【发布时间】：2012-11-10 14:32:57
【问题描述】：
堆栈溢出：
大家好。
我想定义一个具有可变参数的部分应用函数。
为了确认，我准备了几个函数，即 Functions._。 除了 func 之外的函数都有一个可变参数。
如何使用可变参数定义和调用部分应用的函数？ （例如，带有可变参数的 gunc1。）
java版本“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 的评论后编辑]
我想要像下面的 guncN 这样的部分应用函数：
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
但被scalac禁止。
我认为 guncN 每次都应该评估，在时间 (2)、(2, 3) 和 (2, 3, 4, 5, 100) 给出。 下面我们不需要guncM：
val guncM = guncN(2, _: Int*)
println("result: " + guncM(3, 4)) // => 10
【问题讨论】：
标签： scala