[Haskell-beginners] function not working as expected, type issues

[David McBride]

Try using guards:
displayD :: IO () -> String -> Int
displayD string count =
        | count > 0 = hPutStr stdout string
        | otherwise = displayD string (count - 1)

Alternatively use if, then, else.



On Wed, Apr 9, 2014 at 2:03 AM, Alexej Magura <sickhadas at gmail.com> wrote:

> K, so I have a function that I'm writing that takes a *string* and a
> *count* and prints the *string* to STDOUT *count* times:
>
> displayD :: IO () -> String -> Int
> displayD string count = do
>         (count > 0) && hPutStr stdout string
>         (count > 0) && displayD string (count - 1)
>
> However, when I try to compile the file, I get several errors; here's a
> pastebin: http://pastebin.com/DEsuAvfz
>
> What I'm trying to do here is check to see if *count* is greater than 0,
> and then if it is, then I'd like to print *string* to STDOUT until *count*
> equals 0.  I tried doing it via pattern matching, like so:
>
> displayD string (count > 0) = do
>
> But I haven't seen any examples of how to do pattern matching in functions
> using *do*, that is *IO*, so I'm unsure if the above is even anywhere near
> correct.
>
> Still very uncomfortable with declaring function types: I have a hard time
> figuring out which type is returned and what type is expected as an arg.
>  My initial guess is that the *first* type specified is the one returned,
> but I'm not sure.
>
> --
> Alexej Magura
> Sent with Airmail
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