Haskell 尾递归如何工作? [英] How does Haskell tail recursion work?

问题描述

我写了这段代码，我假设 len 是尾递归的，但仍然会发生堆栈溢出.怎么了?

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]

推荐答案

记住 Haskell 是懒惰的.除非绝对必要，否则您的计算 (l+1) 不会发生.

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

简单"的解决方法是使用$!"强制评估:

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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