I didn't look closely enough at what you wrote, you had several issues.  Let me try again:

displayD :: String -> Int -> IO ()
displayD string 0 = return ()
displayD string count = do
  hPutStr stdout string
  displayD string (count - 1)

This is one way to do it.  You use the arguments themselves.  Obviously this won't work with negative counts.

displayD :: String -> Int -> IO ()
displayD string count
  | count <= 0 = return ()
  | otherwise = do
      hPutStr stdout string
      displayD string (count - 1)

And as for figuring out types.  You should try to load your source file in ghci, and then use the :t and :i commands liberally to inspect functions you are using.  You will begin to get the hang of it in no time.


On Wed, Apr 9, 2014 at 2:08 AM, David McBride <toad3k@gmail.com> wrote:
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@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
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