which is then translated into calls to unapply or unapplySeq on the object b.

So, :: in val (head :: tail) refers to the object :: (defined through a case class).

: ::[Int]

This is a type declaration, so ::[Int] is a type. :: itself is a class, and a type constructor as well (because it constructs types given a type parameter -- ::[Int] is one type, ::[String] is another type, etc). It is also a subclass of List, which has only two subclasses: :: and the singleton class of Nil.

This declaration is superfluous, and, generally speaking, one hardly ever uses :: as a type or class. I show it here mostly for completeness.

1 :: Nil

Here, :: is a method. It is a method of List, so, since Nil is a List and 1 is not, it must belong to Nil (or be available through implicit conversion).

The mechanism of note here is that methods ending with :, when used in infix operator notation, bind to the right instead of to the left. Or, in other words, a :: b is equivalent to b.::(a).

This mechanism is rarely used and, I suspect, made mostly to make traditional fp list algorithms more familiar to programmers used to fp. It is used in a few other places on Scala standard library and out of it.

On Scala 2.8, for instance, there's now +:, which serves the same purpose of ::, but is defined for all Seq. It is mirrored by :+, which appends elements, and whose : serves no purpose other than disambiguate it from +, which is overloaded to concatenate strings.

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An operator name ending in : binds to the right. :: [A] is really a case class and a subclass of List[A], as can be seen in the Scala API reference. :: is also a method operating on its right argument, a list, taking its left operand as an argument and returning a ::[A].

So, in your example, the method :: is called on list with 1 as an argument. This constructs an object of type :: [Int] with arguments 1 and list.

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@npiv :: is an operator/method of List –  Vasil Remeniuk Mar 15 '11 at 12:31
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The very quick answer. The method :: is being called on list with 1 as the parameter.

Any method name (and yes, :: is a method) that ends in a colon operates on the right operand.

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To add some additional information to the other answers:

:: is both a subclass of List

final case class ::[B](private var hd: B, private[scala] var tl: List[B]) extends List[B] { ... }

and a method on List

def ::[B >: A] (x: B): List[B] = new scala.collection.immutable.::(x, this)

which is basically implemented in terms of the first one.

So there is no deep theoretical background behind this one, just keep in mind the rule about methods ending with :.

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My understanding of :: (the "cons" operator) is that it is used in this manner because prepending an element to a list (technically, creating a new list with the new element prepended to the old list) is of O(1) , whereas appending it would be of O(N). It's an efficiency thing...

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