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: josepedromagalhaes@gmail.com [mailto:josepedromagalhaes@gmail.com] On Behalf Of José Pedro Magalhães
Sent: 18 June 2016 09:16
To: Simon Peyton Jones <simonpj@microsoft.com>
Cc: Ryan Scott <ryan.gl.scott@gmail.com>; Andres Löh <andres.loeh@gmail.com>; GHC developers <ghc-devs@haskell.org>
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 <simonpj@microsoft.com> 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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