Haskell, error unwrapping Maybe Int -


i wrote code:

is3 :: int -> maybe int  is3 x = is3temp 0 x   is3temp :: int -> int -> maybe int  is3temp p x = if (abs p*p*p) < (abs x) (is3temp (p+(signum x)) x) else (if p*p*p == x (just p) else nothing)   c :: maybe int -> int  c (just x) = 2*x+1  c nothing = 0  --fun::int ->  int --fun = c.is3  c1:: int -> int c1 x = 2*x +1 c1 0 = 0  fun::int ->  int fun x = (is3 x) >>= c1

as can see is3 takes int , returns maybe int.

in fun take result is3 unwrap , try send c1.

and error

error file:1627.hs:47 - type error in application *** expression     : is3 x >>= c1 *** term           : c1 *** type           : int -> int *** not match : -> b c

what wrong here?

let's inspect types come in is3 x >>= c1:

x            :: int is3          :: int -> maybe int  c1           ::                      int ->   int                                      |||     /???\ is3 x        ::        maybe int     |||    ??????? (>>=)        ::        maybe   -> (a   -> maybe b) -> maybe b (>>=) is3 x  ::                     (int -> maybe b) -> maybe b is3 x >>= c1 ::                                         $!%&/#?

as can see, c1 doesn't work right hand side of >>=, since it's return type wrong. must return kind of maybe. 1 can fix return . c1:

return . c1  :: int -> maybe int

or fmap, since monad laws dictate both must have same effect:

fmap c1         :: maybe int -> maybe int fmap c1 (is3 x) ::              maybe int

however, change fun's type. being said, functions name rather bad , c1 typo (if fun's type correct). it's possible wanted use c:

fun :: int -> int fun x = c (is3 x)

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