Re: [GHC] #12490: With RebindableSyntax, ApplicativeDo should eliminate return/pure (was: ApplicativeDo and RebindableSyntax do not desugar as expected)

14 Aug 2016


      #12490: With RebindableSyntax, ApplicativeDo should eliminate return/pure
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        Reporter:  AaronFriel        |                Owner:
            Type:  bug               |               Status:  new
        Priority:  normal            |            Milestone:
       Component:  Compiler          |              Version:  8.0.1
      Resolution:                    |             Keywords:
Operating System:  Unknown/Multiple  |         Architecture:
                                     |  Unknown/Multiple
 Type of failure:  None/Unknown      |            Test Case:
      Blocked By:                    |             Blocking:
 Related Tickets:                    |  Differential Rev(s):
       Wiki Page:                    |
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Description changed by AaronFriel:

@@ -45,1 +45,2 @@
- partially desugars to the specification.
+ partially desugars to the specification, and the final "return" or "pure"
+ remains in the output.

New description:

 In a module with -XApplicativeDo, -XRebindableSyntax, and local
 definitions for everything in the Functor-Applicative-Monad hierarchy, do-
 notation always desugars to "join (... (return ...))" (or
 /s/return/pure/). This forces the result to have at least the constraints
 of join, which in my case is "IxMonad m".

 {{{#!hs
 {-# LANGUAGE RebindableSyntax, ApplicativeDo #-}

 module My where

 -- straightforward definitions of fmap, pure, (<*>), join, return, (>>=),
 -- (>>) and fail in terms of IxFunctor, IxPointed, IxApplicative, IxMonad

 fPure m = do
   a <- m
   b <- m
   pure (a, b)

 fReturn m = do
   a <- m
   b <- m
   return (a, b)
 }}}

 According to -ddump-ds, these desugar to:

 {{{#!hs
 fPure :: IxMonad m => m k1 k1 a -> m k1 k1 (a, a)
 fPure m = My.join ( My.(<*>) (My.fmap (\a b -> My.pure (a, b)) m) m )

 fReturn :: IxMonad m => m k1 k1 a -> m k1 k1 (a, a)
 fReturn m = My.join ( My.(<*>) (My.fmap (\a b -> My.return (a, b)) m) m )
 }}}

 But I would expect:

 {{{#!hs
 fPure m = My.(<*>) (My.fmap (\a b -> (a, b)) m) m

 fReturn m = -- same
 }}}

 It appears that when "return" is not from base, ApplicativeDo only
 partially desugars to the specification, and the final "return" or "pure"
 remains in the output.

--

--
Ticket URL: http://ghc.haskell.org/trac/ghc/ticket/12490#comment:2
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