4 ms·

I like to think of a Monad in terms of the Monad laws, it makes it much less abstract. https://www.haskell.org/haskellwiki/Monad_laws https://www.haskell.org/ha

• AnimalMuppet 12y ago
  And anything that obeys those laws is a monad (if I understand correctly). [Edit: And if the implementing pieces are named correctly.]
  • dllthomas 12y ago
    I think "fmap f m = (m >>= f >>> return)" also needs to obey the functor laws. I can quickly show that "fmap id = id" follows from the monad laws, but I don't know whether "fmap (p . q) = (fmap p) . (fmap q)" does.
    • dllthomas 12y ago
      Ah, it does seem to follow: M1: return a >>= f = f a M2: m >>= return = m M3: (m >>= f) >>= g = m >>= (\ x -> f x >>= g) (fmap f . fmap g) m = (fmap f . fmap g) m by defn of (.) = fmap f (fmap g m) by defn of fmap = fmap f (m >>= return . g) by defn of fmap = (m >>= return . g) >>= return . f by M3 = m >>= (\ x -> (return . g) x >>= return . f) by defn of (.) = m >>= (\ x -> return (g x) >>= return . f) by M1 = m >>= (\ x -> (return . f) (g x)) by defn of (.) = m >>= (\ x -> return (f (g x))) by defn of (.) = m >>= (\ x -> (return . (f . g)) x) eta reduction = m >>= return . (f . g) by defn of fmap = fmap (f . g) m (fmap f . fmap g) m = fmap (f . g) m
• antoaravinth 12y ago
  To be honest, I have tough time reading Haskell code or even laws. I understood the mondas in terms of javascript, as it is written over here: https://blog.jcoglan.com/2011/03/05/translation-from-haskell-to-javascript-of-selected-portions-of-the-best-introduction-to-monads-ive-ever-read/ https://blog.jcoglan.com/2011/03/05/translation-from-haskell...