Re: [Haskell-cafe] Flattening tail recursion?

Will,

countLines [] = 0 countLines (_:ls) = 1 + countLines ls

I would have thought that this was tail recursive and would be flattened into iteration by the compiler. Can anyone explain why? Is it because the call is embedded in an expression?

This is the tail-recursive version: \begin{code} countLines = countLines' 0 where countLines' k [] = k countLines' k (l : ls) = countLines' (k + 1) ls \end{code} See, for example, http://www.haskell.org/hawiki/TailRecursive. HTH, Stefan