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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">Interesting example. It’s a fairly-fundamental limitation of type inference. Here is a simpler version<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal" style="text-indent:36.0pt"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">foo :: forall a. (forall t. (t~a) => t) -> a<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal" style="text-indent:36.0pt"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">-- bar :: Num a => a<o:p></o:p></span></p>
<p class="MsoNormal" style="text-indent:36.0pt"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">bar = foo 1<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">Suppose we try to *<b>infer</b>* the type for bar. We instantiate<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"> foo at type alpha<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"> 1 at type t, giving rise to (Num t)<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">So from the RHS of bar we get the constraint<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"> forall t. (t~alpha) => Num t<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">This is an “implication constraint”; see our paper “Modular type inference with local constraints”. The “forall t. (t~alpha) => ...” part comes directly from
the type of foo’s first argument.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">A possible inferred type for bar would be this, where we simply abstract over the unsolved constraint:<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"> forall alpha. (forall t. (t~alpha) => Num t) => alpha<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">But GHC doesn’t allow types with forall’s in constraints, so when trying to *<b>infer</b>* the type for a let-bound identifier, GHC tries to approximate, to
find a simpler constraint that will do, one that does not involve forall’s in the constraint. This approximation is necessarily a bit ad hoc.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">In this particular case (Num alpha) would do just fine, but in general that is extremely hard to work out, because of the given equalities (here, t~alpha).
So when there are equalities in an implication constraint, GHC doesn’t attempt to extract constraints from the body of the implication, lest we accidentally infer a non-principal type.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">But if you supply a type signature, all is well, as you found.
<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">That explains, I hope, why adding (C) makes your program work.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">You also found that changing foo’s type signature from (B) to (A) made it work, but that’s not true in HEAD; both fail, as indeed they should.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D">Simon<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Verdana","sans-serif";color:#1F497D"><o:p> </o:p></span></p>
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<p class="MsoNormal"><b><span lang="EN-US" style="font-size:10.0pt;font-family:"Tahoma","sans-serif"">From:</span></b><span lang="EN-US" style="font-size:10.0pt;font-family:"Tahoma","sans-serif""> glasgow-haskell-users-bounces@haskell.org [mailto:glasgow-haskell-users-bounces@haskell.org]
<b>On Behalf Of </b>Akio Takano<br>
<b>Sent:</b> 13 May 2013 03:47<br>
<b>To:</b> glasgow-haskell-users@haskell.org<br>
<b>Subject:</b> A type not inferred with RankNTypes<o:p></o:p></span></p>
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<p class="MsoNormal"><o:p> </o:p></p>
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<p class="MsoNormal">Hi,<br>
<br>
The attached program does not typecheck if I don't include a type signature for 'bar' (the line C). I can't figure out if this is a limitation in the type system or a bug in GHC. One thing that confuses me is that replacing the line (B) with (A) makes the program
typecheck.<br>
<br>
Could anyone help me figuring out what is going on?<br>
<br>
I'm using GHC 7.6.2. The error was:<br>
<br>
% ghc forall.hs<br>
[1 of 1] Compiling Foo ( forall.hs, forall.o )<br>
<br>
forall.hs:18:11:<br>
Could not deduce (Fractional a) arising from the literal `0.1'<br>
from the context (Num (Scalar t), Scalar t ~ a)<br>
bound by a type expected by the context:<br>
(Num (Scalar t), Scalar t ~ a) => AD t<br>
at forall.hs:18:7-13<br>
Possible fix:<br>
add (Fractional a) to the context of<br>
a type expected by the context:<br>
(Num (Scalar t), Scalar t ~ a) => AD t<br>
or the inferred type of bar :: a<br>
In the first argument of `foo', namely `0.1'<br>
In the expression: foo 0.1<br>
In an equation for `bar': bar = foo 0.1<br>
<br>
Regards,<br>
Takano Akio<o:p></o:p></p>
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