Defining a Map from String to Function in Scala - scala

I am trying to define a Map literal with key: String, value: (Any)=>String. I tried the following, but get a syntax error:
def foo(x: Int): String = /...
def bar(x: Boolean): String = /...
val m = Map[String, (Any) => String]("hello" -> foo, "goodbye" -> bar)

Funny that no one actually gave a type that would work. Here's one such:
def foo(x: Int): String = x.toString
def bar(x: Boolean): String = x.toString
val m = Map[String, (Nothing) => String]("hello" -> foo, "goodbye" -> bar)
The reason why it works this way is because Function1 is contra-variant on the input, so (Nothing) => String is a superclass of (Int) => String. It is also co-variant on the output, so (Nothing) => Any would be a superclass to any other Function1.
Of course, you can't use it like that. Without manifests, you can't even uncover what the original type of Function1 is. You could try something like this, though:
def f[T : Manifest](v: T) = v -> manifest[T]
val m = Map[String, ((Nothing) => String, Manifest[_])]("hello" -> f(foo), "goodbye" -> f(bar))
val IntManifest = manifest[Int]
val BooleanManifest = manifest[Boolean]
val StringManifest = manifest[String]
m("hello")._2.typeArguments match {
case List(IntManifest, StringManifest) =>
m("hello")._1.asInstanceOf[(Int) => String](5)
case List(BooleanManifest, StringManifest) =>
m("hello")._1.asInstanceOf[(Boolean) => String](true)
case _ => "Unknown function type"
}

Int => String is not a subclass of Any => String, rather, the contrary. You can't put (replace) an Int => String function when a code expects Any => String, since that code can apply the function with, say, "hi".
#Ben suggestion works, but how is it useful? you can't invoke the function once you get it from the Map.
If you really want to do this, maybe define foo as a partial function:
val foo: PartialFunction[Any, String] = {case i: Int => ....}
Obviously, this will fail at runtime if you pass it a string, but you can always test if the function is ok for use with your parameter by using isDefinedAt. (another alternative may be manifests, but I don't see the value here)

If I let the compiler infer it I seem to get an illegal type:
scala> val m = Map("hello" -> foo _, "goodbye" -> bar _)
m: scala.collection.immutable.Map[java.lang.String,(Boolean with Int) => String] =
Map((hello,<function1>), (goodbye,<function1>))
scala> m("hello")(8)
<console>:9: error: type mismatch;
found : Int(8)
required: Boolean with Int
m("hello")(8)
scala> var q = new Boolean with Int
<console>:5: error: illegal inheritance from final class Boolean
var q = new Boolean with Int
Anyway, what you want is not the type Any but a generic of "any type" which is _:
scala> val mm = Map[String, (_) => String]("hello" -> foo _, "goodbye" -> bar _)
mm: scala.collection.immutable.Map[String,Function1[_, String]] =
Map((hello,<function1>), (goodbye,<function1>))
I just posted a question about how to invoke such functions because I don't actually know.

Trait Function1 is contravariant for parameter, so def foo(x: Int): String is not a (Any) => String. So the following would work:
scala> def baz(x: Any): String = "baz"
baz: (x: Any)String
scala> val m2 = Map[String, (String) => String]("hello" -> baz)
m2: scala.collection.immutable.Map[String,(String) => String] = Map((hello,<function1>))

This is how I did it to fulfill a similar requirement.
object MapToMethods {
private def increment(x: Int): Int = x+1
private def decrement(x: Int): Int = x-1
val m: Map[String, Int => Int] =Map("increment" -> increment, "decrement" ->decrement)
println(m("increment")(2)) //prints 3
println(m("decrement")(3)) //prints 2
}

Related

How to use ConcurrentHashMap computeIfAbsent() in Scala

I'm using a ConcurrentHashMap in Scala and I would like to use the computeIfAbsent() method but can't figure out the syntax for the second argument. Can someone show me what would be the proper syntax?
When running the following code
val data = new ConcurrentHashMap[String, LongAdder]
data.computeIfAbsent("bob", k: String => new LongAdder()).increment()
I'm getting the following error
Type mismatch, expected: Function[_ >: String, _ <: LongAdder], actual: (String) => Any
Thanking you in advance
Francis
The problem is that you're using java.util.concurrent.ConcurrentHashMap, which accepts java.util.function.Function as a parameter for computeIfAbsent() instead of scala.Function1 which you pass to it.
Since scala doesn't support lambda conversion for functional interfaces as Java does (at least not without the -Xexperimental flag), you can solve this by implementing a java.util.function.Function explicitly:
val data = new ConcurrentHashMap[String, LongAdder]
val adderSupplier = new java.util.function.Function[String, LongAdder]() {
override def apply(t: String): LongAdder = new LongAdder()
}
data.computeIfAbsent("bob", adderSupplier).increment()
Alternatively, if you need this more often, you may write a utility conversion function or even an implicit conversion:
object FunctionConverter {
implicit def scalaFunctionToJava[From, To](function: (From) => To): java.util.function.Function[From, To] = {
new java.util.function.Function[From, To] {
override def apply(input: From): To = function(input)
}
}
}
import FunctionConverter._
val data = new ConcurrentHashMap[String, LongAdder]()
data.computeIfAbsent("bob", (k: String) => new LongAdder()) // <- implicit conversion applied here
If you enable -Xexperimental flag you can use scala anonymous function notation for this:
scala> val data = new java.util.concurrent.ConcurrentHashMap[String, Int]
data: java.util.concurrent.ConcurrentHashMap[String,Int] = {}
scala> data.computeIfAbsent("bob", _.size)
res0: Int = 3
Note that you still can't pass regular scala Function
scala> val f: String => Int = _.size
f: String => Int = <function1>
scala> data.computeIfAbsent("bob", f)
<console>:13: error: type mismatch;
found : String => Int
required: java.util.function.Function[_ >: String, _ <: Int]
data.computeIfAbsent("bob", f)
^
But eta-expansion will work
scala> def a(s: String): Int = s.size
a: (s: String)Int
scala> data.computeIfAbsent("bob", a)
res3: Int = 3

Scala Function1 Generic parameter and return type

Lets say I want to store a Map[String, Function1] where the parameter and return type of the Function1 can vary. How would I go about storing a Function1[String, String] and Function1[Int, Int] in the same Map.
I've tried Function1[AnyRef, AnyRef] but Function1[String, String] isn't a Function1[AnyRef, AnyRef] so it fails to compile.
If you have only two possible value types, you can wrap values in Either:
val m = Map[String, Either[Int => Int, String => String]]()
If you want to store more than two different types, create your own wrapper, or use something like Coproduct from shapeless.
Does this look like what you're after?
scala> val m = Map[String, Function1[_,_]]()
m: scala.collection.immutable.Map[String,Function1[_, _]] = Map()
scala> val f1 = (i:Int) => i*3
f1: Int => Int = <function1>
scala> val f2 = (b: Boolean) => if (b) "YES" else "NO"
f2: Boolean => String = <function1>
scala> m + ("xcx" -> f2) + ("rtr" -> f1)
res59: scala.collection.immutable.Map[String,Function1[_, _]] = Map(xcx -> <function1>, rtr -> <function1>)

Defining a scala Map of functions with variable return types

Is it possible to generalize the return types of a Map of functions with variable return types to a common signature and then use the actual return type of each function in the Map at runtime?
Explanation:
I'm having a scala Map of string -> functions defined as:
Map[String, (String) => Seq[Any]] = Map("1" -> foo, 2 -> bar, 3 -> baz)
where foo, bar and baz are defined as:
foo(string: String): Seq[A]
bar(string: String): Seq[B]
baz(string: String): Seq[C]
The compilation works fine but at runtime Seq[A or B or C] types returned by the functions are being treated as Seq[Any] thereby giving me a reflection exception.
Lets imagine some Map-alike workaround
Suppose we define type like that
import scala.reflect.runtime.universe._
class PolyMap[K, V[+_]] {
var backing = Map[(K, TypeTag[_]), V[Any]]()
def apply[T: TypeTag](key: K) =
backing.get(key, typeTag[T]).asInstanceOf[Option[V[T]]]
def update[T: TypeTag](key: K, f: V[T]): this.type = {
backing += (key, typeTag[T]) → f
this
}
}
now having
type String2Seq[+X] = String ⇒ Seq[X]
val polyMap = new PolyMap[String, String2Seq]
polyMap("foo") = foo
polyMap("bar") = bar
you could ensure that
polyMap[String]("foo").map(_("x")) == Some(foo("x"))
polyMap[Int]("foo").map(_("x")) == None
polyMap[Int]("bar").map(_("x")) == Some(bar("x"))
I think you can try this variant
def foo(string: String): Seq[String] = {
Seq(string)
}
def bar(string: String): Seq[Int] = {
Seq(1)
}
val map = Map(1 -> foo _, 2 -> bar _)
val res = map(1) match {
case f: (String => Seq[String]) => f
case f: (String => Seq[Int]) => f
case _ => throw new NotImplementedError()
}
println(res("Hello"))
it work's for me.

Overcoming Scala Type Erasure For Function Argument of Higher-Order Function

Essentially, what I would like to do is write overloaded versions of "map" for a custom class such that each version of map differs only by the type of function passed to it.
This is what I would like to do:
object Test {
case class Foo(name: String, value: Int)
implicit class FooUtils(f: Foo) {
def string() = s"${f.name}: ${f.value}"
def map(func: Int => Int) = Foo(f.name, func(f.value))
def map(func: String => String) = Foo(func(f.name), f.value)
}
def main(args: Array[String])
{
def square(n: Int): Int = n * n
def rev(s: String): String = s.reverse
val f = Foo("Test", 3)
println(f.string)
val g = f.map(rev)
val h = g.map(square)
println(h.string)
}
}
Of course, because of type erasure, this won't work. Either version of map will work alone, and they can be named differently and everything works fine. However, it is very important that a user can call the correct map function simply based on the type of the function passed to it.
In my search for how to solve this problem, I cam across TypeTags. Here is the code I came up with that I believe is close to correct, but of course doesn't quite work:
import scala.reflect.runtime.universe._
object Test {
case class Foo(name: String, value: Int)
implicit class FooUtils(f: Foo) {
def string() = s"${f.name}: ${f.value}"
def map[A: TypeTag](func: A => A) =
typeOf[A] match {
case i if i =:= typeOf[Int => Int] => f.mapI(func)
case s if s =:= typeOf[String => String] => f.mapS(func)
}
def mapI(func: Int => Int) = Foo(f.name, func(f.value))
def mapS(func: String => String) = Foo(func(f.name), f.value)
}
def main(args: Array[String])
{
def square(n: Int): Int = n * n
def rev(s: String): String = s.reverse
val f = Foo("Test", 3)
println(f.string)
val g = f.map(rev)
val h = g.map(square)
println(h.string)
}
}
When I attempt to run this code I get the following errors:
[error] /src/main/scala/Test.scala:10: type mismatch;
[error] found : A => A
[error] required: Int => Int
[error] case i if i =:= typeOf[Int => Int] => f.mapI(func)
[error] ^
[error] /src/main/scala/Test.scala:11: type mismatch;
[error] found : A => A
[error] required: String => String
[error] case s if s =:= typeOf[String => String] => f.mapS(func)
It is true that func is of type A => A, so how can I tell the compiler that I'm matching on the correct type at runtime?
Thank you very much.
In your definition of map, type A means the argument and result of the function. The type of func is then A => A. Then you basically check that, for example typeOf[A] =:= typeOf[Int => Int]. That means func would be (Int => Int) => (Int => Int), which is wrong.
One of ways of fixing this using TypeTags looks like this:
def map[T, F : TypeTag](func: F)(implicit ev: F <:< (T => T)) = {
func match {
case func0: (Int => Int) #unchecked if typeOf[F] <:< typeOf[Int => Int] => f.mapI(func0)
case func0: (String => String) #unchecked if typeOf[F] <:< typeOf[String => String] => f.mapS(func0)
}
}
You'd have to call it with an underscore though: f.map(rev _). And it may throw match errors.
It may be possible to improve this code, but I'd advise to do something better. The simplest way to overcome type erasure on overloaded method arguments is to use DummyImplicit. Just add one or several implicit DummyImplicit arguments to some of the methods:
implicit class FooUtils(f: Foo) {
def string() = s"${f.name}: ${f.value}"
def map(func: Int => Int)(implicit dummy: DummyImplicit) = Foo(f.name, func(f.value))
def map(func: String => String) = Foo(func(f.name), f.value)
}
A more general way to overcome type erasure on method arguments is to use the magnet pattern. Here is a working example of it:
sealed trait MapperMagnet {
def map(foo: Foo): Foo
}
object MapperMagnet {
implicit def forValue(func: Int => Int): MapperMagnet = new MapperMagnet {
override def map(foo: Foo): Foo = Foo(foo.name, func(foo.value))
}
implicit def forName(func: String => String): MapperMagnet = new MapperMagnet {
override def map(foo: Foo): Foo = Foo(func(foo.name), foo.value)
}
}
implicit class FooUtils(f: Foo) {
def string = s"${f.name}: ${f.value}"
// Might be simply `def map(func: MapperMagnet) = func.map(f)`
// but then it would require those pesky underscores `f.map(rev _)`
def map[T](func: T => T)(implicit magnet: (T => T) => MapperMagnet): Foo =
magnet(func).map(f)
}
This works because when you call map, the implicit magnet is resolved at compile time using full type information, so no erasure happens and no runtime type checks are needed.
I think the magnet version is cleaner, and as a bonus it doesn't use any runtime reflective calls, you can call map without underscore in the argument: f.map(rev), and also it can't throw runtime match errors.
Update:
Now that I think of it, here magnet isn't really simpler than a full typeclass, but it may show the intention a bit better. It's a less known pattern than typeclass though. Anyway, here is the same example using the typeclass pattern for completeness:
sealed trait FooMapper[F] {
def map(foo: Foo, func: F): Foo
}
object FooMapper {
implicit object ValueMapper extends FooMapper[Int => Int] {
def map(foo: Foo, func: Int => Int) = Foo(foo.name, func(foo.value))
}
implicit object NameMapper extends FooMapper[String => String] {
def map(foo: Foo, func: String => String) = Foo(func(foo.name), foo.value)
}
}
implicit class FooUtils(f: Foo) {
def string = s"${f.name}: ${f.value}"
def map[T](func: T => T)(implicit mapper: FooMapper[T => T]): Foo =
mapper.map(f, func)
}

Some questions about difference between call-by-name and 0-arity functions

There are some discussions here about this, but I have some specific questions I wasn't able to find an answer for. So, by call-by-name, I mean =>T type, and by 0-arity function I mean () => T
I understand (I think) the conceptual difference, but probably I am missing something as I still have lots of questions:
Why do we have the concept of =>T at all if we could always use () => T?
Is there any syntax/functional limitations of each? For now I found only that => cannnot be used as a class field. Is this the only limitation?
Is the generated code always the same for both?
Should I always prefer =>T? And why?
Is it correct to call =>T a type? It looks for me like it does not have any type representation in scala.
1) It's just more handy to use it, especially inside DSLs:
def printAndGet[T](f: => T) = {
val res = f
println(res + " printed")
res
}
scala> :paste
// Entering paste mode (ctrl-D to finish)
val k = printAndGet {
val a = 5
5 * a
}
// Exiting paste mode, now interpreting.
25 printed
k: Int = 25
2) => T can only be a parameter of method or function. And actually => T and () => T aren't interchangable:
scala> def aaa(f: => String) = f
aaa: (f: => String)String
scala> val a: Function1[() => String, String] = aaa _
<console>:8: error: type mismatch;
found : (=> String) => String
required: (() => String) => String
val a: Function1[() => String, String] = aaa _
^
Thanks to #som-snytt, fоund this one:
scala> object O { def f(i: Int) = i; def f(i: => Int) = i + 1 }
defined object O
scala> O.f(5)
res12: Int = 5
scala> O.f(5: (=> Int))
<console>:1: error: no by-name parameter type allowed here
O.f(5: (=> Int))
^
Even this which should work if it compiles - but it doesn't (scala 2.11.2, 2.11.5 REPL just crashes):
scala> val k: (=> Int) => Int = O.f _
k: (=> Int) => Int = <function1>
scala> k(5) //should be 6
res18: Int = 5 //WTF?
Last one seems like a bug
3) Not exactly, if you want the same, just convert => T into () => T:
scala> def aaa(f: => String) = {f _}
aaa: (f: => String)() => String
Bytecode may also differ. For instance, compiler will more likely inline code from => T without generating lambda for it. So, the key difference is that () => T is actually an object (first class citizen), => T isn't.
4) see 1, but sometimes you may need to ensure that user knows that computation might be delayed - () => T is better then.
5) It's part of a type signature, just look at eta-expansion of:
scala> def aaa(f: => String) = {f}
aaa: (f: => String)String
scala> aaa _ //convert method into a function
res7: (=> String) => String = <function1>
scala> val a: ( => String) => String = aaa _
a: (=> String) => String = <function1>
However scala doesn't recognize it as independent type:
scala> val a: Function1[( => String), String] = aaa _
<console>:1: error: no by-name parameter type allowed here
val a: Function1[( => String), String] = aaa _
^