Re: [Haskell-beginners] Got "Non type-variable argument in the constraint" error for a simple function

Thanks Frerich and David! I got it, :-) 2016-02-09 0:43 GMT+08:00 David McBride :

If you are wondering why you are having this problem it is because - can be interpretted as either a one argument negation or a two argument subtraction. If you put parenthesis around (-n) where n is an integer, it will interpret it as unary, something that will not happen in other operators.

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:t (-) (-) :: Num a => a -> a -> a :t (+) (+) :: Num a => a -> a -> a :t (-1) (-1) :: Num a => a :t (+1) (+1) :: Num a => a -> a :t (1-1) (1-1) :: Num a => a :t (1+1) (1+1) :: Num a => a

On Mon, Feb 8, 2016 at 11:24 AM, wizard wrote:

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Dear all,

I just started to learn Haskell with learnyouahaskell.com and at the very beginning, I met a strange issue with following simple function:

-- why does work with "toZero 10" but not for "toZero -10"?toZero :: (Integral t) => t -> [t]toZero 0 = [0]toZero x = if x > 0 then x : toZero (x - 1) else x : toZero (x + 1)

This function works as expected for positive arguments, e.g., "toZero 10" gives me [10,9,8,7,6,5,4,3,2,1,0]. However, GHCI will raise following error if I give it a negative argument, e.g., "toZero -10":

*Main> toZero -10

<interactive>:12:1: Non type-variable argument in the constraint: Num (t -> [t]) (Use FlexibleContexts to permit this) When checking that ‘it’ has the inferred type it :: forall t. (Integral t, Num (t -> [t])) => t -> [t]

This seems strange to me as 10 and -10 has exactly the same type "Num a => a". I've done with chapter 1~10 of learnyouahaskell.com but still has no idea on why this error. Anybody can help to explain this? Thanks a lot.

Regards Zhiyi Xie

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