I read, that in Scala you can overload any operator, not as in Groovy. But I saw no any example how to overload function call.
Is it possible / how to overload function call operator in Scala?
I know that "there is no operators in Scala", but despite this, I need to do what I would name that.
function call is spelled apply. Any object which has a method named apply overloads function call.
class Functionish(val int: Int, val str: String) {
def apply(i: Int): Int = i + int
def apply(s: String): String = s + str + s
}
val f = new Functionish(42, "hello")
println(f(4))
println(f("George, "))
prints
46
George, helloGeorge,
Of note: case class companion objects overload apply. This is why you can create a case class instance with
val instance = CaseClass(foo, bar, baz)
instead of
val instance = new CaseClass(foo, bar, baz)
Every "callable" object is called via apply method:
scala> def foo(i: Int) = i * i
foo: (i: Int)Int
scala> val f = foo _
f: Int => Int = <function1>
scala> f(42)
res2: Int = 1764
scala> f.apply(42)
res3: Int = 1764
Is that what are you looking for?
Related
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
Let's say I want to create an alias for a method:
def foo = bar(_)
This will warn that
Anonymous function convertible to a method value
And I'm not quite sure what that's supposed to mean, because when I try what I think this might mean:
def foo = bar
I get an error
Missing arguments for method bar(a:A)
Cannot resolve reference bar with such signature.
First, If you want to create an "alias" for method, that's enough:
scala> val foo = bar(_) //val instead of def, still warning from Idea
foo: Int => Int = <function1>
Second, this shoud remove Idea's warning:
scala> val foo = bar _
foo: Int => Int
Actually, it's not just alias - your method becomes converted into a function (eta-expansion). You cant' just specify method (compile-time entity) as compiler will expect parameters - you need to convert it into a function (using underscore) first. Sometimes it's done automatically when compiler expects a function:
scala> val foo: Int => Int = bar
foo: Int => Int = <function1>
So this probably what Idea wants from you. In other cases - you have to use eta-expansion operator (_) explicitly.
P.S/1. def foo = bar(_) (def instead of val) makes no sense as it will return new (but same) function every time, val (or lazy val to be safe from NullPointerException) just returns it once.
P.S/2. The difference between (_) and _ is that first is partially applied function (which does _ eta-expansion automatically), which means that for let's say:
scala> def bar(a: Int, b: Int) = a
bar: (a: Int, b: Int)Int
scala> def foo = bar _
foo: (Int, Int) => Int
scala> def foo = bar(_)
<console>:8: error: missing parameter type for expanded function ((x$1) => bar(x$1))
def foo = bar(_)
^
<console>:8: error: not enough arguments for method bar: (a: Int, b: Int)Int.
Unspecified value parameter b.
def foo = bar(_)
^
scala> def foo = bar(_, _)
foo: (Int, Int) => Int
you have to specify bar(_, _) as there are two arguments.
That's a suggestion from IntelliJ. If you click "More" in the suggestion, you'll see the following (sorry it's graphic, I can't copy the text):
You can either ignore or use bar _ instead of bar(_).
I have found currying nicer than partial application or implicits for a certain scenario. However I have not found a way to curry for an argument group that is the first in a list of argument groups.
E.g. take an original function:
def a(x: Int)(y: Int) = x + y
It is easy to fix the first argument a la
def b = a(3)_
But how can you fix the second one, and only the second one, instead?
Type ascription works in this case:
scala> def a(a: Int)(b: Int) = a + b
a: (a: Int)(b: Int)Int
scala> a(_: Int)(3)
res12: Int => Int = <function1>
scala> res12(2)
res13: Int = 5
You could use the ordinary function for that:
def b(x: Int) = a(x)(3)
After having discovered that currying multi parameter-groups method is possible, I am trying to get a partially applied function which requires implicit parameters.
It seams not possible to do so. If not could you explain me why ?
scala> def sum(a: Int)(implicit b: Int): Int = { a+b }
sum: (a: Int)(implicit b: Int)Int
scala> sum(3)(4)
res12: Int = 7
scala> val partFunc2 = sum _
<console>:8: error: could not find implicit value for parameter b: Int
val partFunc2 = sum _
^
I use a singleton object to create this partially applied function and I want to use it in a scope where the implicit int is defined.
That is because you don't have an implicit Int in scope. See:
scala> def foo(x: Int)(implicit y: Int) = x + y
foo: (x: Int)(implicit y: Int)Int
scala> foo _
<console>:9: error: could not find implicit value for parameter y: Int
foo _
^
scala> implicit val b = 2
b: Int = 2
scala> foo _
res1: Int => Int = <function1>
The implicit gets replaced with a real value by the compiler. If you curry the method the result is a function and functions can't have implicit parameters, so the compiler has to insert the value at the time you curry the method.
edit:
For your use case, why don't you try something like:
object Foo {
def partialSum(implicit x: Int) = sum(3)(x)
}
scala> object MySingleton {
| def sum(a: Int)(implicit b: Int): Int = { a+b }
|
|
| def caller(a: Int) = {
| implicit val b = 3; // This allows you to define the partial below
| def pf = sum _ // and call sum()() without repeating the arg list.
| pf.apply(a)
| }
| }
defined module MySingleton
scala> MySingleton.caller(10)
res10: Int = 13
I know this question has come up many times in different ways. But it is still not clear to me. Is there a way to achieve the following.
def foo(a:Int, b:Int) = {}
foo(a,b) //right way to invoke foo
foo(getParams) // is there a way to get this working without explicitly unpacking the tuple??
def getParams = {
//Some calculations
(a,b) //where a & b are Int
}
It's a two step procedure. First turn foo into a function, then call tupled on it to make it a function of a tuple.
(foo _).tupled(getParams)
#dave-griffith is dead on.
You can also call:
Function.tupled(foo _)
If you want to wander into "way more information than I asked for" territory, there are also methods built into partially applied functions (and on Function) for currying. A few input/output examples:
scala> def foo(x: Int, y: Double) = x * y
foo: (x: Int,y: Double)Double
scala> foo _
res0: (Int, Double) => Double = <function2>
scala> foo _ tupled
res1: ((Int, Double)) => Double = <function1>
scala> foo _ curried
res2: (Int) => (Double) => Double = <function1>
scala> Function.tupled(foo _)
res3: ((Int, Double)) => Double = <function1>
// Function.curried is deprecated
scala> Function.curried(foo _)
warning: there were deprecation warnings; re-run with -deprecation for details
res6: (Int) => (Double) => Double = <function1>
Wherein the curried version is invoked with multiple argument lists:
scala> val c = foo _ curried
c: (Int) => (Double) => Double = <function1>
scala> c(5)
res13: (Double) => Double = <function1>
scala> c(5)(10)
res14: Double = 50.0
Finally, you can also uncurry/untuple if needed. Function has builtins for this:
scala> val f = foo _ tupled
f: ((Int, Double)) => Double = <function1>
scala> val c = foo _ curried
c: (Int) => (Double) => Double = <function1>
scala> Function.uncurried(c)
res9: (Int, Double) => Double = <function2>
scala> Function.untupled(f)
res12: (Int, Double) => Double = <function2>
Function.tupled(foo _)(getParams) or the one suggested by Dave.
EDIT:
To respond to your comment:
What if foo happens to be the
constructor of some class?
In that case, this trick won't work.
You can write a factory method in the companion object of your class and then obtain the tupled version of its apply method using one of the aforementioned techniques.
scala> class Person(firstName: String, lastName: String) {
| override def toString = firstName + " " + lastName
| }
defined class Person
scala> object Person {
| def apply(firstName: String, lastName: String) = new Person(firstName, lastName)
| }
defined module Person
scala> (Person.apply _).tupled(("Rahul", "G"))
res17: Person = Rahul G
With case classes you get a companion object with an apply method for free, and thus this technique is more convenient to use with case classes.
scala> case class Person(firstName: String, lastName: String)
defined class Person
scala> Person.tupled(("Rahul", "G"))
res18: Person = Person(Rahul,G)
I know that's a lot of code duplication but alas... we don't have macros (yet)! ;)
I appreciate some of the other answers which were closer to what you asked for, but I found it easier for a current project to add another function which converts tuple parameters into the split parameters:
def originalFunc(a: A, b: B): C = ...
def wrapperFunc(ab: (A, B)): C = (originalFunc _).tupled(ab)
Now, you can implement foo and make it take a param of the Tuple2 class like so.
def foo(t: Tuple2[Int, Int]) = {
println("Hello " + t._1 + t._2)
"Makes no sense but ok!"
}
def getParams = {
//Some calculations
val a = 1;
val b = 2;
(a, b) //where a & b are Int
}
// So you can do this!
foo(getParams)
// With that said, you can also do this!
foo(1, 3)