无限斐波那契序列 [英] Infinite fibonacci sequence

问题描述

我试图在F＃中使用序列模仿哈斯克尔着名的无限斐波那契列表。为什么不按预期评估以下顺序？如何评估它？

  let rec fibs = lazy（Seq.ofpend 
（Seq.ofList [0; 1））
（（Seq.map2（+）（fibs.Force（））
（Seq.skip 1（fibs.Force（））））））
  

解决方案

问题是您的代码仍然不够懒： Seq.append 在可以访问结果之前计算，但计算第二个参数（ Seq.map2 ... ）需要评估它自己的参数，它强制定义相同的惰性值。这可以通过使用 Seq.delay 函数解决
。您也可以放弃懒惰包装，并且 list s已经 seq s，所以你不需要 Seq.ofList ：

  let rec fibs = 
 Seq.append [0; 1] 
（Seq.delay（fun（） - > Seq.map2（+）fibs（Seq.skip 1 fibs）））$然而，我个人建议使用一个序列表达式，我发现这个序列表达式更令人愉快（阅读并且更易于正确写入）： 
 
 
  let rec fibs = seq {
 yield 0 
 yield 1 
产量！ Seq.map2（+）fibs（fibs |> Seq.skip 1）
} 
  
 
I'm trying to imitate Haskell's famous infinite fibonacci list in F# using sequences. Why doesn't the following sequence evaluate as expected? How is it being evaluated?
let rec fibs = lazy (Seq.append 
                        (Seq.ofList [0;1]) 
                        ((Seq.map2 (+) (fibs.Force()) 
                                       (Seq.skip 1 (fibs.Force())))))

解决方案
The problem is that your code still isn't lazy enough: the arguments to Seq.append are evaluated before the result can be accessed, but evaluating the second argument (Seq.map2 ...) requires evaluating its own arguments, which forces the same lazy value that's being defined.  This can be
worked around by using the Seq.delay function.  You can also forgo the lazy wrapper, and lists are already seqs, so you don't need Seq.ofList:
let rec fibs = 
    Seq.append [0;1]
        (Seq.delay (fun () -> Seq.map2 (+) fibs (Seq.skip 1 fibs)))
However, personally I'd recommend using a sequence expression, which I find to be more pleasant to read (and easier to write correctly):
let rec fibs = seq {
    yield 0
    yield 1
    yield! Seq.map2 (+) fibs (fibs |> Seq.skip 1)
}


                        
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