The Universe of Disco

Getting Applicatives from Monads and “>>=” from “join”

I complained recently about GHC not being able to infer an Applicative instance from a type that already has a Monad instance, and there is a related complaint that the Monad instance must define >>=. In some type classes, you get a choice about what to define, and then the rest of the functions are built from the ones you provided. To take a particular simple example, with Eq you have the choice of defining == or /=, and if you omit one Haskell will construct the other for you. It could do this with >>= and join, but it doesn't, for technical reasons I don't understand [1] [2] [3].

But both of these problems can be worked around. If I have a Monad instance, it seems to work just fine if I say:

    instance Applicative Tree where
      pure = return
      fs <*> xs = do
          f <- fs
          x <- xs
          return (f x)

Where this code is completely canned, the same for every Monad.

And if I know join but not >>=, it seems to work just fine if I say:

    instance Monad Tree where
      return = ...
      x >>= f  = join (fmap f x) where
        join tt = ...

I suppose these might faul foul of whatever problem is being described in the documents I linked above. But I'll either find out, or I won't, and either way is a good outcome.

[ Addendum: Vaibhav Sagar points out that my definition of <*> above is identical to that of Control.Monad.ap, so that instead of defining <*> from scratch, I could have imported ap and then written <*> = ap. ]

[ Addendum 20221021: There are actually two definitions of <*> that will work. [1] [2] ]

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