Anonymous recursive function in Scala

问题

Is there a way to write an anonymous function that is recursive in Scala? I'm thinking of something like this:

((t: Tree) => {
    print(t.value);
    for (c <- t.children)
        thisMethod(c)
})(root)

(Related question: Which languages support *recursive* function literals / anonymous functions?)

回答1:

As described in the link you posted. You can use Y-combinator. Here is example:

scala> def fix[A,B](f: (A=>B)=>(A=>B)): A=>B = f(fix(f))(_)
fix: [A,B](f: ((A) => B) => (A) => B)(A) => B

scala> val fact = fix[Int,Int](f => a => if(a<=0) 1 else f(a-1) * a)
fact: (Int) => Int = <function1>

scala> fact(12)
res0: Int = 479001600

Note it doesn't work with big numbers. Be careful with tail call optimization.

回答2:

If you don't want to hit the "Amazing mathematics" you could just revert to the object aspects of scala.

val fact = new Function1[Int,Int]{
    def apply(x:Int):Int = if(x==1) x else x * apply(x-1)
}

回答3:

in order to make it look geekier you can also use this code style:

val fact = new ((Int) => Int){
  def apply(x:Int):Int = if(x==1) x else x * apply(x-1)
}

回答4:

Adding to the many good responses here in this thread, the fact that Scala is not giving us tail call optimizable Fixed-point combinator has been bothering me so much so that I've decided to write a macro to translate Y-combinator-like call to an ordinary, idiomatic recursive call (with tail call optimization, of course). The idea is that a call like

fix[Int,Int]((next) => (y) => ...body...)

is readily translatable into

({(input) =>
  object next {
    def apply(y:Int):Int = ...body...
  }
  next(input)
})

I've put up macro implementation targeting Scala 2.11 (with minor tweak should also work with 2.10) into this gist.

With this macro, we can perform ordinary recursive tasks in anonymous manner without fearing stack overflow e.g.

import asia.blip.ymacro.YMacro._
(y[BigInt,BigInt]((xx) => (y) => if(y==1) 1 else y * xx(y-1)))(2000)

gives

res0: BigInt = 33162750924506332411753933805763240382811...

回答5:

A very simple approach:

val fact = { (x: Int) =>
  def f(x: Int): Int = if (x == 0) 1 else x * f(x-1)
  f(x)
}

// Use as anonymous function below
(1 to 5).map { (x: Int) =>
  def f(x: Int): Int = if (x == 0) 1 else x * f(x-1)
  f(x)
}

// res0: scala.collection.immutable.IndexedSeq[Int] = Vector(1, 2, 6, 24, 120)

来源：https://stackoverflow.com/questions/5337464/anonymous-recursive-function-in-scala

标签

scala

recursion

anonymous-function