YCombinator在Swift中不起作用 [英] YCombinator not working in Swift

问题描述

我正在尝试创建一个lambda函数，以获取阶乘函数，但这会引发分段错误和错误.我如何在Swift中使用它.请观看此视频，以获取我想做的事情的参考

I am trying to create a lambda function as such to get a factorial function but this throws a segmentation fault and errors out. How do I get this working in Swift. Please look at this video for reference on what I am trying to do http://www.confreaks.com/videos/1287-rubyconf2012-y-not-adventures-in-functional-programming

typealias f = () -> ()
typealias g = (Int) -> (Int)
typealias F = Any -> g

let y = { (gen: Any) -> g in
    (gen as F)(gen)
}
let fact = y({ (gen: Any) -> g in
    { (n: Int) -> Int in
        if n == 0 {
            return 1
        } else {
            return n * (gen as F)(gen)(n - 1)
        }
    }
})

fact(10)

推荐答案

您可以使用递归类型实现真正的(没有显式递归)Y组合器，而无需任何不安全的技巧(贷记

You can implement a real (without explicit recursion) Y combinator using a recursive type, without any unsafe tricks (credits to Rosetta Code):

struct RecursiveFunc<F> {
  let o : RecursiveFunc<F> -> F
}

func Y<A, B>(f: (A -> B) -> A -> B) -> A -> B {
  let r = RecursiveFunc<A -> B> { w in f { w.o(w)($0) } }
  return r.o(r)
}

let factorial = Y { (f: Int -> Int) -> Int -> Int in
  { $0 <= 1 ? 1 : $0 * f($0-1) }
}
println(factorial(10))

Any并没有真正的帮助，因为

Any doesn't really help because Any cannot represent function types.

更新:从Xcode 6.1 beta 3开始，Any可以表示函数类型，并且您的代码可以编译并正常工作.

Update: Starting in Xcode 6.1 beta 3, Any can represent function types, and your code compiles and works correctly.

这篇关于YCombinator在Swift中不起作用的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！