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

Alexej Magura sickhadas at gmail.com
Wed Apr 9 06:03:54 UTC 2014

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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