If you apply return to 'a' you get a value of type m Char, belonging to the Monad class: :t return 'a' → return 'a' :: Monad m => m Char. If you apply the resulting value to some other value, say, 42, you get 'a': (return 'a') 42 → 'a'. The same expression with a constraint – (return 'a' :: [Char]) 7 – causes an error. What kind of monadic value, if it is a monadic value, is it? If it is used, what is it used for?

If you apply return to 'a' you get a value of type m Char, belonging to the Monad class: :t return 'a' → return 'a' :: Monad m => m Char. If you apply the resulting value to some other value, say, 42, you get 'a': (return 'a') 42 → 'a'. The same expression with a constraint – (return 'a' :: [Char]) 7 – causes an error. What kind of monadic value, if it is a monadic value, is it? If it is used, what is it used for?

• haskell
• return
• constraints
• monads

Share Improve this question Follow asked Mar 22 at 20:14 SkabSkab 7355 bronze badges 1

• 1 I'm sure there's a better duplicate somewhere, but stackoverflow/questions/69325169/… is very closely related and should be helpful. – amalloy Commented Mar 22 at 20:25

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Sorted by: Reset to default 3

You make a function call with (return 'a') 42 where (return 'a') is thus the function. This means you use the (->) a monad instance. Indeed ^{[Haskell-src]}:

-- | @since base-2.01
instance Functor ((->) r) where
  fmap = (.)

-- | @since base-2.01
instance Applicative ((->) r) where
  pure = const
  f g x = f x (g x)
  liftA2 q f g x = q (f x) (g x)

-- | @since base-2.01
instance Monad ((->) r) where
  f >>= k = \ r -> k (f r) r

So here return = pure = const. Therefore return 'a' is in this context a function that ignores the parameter, and returns 'a', so (return 'a') 42 will return 'a'.

what is it used for?

To combine functions. For example:

myFunc :: Int -> Int
myFunc = do
  a <- (*2)
  b <- (3+)
  return (a+b)

Is a function that takes a value x, then calculates a = (x*2) and b = (3+x), and finally returns a+b, so it maps x to 2*x+x+3.