Hello there here is my issue. This is my running code, which is fine:
showBalls = do howMany <- getInt return . take HowMany $ repeat 9 where my getInt does several checks in order to retrieve user input Int. However, my question is, is there a way to rewrite this part of code with the use of monads?
My aim is to use >>= and have a final one-line functions such as:
showBalls = fmap (take <$> (repeat 9)) getLine however it doesnt work (as expected). Any suggestion?
You can use flip to reverse the arguments of take, which makes it possible to apply repeat 9 to it and get function:
> :t flip take (repeat 9) flip take (repeat 9) :: Num a => Int -> [a] To use >>= with getInt we need Int -> IO a function:
> :t (getInt >>=) (getInt >>=) :: (Int -> IO b) -> IO b To get that, compose with return (it's m [a] because it will work for every monad, but is also restricted to lists; from the above type, m becomes IO and b becomes [a]):
> :t return . flip take (repeat 9) return . flip take (repeat 9) :: (Monad m, Num a) => Int -> m [a] The final expression is:
> :t getInt >>= return . flip take (repeat 9) getInt >>= return . flip take (repeat 9) :: Num a => IO [a] But, as I mentioned, I'm not really convinced it's strictly better than the original. As a more useful bit of knowledge, playing around with expressions and their types in GHCi is a good way of inventing transformations like that.
x >>= return . f is fmap f x or f <$> x, this can be further simplified to flip take (repeat 9) <$> getInt
fmap take (repeat 9) <$> getInt?
flip is correct. The <$> is already doing the fmap.I got it my self too but using lambda notation :)
showBalls = getInt >>= (\a -> return . take a $ repeat 9 ) 
I believe this should work:
showBalls = (liftM . flip take $ repeat 9) getInt There is probably point free way to do it as well, but I couldn't figure it out.
$ instead of .