Re: [Haskell-cafe] flexible contexts problem

[Brandon S. Allbery KF8NH]

On Sep 12, 2009, at 20:17 , Sean McLaughlin wrote:

However, if you replace the function f with

f :: (MonadState s m, Has Int s) => m Int
f = do x <- project
          y <- project
          return x

then it fails with the error

    Could not deduce (Has α s)
      from the context (MonadState s m, Has Int s)
      arising from a use of `project'
                   at /Users/sean/uploads/Weird.hs:16:12-18
    Possible fix:
      add (Has α s) to the context of the type signature for `f'
    In a stmt of a 'do' expression: y <- project
    In the expression:
        do x <- project
           y <- project
           return x
    In the definition of `f':
        f = do x <- project
               y <- project
               return x

--
brandon s. allbery [solaris,freebsd,perl,pugs,haskell] allb...@kf8nh.com
system administrator [openafs,heimdal,too many hats] allb...@ece.cmu.edu
electrical and computer engineering, carnegie mellon university    KF8NH

I think what this is really telling you is that there's no way for ghc to tell what type y has. x is fine because it's in the result (hence fixed by the type (m Int)), but y is just "there" and all ghc knows is that it has (Has α s) and (MonadState s m, Has Int s) but no way to get from one to the other (more specifically: it has no clue what α might be). Note that project is explicitly polymorphic, so the fact that you used it to get an (Int) does not in any sense guarantee that the subsequent use *also* produces an Int. If you want this, you'll have to constrain the call somehow.

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