I notice that List class define the method ::, which adds an element at the beginning of the list
def ::(x: A): List[A]
Example:
1 :: List(2, 3) = List(2, 3).::(1) = List(1, 2, 3)
However, I am confused at How does scala compiler recognize such conversion? Because as far as I am concerned,
1 :: List(2,3)
should raise an error: :: is not a member of Int
Do I miss something about operator definition of scala?
Methods whose names end with : are right-associative when called using infix operator notation. I.e.
a foo_: b
is the same as
b.foo_:(a)
This rule exists specifically for the case of methods like this, which are commonly (in other languages such as Haskell and ML) operators like : or ::.
Related
I am little bit confused about +: and :: operators that are available.
It looks like both of them gives the same results.
scala> List(1,2,3)
res0: List[Int] = List(1, 2, 3)
scala> 0 +: res0
res1: List[Int] = List(0, 1, 2, 3)
scala> 0 :: res0
res2: List[Int] = List(0, 1, 2, 3)
For my novice eye source code for both methods looks similar (plus-colon method has additional condition on generics with use of builder factories).
Which one of these methods should be used and when?
+: works with any kind of collection, while :: is specific implementation for List.
If you look at the source for +: closely, you will notice that it actually calls :: when the expected return type is List. That is because :: is implemented more efficiently for the List case: it simply connects the new head to the existing list and returns the result, which is a constant-time operation, as opposed to linear copying the entire collection in the generic case of +:.
+: on the other hand, takes CanBuildFrom, so you can do fancy (albeit, not looking as nicely in this case) things like:
val foo: Array[String] = List("foo").+:("bar")(breakOut)
(It's pretty useless in this particular case, as you could start with the needed type to begin with, but the idea is you can prepend and element to a collection, and change its type in one "go", avoiding an additional copy).
Alright, Scala has me feeling pretty dense. I'm finding the docs pretty impenetrable -- and worse, you can't Google the term "Scala ++:" because Google drops the operator terms!
I was reading some code and saw this line:
Seq(file) ++: children.flatMap(walkTree(_))
But couldn't figure it out. The docs for Seq show three things:
++
++:
++:
Where the latter two are over loaded to do.. something. The actual explanation in the doc says that they do the same thing as ++. Namely, add one list to another.
So, what exactly is the difference between the operators..?
++ and ++: return different results when the operands are different types of collection. ++ returns the same collection type as the left side, and ++: returns the same collection type as the right side:
scala> List(5) ++ Vector(5)
res2: List[Int] = List(5, 5)
scala> List(5) ++: Vector(5)
res3: scala.collection.immutable.Vector[Int] = Vector(5, 5)
There are two overloaded versions of ++: solely for implementation reasons. ++: needs to be able to take any TraversableOnce, but an overloaded version is provided for Traversable (a subtype of TraversableOnce) for efficiency.
Just to make sure:
A colon (:) in the end of a method name makes the call upside-down.
Let's make two methods and see what's gonna happen:
object Test {
def ~(i: Int) = null
def ~:(i: Int) = null //putting ":" in the tail!
this ~ 1 //compiled
1 ~: this //compiled
this.~(1) //compiled
this.~:(1) //compiled.. lol
this ~: 1 //error
1 ~ this //error
}
So, in seq1 ++: seq2, ++: is actually the seq2's method.
edited: As #okiharaherbst mentions, this is called as right associativity.
Scala function naming will look cryptic unless you learn a few simple rules and their precedence.
In this case, a colon means that the function has right associativity as opposed to the more usual left associativity that you see in imperative languages.
So ++: as in List(10) ++: Vector(10) is not an operator on the list but a function called on the vector even if it appears on its left hand-side, i.e., it is the same as Vector(10).++:(List(10)) and returns a vector.
++ as in List(10) ++ Vector(10) is now function called on the list (left associativity), i.e., it is the same as List(10).++(Vector(10)) and returns a list.
what exactly is the difference between the operators..?
The kind of list a Seq.++ operates on.
def ++[B](that: GenTraversableOnce[B]): Seq[B]
def ++:[B >: A, That](that: Traversable[B])(implicit bf: CanBuildFrom[Seq[A], B, That]): That
def ++:[B](that: TraversableOnce[B]): Seq[B]
As commented in "What is the basic collection type in Scala?"
Traversable extends TraversableOnce (which unites Iterator and Traversable), and
TraversableOnce extends GenTraversableOnce (which units the sequential collections and the parallel.)
Wasn't sure if I should ask this here or on Programmers, but anyway
In Scala it's possible to write method calls using infix syntax, i.e. omitting dots and parens.
As an example you could do this:
lst foreach println // equivalent to lst.foreach(println)
Naturally one would assume that lst map _.toString would be evaluated to lst.map(_.toString), which is equivalent to lst.map(x$1 => x$1.toString)
But dropping lst map _.toString into the repl yields a surprising result, it's evaluated as ((x$1) => sList.map(x$1.toString)) causing the method call to malfunction.
So why is that? Why is it that the simple rule of a.f(b) being equivalent to a f b no longer applies when writing a f _.b?
Because the expression is ambiguous.
From Scala's (somewhat outdated) spec P94: http://www.scala-lang.org/docu/files/ScalaReference.pdf
An expression(of syntactic category Expr) may contain embedded underscore symbols _ at places where identifiers are legal. Such an expression represents an anonymous function where subsequent occurrences of underscores denote successive parameters.
Since lst map _.toString is a legal expression, it can naturally be evaluated as an anonymous function like (x) => lst.map(x.toString).
You can still use infix expression by curly brackets that make Scala compiler evaluate placeholder function first.
scala> val lst = List(1,2,3,4,5)
lst: List[Int] = List(1, 2, 3, 4, 5)
scala> lst map { _.toString }
res43: List[String] = List(1, 2, 3, 4, 5)
I'm looking at someone else's source code (Scala), where I see the operator :+= being called on a variable of type IndexedSeq. I am looking all over the scaladocs page for that class to figure out what that operator does, but I do not see it. I'm thinking that either it's defined in a class outside of IndexedSeq's inheritance hierarchy, or else the javascript on the scaladocs page is hiding it somewhere I can't see it. (Actually it's neither; see answer below.)
I've hit every button on the scaladocs page trying to unhide everything. I've looked in the web-page's HTML code. There has got to be a way to look up an operator from the documentation of a class to which it can be applied. Hasn't there?
(N.B.: I looked up that operator using symbolhound, so I know what that operator means now. This question is about scala documentation in general, not that particular operator.)
All operators in Scala are normal methods.
You cannot find it because it is compiler magic for re-assignement, it is not an operator. Or to say it another way: it looks like an operator of its own, but it is actually "an operator followed by the = character".
The compiler will magically turn that into a assignment if the operator (here :+) returns the proper type, and the original value was a var, obviously.
Since it is not provided by any implicit nor explicit method on Seq[T] or whatever, it does not appear anywhere in the generated scaladoc.
So to answer the general question:
It is a language construct, so the only place where it is documented is the specification, sadly,
but, if you find some "<?>=" unknown operator somewhere, look for the definition of "<?>", that one is sure to be documented.
Edit: I finally found where this is defined in the SLS:
§6.12.4:
An assignment operator is an operator symbol (syntax category op in (§1.1)) that
ends in an equals character “=”, with the exception of operators for which one of the
following conditions holds:
(1) the operator also starts with an equals character, or
(2) the operator is one of (<=), (>=), (!=).
It also says later on that it only happens when all other options have been tried (including potential implicits).
Is this value assigned to a variable? If it's the case I think this syntax sugar:
scala> var x = IndexedSeq(1,2,3)
x: IndexedSeq[Int] = Vector(1, 2, 3)
scala> x :+= 10
scala> x
res59: IndexedSeq[Int] = Vector(1, 2, 3, 10)
scala> val y = IndexedSeq(1,2,3)
y: IndexedSeq[Int] = Vector(1, 2, 3)
scala> y :+= 10
<console>:16: error: value :+= is not a member of IndexedSeq[Int]
y :+= 10
^
It is syntax sugar for "operation and assignment", like +=:
scala> var x = 10
x: Int = 10
scala> x += 1
scala> x
res63: Int = 11
Which de-sugars to x = x + 1.
What i know:
scala> def fx(s: String *) = s.foreach(println)
fx: (s: String*)Unit
scala> val lst = List("1","2","3")
lst: List[java.lang.String] = List(1, 2, 3)
scala> fx(lst:_*)
1
2
3
What i want to know:
How can I implement :_*? by map?
Is there any other way that replace it?
How :_* defined in Scala?
Thank you
It is only a syntactic sugar to indicates to the compiler that you already provide a sequence of elements, there is no other "implementation" of it. For more information, you can refer to the Scala Language Specification (§6.6, p. 78)
For a method that takes variable arguments :_* means you want to pass the members of a collection as the variable arguments.
The corresponding varargs example to yours above:
scala> fx("a", "b", "c")
a
b
c
You can't implement -- it is a language feature. It doesn't mean anything outside the context of calling a varargs method.