Reader monad example

reader.hs

import Control.Monad.Reader

hello :: Reader String String
hello = do
    name <- ask
    return ("hello, " ++ name ++ "!")

bye :: Reader String String
bye = do
    name <- ask
    return ("bye, " ++ name ++ "!")

convo :: Reader String String
convo = do
    c1 <- hello
    c2 <- bye
    return $ c1 ++ c2

main = print . runReader convo $ "adit"

Out of curiosity, is there any reason not to write the last line as the (imo simpler to me as a relative Haskell newb) as:

main = print $ runReader convo "adit"

?

I ask because $ is typically only used to allow the evaluation of its right hand side, when you need to call more functions to produce the argument you want to pass. Using it with a literal RHS because you needed to do some work to produce the function was confusing to me at first.

Similarly, line 17 uses $ while lines 6 and 11 use parens. That's a more typical usage of $, so I grasped it quickly enough, but the style difference for no apparent reason feels like bad form for example code. Using one or the other for all three would be nice. ^_^

Out of curiosity, is there any reason not to write the last line as the (imo simpler to me as a relative Haskell newb) as:

main = print $ runReader convo "adit"

?

I ask because $ is typically only used to allow the evaluation of its right hand side, when you need to call more functions to produce the argument you want to pass. Using it with a literal RHS because you needed to do some work to produce the function was confusing to me at first.

Similarly, line 17 uses $ while lines 6 and 11 use parens. That's a more typical usage of $, so I grasped it quickly enough, but the style difference for no apparent reason feels like bad form for example code. Using one or the other for all three would be nice. ^_^

ok I see that shared data, but the readers are literally being unwrapped concatenated and then rewrapped... what about <$> and >>=?

ok I see that shared data, but the readers are literally being unwrapped concatenated and then rewrapped... what about <$> and >>=?

answered my own question:

convo = hello >>= \h -> (bye >>= \b -> return $ h ++ b)

so the reader makes sure hello bye and the returned functions are all called with the same input

alternatively with fmap (<$>):

convo = hello >>= \h -> (\b -> h ++ b) <$> bye

or with apply (<*>)

import Control.Monad.Reader

hello :: Reader String String
hello = asks $ \name -> ("hello, " ++ name ++ "!")

bye :: Reader String String
bye = asks $ \name -> ("bye, " ++ name ++ "!")

convo :: Reader String String
convo = asks (const (++)) <*> hello <*> bye

main = print . runReader convo $ "adit"

answered my own question:

convo = hello >>= \h -> (bye >>= \b -> return $ h ++ b)

so the reader makes sure hello bye and the returned functions are all called with the same input

alternatively with fmap (<$>):

convo = hello >>= \h -> (\b -> h ++ b) <$> bye

or with apply (<*>)

import Control.Monad.Reader

hello :: Reader String String
hello = asks $ \name -> ("hello, " ++ name ++ "!")

bye :: Reader String String
bye = asks $ \name -> ("bye, " ++ name ++ "!")

convo :: Reader String String
convo = asks (const (++)) <*> hello <*> bye

main = print . runReader convo $ "adit"