尾函数的Haskell递归

Question

我有这个功能

func1 :: Double -> [(Double,Double)] -> Maybe [(Double,Double)]
...............
func2 :: Double -> [(Double,Double)] -> [(Double,Double)]
func2 d [] =  []
func2 d list = 
  let dsegs1 = func1 d list
      dsegs2 = func2 d (tail list)
  in  fromJust dsegs1 ++ dsegs2

我想在func2实现的简单流程如下：

let x = func2 3.0 list
let y = func2 3.0 (tail list)
let z = func2 3.0 (tail (tail list))
let a = func2 3.0 (tail (tail (tail list)))

调用func2 n 次，直到它最后什么都不返回并连接x,y,z,...,a。

我该怎么做？

Answer 1

看起来你想要map (func2 3.0) (tails list)。