Haskell function checking if number is odd, without using the odd function

Question

Can anyone help me with that? I am trying to write a function checking if an x is odd, without using the odd function. Like this it does not work but i don't know why.

   ugerade :: Integral a => a -> Bool
   ugerade x
    |x elem oddList  = True
    |otherwise = False
    where
     oddList=[x | x<-[1,3..]]

Error

    Could not deduce (Num t0) arising from the literal ‘1’
from the context (Integral a)
  bound by the type signature for ugerade :: Integral a => a -> Bool
  at /Users/Mauritius/Desktop/Haskell/u02/2-2/funktionen.hs:24:11-33
The type variable ‘t0’ is ambiguous
Relevant bindings include
  oddList :: [t0]
    (bound at /Users/Mauritius/Desktop/Haskell/u02/2-2/funktionen.hs:29:4)
Note: there are several potential instances:
  instance Integral a => Num (GHC.Real.Ratio a)
    -- Defined in ‘GHC.Real’
  instance Num Integer -- Defined in ‘GHC.Num’
  instance Num Double -- Defined in ‘GHC.Float’
  ...plus three others
In the expression: 1
In the expression: [1, 3 .. ]
In a stmt of a list comprehension: x <- [1, 3 .. ]

no it isn't. but i wanted to do it with a list-comprehension. i don't get the mistake — Mauritius, Apr 13 '16 at 09:19
`not . even` ? :-) More in the spirit of your effort so far would be `ungerade x = Data.Maybe.fromJust (lookup x (zip [1..] (cycle [True,False])))` which at least terminates for positive `x`. — yatima2975, Apr 13 '16 at 15:41
i solved it with modulo. did not think about that first but is obviously the easiest way if you do not want to use even — Mauritius, Apr 14 '16 at 17:25

Chris Taylor · Accepted Answer · 2016-04-13T09:28:40.287

The problem is in the line

x elem oddList

which should either say

elem x oddList

since elem is a function elem :: Eq a => a -> [a] -> Bool, or

x `elem` oddList

where you are using backticks to indicate infix function application.

Note that your function does not work as you intended. For odd numbers it will eventually return True (although it takes a long time for large arguments) but for even numbers it will never return, because the function cannot prove that an even number is never in the list oddList.

Also note that writing

oddList = [ x | x <- [1,3..] ]

is redundant, you could just write

oddList = [1,3..]

instead, and also writing

f x | x `elem` oddList = True
    | otherwise        = False

is redundant, where you could just write

f x = x `elem` oddList

or even

f x = x `elem` [1,3..]

or

f = (`elem` [1,3..])

thanks. That works! At least no error, but it is not terminating either — Mauritius, Apr 13 '16 at 09:26

Haskell function checking if number is odd, without using the odd function

1 Answers1