Is there an implicit in this call to flatMap? - scala

In this code :
import java.io.File
def recursiveListFiles(f: File): Array[File] = {
val these = f.listFiles
these ++ these.filter(_.isDirectory).flatMap(recursiveListFiles)
}
taken from : How do I list all files in a subdirectory in scala?
Why does flatMap(recursiveListFiles) compile ? as recursiveListFiles accepts a File parameter ? Is file parameter implicitly passed to recursiveListFiles ?

No because the expanded flatMap looks like:
flatMap(file => recursiveListFiles(file))
So each file in these is getting mapped to an Array[File], which gets flattened in flatMap. No implicit magic here (in the way that you're asking).

Sort of. flatMap quite explicitly passes an argument to its own argument. This is the nature of a higher-order function -- you basically give it a callback, it calls the callback, and then it does something with the result. The only implicit thing happening is conversion of a method to a function type.
Any method can be converted to an equivalent function type. So def recursiveListFiles(f: File): Array[File] is equivalent to File => Array[File], which is great, because on an Array[File], you have flatMap[B](f: File => Array[B]): Array[B], and your method's function type fits perfectly: the type parameter B is chosen to be File.
As mentioned in another answer, you could explicitly create that function by doing:
these.filter(_.isDirectory).flatMap(file => recursiveListFiles(file)) or
these.filter(_.isDirectory).flatMap(recursiveListFiles(_)) or
these.filter(_.isDirectory).flatMap(recursiveListFiles _)
In more complex circumstances, you might need to work with one of those more verbose options, but in your case, no need to bother.

flatMap takes a function f: (A) ⇒ GenTraversableOnce[B] to return List[B].
In your case, it is taking recursiveListFiles which is a File ⇒ Array[File] to therefore return a List[File]. This resulting List[File] is then concatenated to these.

Related

Trying to skip implicit parameter list

I'd like to call a function returned by a function with an implicit parameter, simply and elegantly. This doesn't work:
def resolveA(implicit a: A): String => String = { prefix =>
s"$prefix a=$a"
}
case class A(n: Int)
implicit val a = A(1)
println(resolveA("-->")) // won't compile
I've figured out what's going on: Scala sees the ("-->") and thinks it's an attempt to explicitly fill in the implicit parameter list. I want to pass that as the prefix argument, but Scala sees it as the a argument.
I've tried some alternatives, like putting an empty parameter list () before the implicit one, but so far I've always been stopped by the fact that Scala thinks the argument to the returned function is an attempt to fill in the implicit parameter list of resolveA.
What's a nice way to do what I'm trying to do here, even if it's not as nice as the syntax I tried above?
Another option would be to use the apply method of the String => String function returned by resolveA. This way the compiler won't confuse the parameter lists, and is a little shorter than writing implicltly[A].
scala> resolveA[A].apply("-->")
res3: String = --> a=A(1)

Strange implicit def with function parameter behaviour in Scala

I've written a simple code in Scala with implicit conversion of Function1 to some case class.
object MyApp extends App{
case class FunctionContainer(val function:AnyRef)
implicit def cast(function1: Int => String):FunctionContainer = new FunctionContainer(function1)
def someFunction(i:Int):String = "someString"
def abc(f : FunctionContainer):String = "abc"
println(abc(someFunction))
}
But it doesn't work. Compiler doesn't want to pass someFunction as an argument to abc. I can guess its reasons but don't know exactly why it doesn't work.
When you use a method name as you have, the compiler has to pick how to convert the method type to a value. If the expected type is a function, then it eta-expands; otherwise it supplies empty parens to invoke the method. That is described here in the spec.
But it wasn't always that way. Ten years ago, you would have got your function value just by using the method name.
The new online spec omits the "Change Log" appendix, so for the record, here is the moment when someone got frustrated with parens and introduced the current rules. (See Scala Reference 2.9, page 181.)
This has not eliminated all irksome anomalies.
Conversions
The rules for implicit conversions of methods to functions (§6.26) have been tightened. Previously, a parameterized method used as a value was always implicitly converted to a function. This could lead to unexpected results when method arguments were forgotten. Consider for instance the statement below:
show(x.toString)
where show is defined as follows:
def show(x: String) = Console.println(x)
Most likely, the programmer forgot to supply an empty argument list () to toString. The previous Scala version would treat this code as a partially applied method, and expand it to:
show(() => x.toString())
As a result, the address of a closure would be printed instead of the value of s. Scala version 2.0 will apply a conversion from partially applied method to function value only if the expected type of the expression is indeed a function type. For instance, the conversion would not be applied in the code above because the expected type of show’s parameter is String, not a function type. The new convention disallows some previously legal code. Example:
def sum(f: int => double)(a: int, b: int): double =
if (a > b) 0 else f(a) + sum(f)(a + 1, b)
val sumInts = sum(x => x) // error: missing arguments
The partial application of sum in the last line of the code above will not be converted to a function type. Instead, the compiler will produce an error message which states that arguments for method sum are missing. The problem can be fixed by providing an expected type for the partial application, for instance by annotating the definition of sumInts with its type:
val sumInts: (int, int) => double = sum(x => x) // OK
On the other hand, Scala version 2.0 now automatically applies methods with empty parameter lists to () argument lists when necessary. For instance, the show expression above will now be expanded to
show(x.toString())
Your someFunction appears as a method here.
You could try either
object MyApp extends App{
case class FunctionContainer(val function:AnyRef)
implicit def cast(function1: Int => String):FunctionContainer = new FunctionContainer(function1)
val someFunction = (i:Int) => "someString"
def abc(f : FunctionContainer):String = "abc"
println(abc(someFunction))
}
or
object MyApp extends App{
case class FunctionContainer(val function:AnyRef)
implicit def cast(function1: Int => String):FunctionContainer = new FunctionContainer(function1)
def someFunction(i:Int): String = "someString"
def abc(f : FunctionContainer):String = "abc"
println(abc(someFunction(_: Int)))
}
By the way: implicitly casting such common functions to something else can quickly lead to problems. Are you absolutely sure that you need this? Wouldn't it be easier to overload abc?
You should use eta-expansion
println(abc(someFunction _))

Convert a list to a map and get item in one line

Try:
(List(('c', 1)).toMap)('c')
Error:
found : Char('c')
required: <:<[(Char, Int),(?, ?)]
However, this works fine:
val m = List(('c', 1)).toMap
m('c') // gives 1 as expected
Why would it matter whether to store it first in a variable or not? Could this be a bug?
The problem is that the full signature of toMap is the following:
def toMap[T, U](implicit ev: <:<[A, (T, U)]): Map[T, U]
The collections API designers didn't want List(1).toMap to compile, so they require you to provide implicit evidence that the contents of the list are tuples.
Normally you don't have to think about this—you just call toMap without an argument and the compiler finds the implicit evidence. When you write something that looks like you're calling toMap with an argument, though, you get into trouble, because it's always possible to provide implicit arguments explicitly.
The easiest workaround (assuming you don't want to define a separate variable) is to call apply explicitly. whatever(foo) is just syntactic sugar for whatever.apply(foo) (unless whatever is a method), and if you write the following your code will compile:
scala> List(('c', 1)).toMap.apply('c')
res0: Int = 1
Now it's perfectly clear that you're not trying to call the toMap method with c as an argument.

Why do I need a curried function to be able to pass function literals with short placeholder syntax?

Given this definition:
class Foo(var x: String) {}
object Helper {
def model[T](get: ⇒ T, set: T ⇒ Unit) : Model[T] = new Model[T] {
override def getObject(): T = get
override def setObject(obj: T) { set(obj) }
}
}
I try to call model like this:
val f = new Foo("initial")
val stringModel = model(f.x, f.x = _)
But that doesn't work, the compiler gives me this, complaining about the underscore:
missing parameter type for expanded function ((x$1) => f.x = x$1)
If I change the definition of model to use two parameter lists like this:
def model[T](get: ⇒ T)(set: T ⇒ Unit) // rest is unchanged
Then I can call it like this:
model(f.x)(f.x = _)
Which I find nice and concise. I don't really mind doing it like this, though it makes method overloading harder. I would like to understand, however, why the second variant works and the first one doesn't?
The second variant works because Scala refines its types parameter-block by parameter-block. If you don't specify the type of your input parameter for the function, it's possible that it would change the type T that it has inferred based on the first parameter. If you push it to a separate parameter block, Scala's already decided what T must be by the time it hits that block, so it fills in the only possible value for the function argument type.

trouble understanding with Map and map in scala

I tried to make some sort of convenient class (below) to hold folder and get file by using filename (string). This work as expect but one thing I don't understand is map part Map(folder.listFiles map {file => file.getName -> file}:_*).
I place :_* there to prevent some kind of type incompatible but I don't know what does it really do. Also, what is _* and could I replace is with anything more specific ?
thanks
class FolderAsMap (val folderName:String){
val folder = new File(folderName)
private val filesAsMap: Map[String, File] = Map(folder.listFiles map
{file => file.getName -> file}:_*)
def get(fileName:String): Option[File] = {
filesAsMap.get(fileName)
}
}
: _* is correct. Alternatively, you can use toMap:
folder.listFiles map {file => file.getName -> file}.toMap
Map(...) is method apply in object Map: def apply [A, B] (elems: (A, B)*): Map[A, B]. It has a repeated parameter. It is expected to be called with multiple parameters. The : _* is used to signal you are passing all the parameters as just one Seq argument.
It avoids ambiguities. In java, (where equivalent varargs are arrays instead of Seqs) there is a possible ambiguity, if a method f(Object... args) and you call it with f(someArray), it could mean that args has just one item, with is someArray (so f receives an array of just one element, which its someArray), or args is someArray and f receives someArray directly). Java choose the second version. In scala, with a richer type system and Seq rather than Array the ambiguity may arise much more often, and the rule is that you always have to write : _* when passing all arguments as one, even when no ambiguity is possible, as in here, rather than a complex rule to tell when there is an actual ambiguity.
The _* makes the compiler pass each element of folder.listFiles map { file => file.getName -> file} as an own argument to Map instead of all of it as one argument.
In this case the map function creates a Array (because folder.listFiles returns that type). So if you write:
val files = folder.listFiles map { file => file.getName -> file }
...the returned type will be Array[(String, File)]. To convert this to a Map you will need to pass files one by one to the maps constructor using the _* (or use the method toMap like #didierd wrote):
val filesAsMap = Map(files : _*)