Haskell recursion of tail function

Question

I have this function

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

The simple flow I would like to achieve in func2 is as follows:

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)))

call func2 n times until it returns nothing at the end and concat x,y,z,..., a.

How would I do that?

Louis Wasserman · Accepted Answer · 2016-04-06 23:09:51Z

1

It looks like you want map (func2 3.0) (tails list).

answered Apr 6, 2016 at 23:09

Louis Wasserman

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1 Comment

Sunil Gautam Over a year ago

thanks.. i solved it anyways but has 5 lines of code.. your one is so simple.. thanks man

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