Scala Functional Literals with Implicits

问题:

Forgive me if this has already been asked elsewhere. I have a Scala syntax question involving function-values and implicit parameters.

I'm comfortable using implicits with Scala's currying feature. For instance if I had a sum function and wanted to make the second argument an implicit:

scala> def sum(a: Int)(implicit b: Int) = a + b
sum: (a: Int)(implicit b: Int)Int

Is there a way to do this using the function-value syntax? Ignoring the implicit for a moment, I typically write curried function-values like this:

scala> val sum2 = (a: Int) => (b: Int) => a + b
sum: (Int) => (Int) => Int = <function1>

However, the function signature in the second approach is much different (the currying is being expressed explicitly). Just adding the implicit keyword to b doesn't make much sense and the compiler complains as well:

scala> val sum2 = (a: Int) => (implicit b: Int) => a + b
<console>:1: error: '=>' expected but ')' found.
       val sum2 = (a: Int) => (implicit b: Int) => a + b
                                              ^

Furthermore partially-applying sum from the very first approach to get a function-value causes problems as well:

scala> val sumFunction = sum _
<console>:14: error: could not find implicit value for parameter b: Int
       val sumFunction = sum _
                         ^

This leads me to believe that functions that have implicit parameters must have said parameters determined when the function-value is created, not when the function-value is applied later on. Is this really the case? Can you ever use an implicit parameter with a function-value?

Thanks for the help!

回答1:

scala>  val sum2 = (a: Int) => {implicit b: Int => a + b}
sum2: (Int) => (Int) => Int = <function1>

This will just make b an implicit value for the scope of the function body, so you can call methods that expect an implicit Int.

I don't think you can have implicit arguments for functions since then it is unclear what the function is. Is it Int => Int or () => Int?

The closest I found is:

scala> case class Foo(implicit b: Int) extends (Int => Int) {def apply(a: Int) = a + b}
defined class Foo

scala> implicit val b = 3
b: Int = 3

scala> Foo()
res22: Foo = <function1>

scala> res22(2)
res23: Int = 5

回答2:

In this snippet

scala> val sum2 = (a: Int) => (b: Int) => a + b
sum: (Int) => (Int) => Int = <function1>

Note that the precise type of sum2 is Function1[Int, Function1[Int, Int]]. It could also be written as

val sum2 = new Function1[Int, Function1[Int, Int]] {
    def apply(a: Int) = new Function1[Int, Int] {
        def apply(b: Int) = a + b
    }
}

Now, if you try to make b implicit, you get this:

scala>     val sum2 = new Function1[Int, Function1[Int, Int]] {
     |         def apply(a: Int) = new Function1[Int, Int] {
     |             def apply(implicit b: Int) = a + b
     |         }
     |     }
<console>:8: error: object creation impossible, since method apply in trait Function1 of type (v1: Int)Int is not defined
               def apply(a: Int) = new Function1[Int, Int] {
                                       ^

Or, in other words, Function's interfaces do not have implicit parameters, so anything with an implicit parameter is not a Function.

回答3:

Try overloading the apply method.

scala> val sum = new Function1[Int, Function1[Int, Int]] {
         |      def apply(a: Int) = (b: Int) => a + b
         |      def apply(a: Int)(implicit b: Int) = a + b
         |}
sum: java.lang.Object with (Int) => (Int) => Int{def apply(a:Int)(implicit b: Int): Int} = <function1>

scala> sum(2)(3)
res0: Int = 5

scala> implicit val b = 10
b: Int = 10

scala> sum(2)
res1: Int = 12