Haskell如何进行尾递归工作？ [英] How does Haskell tail recursion work?

问题描述

我写了这段代码，我假设 len 是尾递归，但仍然会发生堆栈溢出。什么是错的？

  myLength :: [a]  - >整数
 
 myLength xs = len xs 0 
其中len [] l = l 
 len（x：xs）l = len xs（l + 1）
 
 main = print $ myLength [1..10000000] 
  

解决方案

记住Haskell是懒惰的。你的计算（1 + 1）不会发生，直到绝对必要。

'easy'修正是使用'$！'来强制评估：

  myLength :: [a]  - >整数
 myLength xs = len xs 0 
其中len [] l = l 
 len（x：xs）l = len xs $！ （l + 1）
 
 main = print $ myLength [1..10000000] 
  

I wrote this snippet of code and I assume len is tail-recursive, but a stack overflow still occurs. What is wrong?

myLength :: [a] -> Integer

myLength xs = len xs 0
    where len [] l = l
          len (x:xs) l = len xs (l+1)

main = print $ myLength [1..10000000]

解决方案

Remember that Haskell is lazy. Your computation (l+1) will not occur until it's absolutely necessary.

The 'easy' fix is to use '$!' to force evaluation:

myLength :: [a] -> Integer
myLength xs = len xs 0
where len [] l = l
      len (x:xs) l = len xs $! (l+1)

      main = print $ myLength [1..10000000]

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