#1537: do notation translation
---------------------------+------------------------------------------------
Reporter: Isaac Dupree | Owner:
Type: bug | Status: new
Priority: normal | Milestone:
Component: Compiler | Version: 6.6.1
Severity: normal | Keywords:
Difficulty: Unknown | Os: Unknown
Testcase: | Architecture: Unknown
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Really there are two things:
Normally, (do True) fails to compile in GHC because there is no Monad
Bool, whereas Haskell-98 translates it to (True).
With `-fno-implicit-prelude`, ( according to
http://comonad.com/reader/2007/parameterized-monads-in-haskell/ ):
"""
Caveat: It appears that GHC enforces the fact that the arguments and
results of (>>=) must have a signature like
(>>=) :: forall m a. (...) => m a -> (a -> m b) -> m b
insofar as when you use the do-sugar, your types will not be able to vary.
Ideally it should be able to get by with a more liberal signature, but it
seems like no one has needed it before now.
"""
I think do-notation will be the simplest sugar (from one point of view at
least) when it just translates to (>>=), (>>), fail, and let..in.. as
specified in Haskell98 (or non-Prelude-qualified when `-fno-implicit-
prelude`, of course).
It appears #303 was an older similar problem. Also, maybe the behavior
(in the new version) should be explicitly documented somewhere in the
User's Guide.
--
Ticket URL: <http://hackage.haskell.org/trac/ghc/ticket/1537>
GHC <http://www.haskell.org/ghc/>
The Glasgow Haskell Compiler
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