为什么“let”语句强制“applicative do”块需要monad约束？

Question

Consider this example:

{-# language ApplicativeDo #-}

module X where

data Tuple a b = Tuple a b deriving Show

instance Functor (Tuple a) where
    fmap f (Tuple x y) = Tuple x (f y)

instance Foldable (Tuple a) where
    foldr f z (Tuple _ y) = f y z

instance Traversable (Tuple a) where
    traverse f (Tuple x y) = do
        y' <- f y
        let t' = Tuple x y'
        return $ t'

Looks nice: But no:

[1 of 1] Compiling X                ( X.hs, interpreted )

X.hs:15:9: error:
    • Could not deduce (Monad f) arising from a do statement
      from the context: Applicative f
        bound by the type signature for:
                   traverse :: forall (f :: * -> *) a1 b.
                               Applicative f =>
                               (a1 -> f b) -> Tuple a a1 -> f (Tuple a b)
        at X.hs:14:5-12
      Possible fix:
        add (Monad f) to the context of
          the type signature for:
            traverse :: forall (f :: * -> *) a1 b.
                        Applicative f =>
                        (a1 -> f b) -> Tuple a a1 -> f (Tuple a b)
    • In a stmt of a 'do' block: y' <- f y
      In the expression:
        do y' <- f y
           let t' = Tuple x y'
           return $ t'
      In an equation for ‘traverse’:
          traverse f (Tuple x y)
            = do y' <- f y
                 let t' = ...
                 return $ t'
   |
15 |         y' <- f y
   |         ^^^^^^^^^
Failed, no modules loaded.

Even this fails:

instance Traversable (Tuple a) where
    traverse f (Tuple x y) = do
        y' <- f y
        let unrelated = 1
        return $ Tuple x y'

So, introducing any let statement removes the "applicative" from the "applicative do" . Why?

Answer 1

It would translate into

let unrelated = 1 in return $ Tuple x y'

which doesn't have the form return <something> , and applicative do requires the last statement to be a return or pure :

In general, the rule for when a do statement incurs a Monad constraint is as follows. If the do-expression has the following form:

 do p1 <- E1; ...; pn <- En; return E

where none of the variables defined by p1...pn are mentioned in E1...En , and p1...pn are all variables or lazy patterns, then the expression will only require Applicative . Otherwise, the expression will require Monad . The block may return a pure expression E depending upon the results p1...pn with either return or pure .

Note: the final statement must match one of these patterns exactly:

 return E return $ E pure E pure $ E

otherwise GHC cannot recognise it as a return statement, and the transformation to use <$> that we saw above does not apply. In particular, slight variations such as return. Just $ x return. Just $ x or let x = e in return x would not be recognised.

If you look at the description of desugaring in https://gitlab.haskell.org/ghc/ghc/wikis/applicative-do , it also doesn't support let in any way.

Answer 2

What applicative expression would you like this to desugar to? A monadic expression is a series of chained scopes, so it makes sense for a let to introduce a binding that extends over all the remaining scopes, but with applicative the various expressions can't really depend on each other, so there's no scope that it makes sense to desugar the let into.