[replying to haskell-cafe]
On 9/15/07, Norman Ramsey <[EMAIL PROTECTED]> wrote:
> Dear Haskellers,
>
> I've had a Haskell program rejected for reasons I don't understand.
> Here's the relevant bit of code; the problem is that I expected the
> type-inference engine to generalize the abbreviation 'fold' to an
> overloaded function, but it doesn't---to make the code work, I had to
> expand 'fold' into 'foldRegsUsed' everywhere it appears. I'm baffled
> because 'fold' isn't mutually recursive with anything, so I would have
> expected ordinary Hindley-Milner style inference to generalize it to
> something of the type
>
> UserOfLocalRegs a => (b -> LocalReg -> b) -> b -> a -> b
>
> But that's not what happens. This code fails to compile because the
> compiler is willing to use 'fold' at only one type (CmmExpr as it happens):
>
> class UserOfLocalRegs a where
> foldRegsUsed :: (b -> LocalReg -> b) -> b -> a -> b
>
> instance UserOfLocalRegs Middle where
> foldRegsUsed f z m = middle m
> where middle (MidComment {})= z
> middle (MidAssign _lhs expr) = fold f z expr
> middle (MidStore addr rval) = fold f (fold f z addr)
> rval
> middle (MidUnsafeCall tgt _ress args) = fold f (fold f z tgt)
> args
> middle (CopyIn _ _formals _) = z
> middle (CopyOut _ actuals)= fold f z actuals
> fold = foldRegsUsed
>
>
> What rule of the language have I overlooked?
>
Monomorphism restriction? Replacing fold with foldRegsUsed would work
because there's a type signature for foldRegsUsed.
Cheers,
Tim
--
Tim Chevalier * catamorphism.org * Often in error, never in doubt
"...There is no mystery; there is only paradox, the incontrovertible
union of contradictory truths." -- Edward Abbey
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