Yes, but none of that has anything to do with a walk over the data type, as 
deriving(Functor) does!

You are right that what we need is the result of simplifying the instantiated 
constraint

                (Generic [a], GC (Rep [a]))

Simplify that constraint (simplifyDeriv does that), including reducing 
type-function applications, and that’s your context.

But no need to look at the data type’s constructors, as deriving(Functor) does.

Simon

From: [email protected] [mailto:[email protected]] On 
Behalf Of José Pedro Magalhães
Sent: 18 June 2016 09:16
To: Simon Peyton Jones <[email protected]>
Cc: Ryan Scott <[email protected]>; Andres Löh <[email protected]>; 
GHC developers <[email protected]>
Subject: Re: Inferring instance constraints with DeriveAnyClass

I still don't think you can do it just from the default method's type. A 
typical case is the following:

class C a where
  op :: a -> Int
  default op :: (Generic a, GC (Rep a)) => a -> Int

When giving an instance C [a], you might well find out that you need C a =>, 
but this is not something
you can see in the type of the default method; it follows only after the 
expansion of Rep [a] and resolving
the GC constraint a number of times.


Best regards,
Pedro

On Fri, Jun 17, 2016 at 12:43 PM, Simon Peyton Jones 
<[email protected]<mailto:[email protected]>> wrote:
| My question is then: why does DeriveAnyClass take the bizarre approach
| of co-opting the DeriveFunctor algorithm? Andres, you originally
| proposed this in #7346 [2], but I don't quite understand why you
| wanted to do it this way. Couldn't we infer the context simply from
| the contexts of the default method type signatures?

That last suggestion makes perfect sense to me.  After all, we are going to 
generate an instance looking like

instance .. => C (T a) where
  op1 = <default-op1>
  op2 = <default-op2>

so all we need in ".." is enough context to satisfy the needs of <default-op1> 
etc.

Well, you need to take account of the class op type sig too:

class C a where
  op :: Eq a => a -> a
  default op :: (Eq a, Show a) => a -> a

We effectively define
  default_op :: (Eq a, Show a) => a -> a

Now with DeriveAnyClass for lists, we effectively get

instance ... => C [a] where
   op = default_op

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.

Simon
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