Hi, There are several issues with the code: - type annotation states that the function returns 'IO ()', while on the else branch 'IO Int' is returned. To fix it you should either change the returned type to 'IO Int' or replace 'return n' with 'return ()'; - in do notation you have to use let syntax, so 'x n = n * n' should be 'let x n = n * n'; - this brings us to another issue in line 'putStrLn (show x n)', which is equivalent to 'putStrLn ((show x) n)'. You should put parentheses explicitly to fix it, or change x definition to 'let x = n * n', so that you can just write 'putStrLn (show x)'. Cheers! Mateusz On Sat, May 10, 2014 at 04:32:56PM +0200, Roelof Wobben wrote:
Hello,
I have to make a loop where on the number 1 - 100 the square is calculated.
So I did this :
loop :: Int -> IO () loop n = do if n <= 100 then do x n = n * n; putStrLn (show n ); putChar ' '; putStrLn (show x n); loop (n + 1); else return n ;
main :: IO () main = loop 1
but now I see this error message :
* Main.hs@5:13-5:14 parse error on input `=' * * undefine :: IO ()
So somthing is wrong at this line x n = n * n But its looking well to me.
Where did I take the wrong lane or were did I misunderstood something. Give me a tip or a explanation please and not a answer. Otherwise I do not learn anything .
Roelof
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
-- mlen