How to solve longEnough?. In Haskell

Question

longEnough n xs: checks if a list has more than n elements.

Examples:

• longEnough 2 [1..5] == True
• longEnough 3 [1,2,3] == False
• longEnough 0 [] == False
• longEnough 20 [1..] == True

Answer 1

I guess this is homework and you are still learning the basics so I'll start by giving you some hints using the recursion instead of foldr (as @dfeuer proposed):

First start by noting some obvious cases:

• if xs = [] then the result is always False (assuming you don't care about negative n in some strange ways)
• if n = 0 and xs is non-empty then it's always True
• in all other cases you have
  • n > 0
  • xs has more than one element

Maybe you have some recursive idea to break the last case down?

Here is a skeleton:

longEnough :: Int -> [a] -> Bool
longEnough _ []     = False
longEnough 0 _      = True
longEnough n (_:xs) = let n' = (n-1) in undefined

For those cases - if you look closely you'll see that I even added more hints on the solution.

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PS

• maybe you want to think about negative n and what should happen to those ... I did not here
• if you know what foldr is all about you should probably try to implement this using foldr too

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Solution

Seems there is no more feedback coming from the OP so I guess I can as well post the solution as I would start with:

longEnough :: Int -> [a] -> Bool
longEnough _ []     = False
longEnough 0 _      = True
longEnough n (_:xs) = longEnough (n-1) xs

(Not really much left to do...)

Here are the mentioned test-cases:

λ> longEnough 2 [1..5]
True
λ> longEnough 3 [1,2,3]
False
λ> longEnough 0 []
False
λ> longEnough 20 [1..]
True