Scala 中的匿名递归函数 [英] Anonymous recursive function in Scala

问题描述

有没有办法在 Scala 中编写一个递归的匿名函数?我在想这样的事情:

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)

(相关问题:哪些语言支持*递归*函数字面量/匿名函数?)

推荐答案

如您发布的链接中所述.您可以使用 Y 组合器.示例如下:

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

请注意，它不适用于大数字.小心尾调用优化.

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

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