Question

假设我有一个函数/方法：

class A {

def function() = {
   subroutine("A")
   ...
   subroutine("B")
   ...
   subroutine("C")
}

def subroutine(a: String) = { ... }

}

是否可以使用Scala反射编程在函数（）中查找子程序（a：String）的所有3个调用，而不调用function（）本身？（这可能需要很长的时间）

Answer 1

您可以尝试通过一些很酷的功能工具来解决这个问题，而不需要内省。

让我们使用scalaz library它有特殊的类型Arrow，它是import scalaz._ import scalaz.syntax.tree._ import scalaz.std.function._ import scalaz.syntax.arrow._ import scalaz.std.string._ case class Subroutine[-A, +B](hier: Seq[Tree[String]], run: A => B) { def named(name: String) = Subroutine(Seq(name.node(hier: _*)), run) def printHier = hier.map(_.drawTree).mkString("\n" + "V" * 15 + "\n") } object Subroutine { def named[A, B](tag: String)(run: A => B) = Subroutine(Seq(tag.leaf), run) implicit def anon[A, B](run: A => B) = Subroutine(Seq.empty, run) implicit object subroutineArrow extends Arrow[Subroutine] { def arr[A, B](f: (A) => B): Subroutine[A, B] = anon(f) def first[A, B, C](f: Subroutine[A, B]): Subroutine[(A, C), (B, C)] = Subroutine(f.hier, f.run.first[C]).named("$1->") override def second[A, B, C](f: Subroutine[A, B]): Subroutine[(C, A), (C, B)] = Subroutine(f.hier, f.run.second[C]).named("$2->") def id[A]: Subroutine[A, A] = anon(identity) def compose[A, B, C](f: Subroutine[B, C], g: Subroutine[A, B]): Subroutine[A, C] = Subroutine(g.hier ++ f.hier, f.run compose g.run) } }的抽象。
它允许你做任何你通常用函数做的事情因此，让我们定义一个特殊的类型，它不仅包含函数，而且以一种可读的方式调用层次结构。

import Subroutine._

val square = { (x: Double) => x * x } named "square"
val sqrt = math.sqrt _ named "sqrt"

val sum = Subroutine.named[(Double, Double), Double]("sum"){ case (x, y) => x + y}

val abs = ((square *** square) >>> sum >>> sqrt) named "abs"

现在让我们定义一些子程序

abs.run(3,4)

从这里你可以验证

5.0

给出结果abs.printHier

而

"abs"
|
+- "$1->"
|  |
|  `- "square"
|
+- "$2->"
|  |
|  `- "square"
|
+- "sum"
|
`- "sqrt"

给出有趣的通话顺序，如

def pack22[X] = Subroutine.anon[(X, X, X, X), ((X, X), (X, X))] { case (a, b, c, d) => ((a, b), (c, d)) }

val abs4 = ((abs *** abs) >>> abs <<< pack22[Double]) named "abs4"

甚至定义

abs4.run(15, 20, 36, 48)

并评估

abs4.printHier

和

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在scala中，如何查找函数中定义的子例程的调用

1 个答案: