haskell - Haskell 尾递归如何工作？

Question

我编写了这段代码，并且我假设 len 是尾递归的，但仍然会发生堆栈溢出。怎么了？

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]

Answer 1

请记住，Haskell 很懒。除非绝对必要，否则您的计算 (l+1) 不会发生。

“简单”的修复方法是使用“$!”强制评价:

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]