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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US">Yes, but none of that has anything to do with a walk over the data type, as deriving(Functor) does!<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US">You are right that what we need is the result of simplifying the instantiated constraint<o:p></o:p></span></p>
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</span><span style="font-family:"Courier New"">(Generic [a], GC (Rep [a]))</span><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US"><o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US">Simplify that constraint (simplifyDeriv does that), including reducing type-function applications, and that’s your context.<o:p></o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US"><o:p> </o:p></span></p>
<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US">But no need to look at the data type’s constructors, as deriving(Functor) does.<o:p></o:p></span></p>
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<p class="MsoNormal"><span style="font-size:11.0pt;font-family:"Calibri",sans-serif;color:#1F497D;mso-fareast-language:EN-US">Simon<o:p></o:p></span></p>
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<p class="MsoNormal"><b><span lang="EN-US" style="font-size:11.0pt;font-family:"Calibri",sans-serif">From:</span></b><span lang="EN-US" style="font-size:11.0pt;font-family:"Calibri",sans-serif"> josepedromagalhaes@gmail.com [mailto:josepedromagalhaes@gmail.com]
<b>On Behalf Of </b>José Pedro Magalhães<br>
<b>Sent:</b> 18 June 2016 09:16<br>
<b>To:</b> Simon Peyton Jones <simonpj@microsoft.com><br>
<b>Cc:</b> Ryan Scott <ryan.gl.scott@gmail.com>; Andres Löh <andres.loeh@gmail.com>; GHC developers <ghc-devs@haskell.org><br>
<b>Subject:</b> Re: Inferring instance constraints with DeriveAnyClass<o:p></o:p></span></p>
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I still don't think you can do it just from the default method's type. A typical case is the following:<o:p></o:p></p>
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<o:p> </o:p></p>
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<span style="font-family:"Courier New"">class C a where</span><o:p></o:p></p>
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<span style="font-family:"Courier New""> op :: a -> Int</span><o:p></o:p></p>
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<span style="font-family:"Courier New""> default op :: (Generic a, GC (Rep a)) => a -> Int</span><o:p></o:p></p>
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<o:p> </o:p></p>
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When giving an instance C [a], you might well find out that you need C a =>, but this is not something<o:p></o:p></p>
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you can see in the type of the default method; it follows only after the expansion of Rep [a] and resolving<o:p></o:p></p>
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the GC constraint a number of times.<o:p></o:p></p>
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<o:p> </o:p></p>
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<o:p> </o:p></p>
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Best regards,<o:p></o:p></p>
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Pedro <o:p></o:p></p>
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<o:p> </o:p></p>
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On Fri, Jun 17, 2016 at 12:43 PM, Simon Peyton Jones <<a href="mailto:simonpj@microsoft.com" target="_blank">simonpj@microsoft.com</a>> wrote:<o:p></o:p></p>
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| My question is then: why does DeriveAnyClass take the bizarre approach<br>
| of co-opting the DeriveFunctor algorithm? Andres, you originally<br>
| proposed this in #7346 [2], but I don't quite understand why you<br>
| wanted to do it this way. Couldn't we infer the context simply from<br>
| the contexts of the default method type signatures?<br>
<br>
That last suggestion makes perfect sense to me. After all, we are going to generate an instance looking like<br>
<br>
instance .. => C (T a) where<br>
op1 = <default-op1><br>
op2 = <default-op2><br>
<br>
so all we need in ".." is enough context to satisfy the needs of <default-op1> etc.<br>
<br>
Well, you need to take account of the class op type sig too:<br>
<br>
class C a where<br>
op :: Eq a => a -> a<br>
default op :: (Eq a, Show a) => a -> a<br>
<br>
We effectively define<br>
default_op :: (Eq a, Show a) => a -> a<br>
<br>
Now with DeriveAnyClass for lists, we effectively get<br>
<br>
instance ... => C [a] where<br>
op = default_op<br>
<br>
What is ..? Well, we need (Eq [a], Show [a]); but we are given Eq [a] (because that's op's instantiated type. So Show a is all we need in the end.<br>
<br>
Simon<br>
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