Question

Stack Over Flow：

大家好。

我想定义一个部分应用的函数，它有varargs。

为了确认，我准备了几个函数，即Functions._。除func之外的函数都有一个varargs参数。

如何使用varargs定义和调用部分应用的函数？（例如gunc1 with varargs。）

java版“1.7.0_07” Scala代码运行器版本2.9.1.final

object Functions {
    def func(x: Int, y: Int) = x + y
    def gunc(x: Int*) = x.sum
    def hunc(x: Int, y: Int*) = x + y.sum
    def iunc(x: Int)(y: Int*) = x + y.sum // curried
    def junk(x: String, y: Int*) = x + y.sum
}

object PartiallyApplied extends App {
    import Functions._

    val func0 = func(1, _: Int) // I can.
    println("result: " + func0(2))

    val gunc0 = gunc(1, _: Int) // I can invoke if I specify the varargs.size. But gunc0 no longer has varargs...
    println("result: " + gunc0(2))

//    val gunc1 = gunc(1, _: Int*) => compile error: ')' expected but identifier found.
//    val gunc1 = gunc(1, _: Seq[Int]) => compile error: type mismatch | found: Seq[Int] | required: Int
    val gunc1 = gunc(1, _: Int) // I can invoke if I specify the varargs.size. But gunc1 no longer has varargs...
    println("result: " + gunc1(2))

//    val hunc0 = hunc(1)_ => compile error: _ must follow method; cannot follow Int
//    val hunc0 = hunc(1, _: Int*) => compile error: ')' expected but identifier found.
//    val hunc0 = hunc(1, _: Seq[Int]) => compile error: type mismatch | found: Seq[Int] | required: Int
    val hunc0 = hunc(1, _: Int) // I can invoke if I specify the varargs.size. But hunc0 no longer has varargs...
    println("result: " + hunc0(2))
//    println("result: " + hunc0(2, 3)) => compile error: too many arguments for method apply: (v1: Int)Int in trait Function1
//    println("result: " + hunc0((2, 3): _*)) => compile error: type mismatch | found: (Int, Int) | required: Seq[Int]
//    println("result: " + hunc0(Seq(2, 3))) => compile error: type mismatch | found: Seq[Int] | required: Int
//    println("result: " + hunc0(Seq(2, 3): _*)) => compile error: no `: _*' annotation allowed here (such annotations are only allowed in arguments to *-parameters)

    val hunc1 = hunc(1, _: Int, _: Int) // I can invoke if I specify the varargs.size. But hunc1 no longer has varargs...
    println("result: " + hunc1(2, 3))

    val hunc2 = hunc(1, _: Int, _: Int, _:Int) // I can invoke if I specify the varargs.size. But hunc2 no longer has varargs...
    val hunc3 = hunc2(2, _: Int, _: Int)
    val hunc4 = hunc3(3, _: Int)
    println("result: " + hunc4(4))

    println("result: " + hunc5(2, 3))

//    val iunc0 = iunc(1)(_: Int*) => compile error: ')' expected but identifier found.
//    val iunc0 = iunc(1)(_: Seq[Int]) => compile error: type mismatch | found: Seq[Int] | required: Int
    val iunc0 = iunc(1)(_: Int)
    println("result: " + iunc0(2))

    val iunc1 = iunc(1)(_: Int, _: Int)
    println("result: " + iunc1(2, 3))
}

object NotPartiallyApplied extends App {
    import Functions._

    println("result: " + gunc(1))
    println("result: " + gunc(1, 2, 3))
    println("result: " + gunc(Seq(1, 2, 3): _*))

    println("result: " + hunc(1))
    println("result: " + hunc(1, 2, 3))
    println("result: " + hunc(1, Seq(2, 3): _*))

    println("result: " + iunc(1)(2, 3))
    println("result: " + iunc(1)(Seq(2, 3): _*))

    println("result: " + junk("x"))
    println("result: " + junk("x", 2, 3))
    println("result: " + junk("x", Seq(2, 3): _*))
}

[在Rex Kerr的评论之后编辑]

我想要部分应用的功能如下面的guncN：

val guncN = gunc(1, _: Int*)
println("result: " + guncN(2)) // => 3
println("result: " + guncN(2, 3)) // => 6
println("result: " + guncN(2, 3, 4, 5, 100)) // => 115

但它被scalac禁止。

我认为应该每次评估guncN，时间（2），（2,3）和（2,3,4,5,100）。我们下面不需要guncM：

val guncM = guncN(2, _: Int*)
println("result: " + guncM(3, 4)) // => 10

Answer 1

如果你只想要varargs，你可以在2.9：

Welcome to Scala version 2.9.1.final
(Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_31).
...
scala>     def sum(xs: Int*) = xs.sum
sum: (xs: Int*)Int

scala>     val summer = sum _
summer: Int* => Int = <function1>

scala> summer(1,2,3)
res0: Int = 6

但不是现在的2.10，而且没有明显的改变计划：https://issues.scala-lang.org/browse/SI-4176

Welcome to Scala version 2.10.0-RC2
(Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_31).
...

scala> def sum(xs: Int*) = xs.sum
sum: (xs: Int*)Int

scala> val summer = sum _
summer: Seq[Int] => Int = <function1>

scala> summer(1,2,3)
<console>:10: error: too many arguments for method apply:
(v1: Seq[Int])Int in trait Function1
              summer(1,2,3)

如果您有2.9，那么您可以

scala> def temp(early: Int*)(rest: Int*) = sum((early ++ rest): _*)
temp: (i: Int, j: Int)(rest: Int*)Int

scala> val presummed = temp(1,2) _
presummed: Int* => Int = <function1>

scala> presummed(4,5)
res1: Int = 12

但也许您不应该依赖于varargs函数，因为它似乎已经被弃用（或者至少在-Yeta-expand-keeps-star标志后移动，-Y选项可能无法正常工作或者随时删除。）

Answer 2

我找到的解决方案是编写功能：

Welcome to Scala version 2.10.0-RC1 (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_37).
...
scala> 1 :: (_:Int) :: Nil : Seq[Int]
res42: Int => Seq[Int] = <function1>

scala> gunc _
res43: Seq[Int] => Int = <function1>

scala> res42 andThen res43
res54: Int => Int = <function1>

scala> res54(5)
res56: Int = 6

Answer 3

也许你不应该依赖varargs函数，因为它似乎已被弃用（或者至少被移到-Yeta-expand-keeps-star标志之后）

SI-6816确认弃用，在Scala 2.12.x，2016年12月，PR 5558，该标志明确消失（discussion here）。

我想要一个部分功能，我们不知道args.size BEFOREHAND

这似乎仍然不可能（在Scala 2.12.x，2016年12月）

如何使用varargs定义和调用部分应用的函数？

3 个答案: