[GHC] #14333: GHC doesn't use the fact that Coercible is symmetric
GHC
ghc-devs at haskell.org
Mon Oct 9 19:06:41 UTC 2017
#14333: GHC doesn't use the fact that Coercible is symmetric
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Reporter: Iceland_jack | Owner: (none)
Type: bug | Status: new
Priority: normal | Milestone:
Component: Compiler | Version: 8.2.1
Resolution: | Keywords: TypeFamilies,
| Roles
Operating System: Unknown/Multiple | Architecture:
| Unknown/Multiple
Type of failure: None/Unknown | Test Case:
Blocked By: | Blocking:
Related Tickets: | Differential Rev(s):
Wiki Page: |
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Comment (by goldfire):
Could you include the full code of the concrete use case? I followed the
link, but that's a non-trivial and not-self-contained file.
GHC accepts
{{{#!hs
f :: Coercible b a => a x -> b x
f = coerce
}}}
So, if `SameRep a b` implies `Coercible (Rep a) (Rep b)`, then the final
implication above should hold. If, on the other hand, `SameRep a b`
implies `forall x. Coercible (Rep a x) (Rep b x)`, then you're in trouble
(at least with the current implementation).
It occurs to me that you do not necessarily need to do anything clever
with `a b ~R# c d`, but instead `F a b ~R# F c d`, where `F` has an arity
of 1. Tackling the special case where we have a type function at the head,
instead of any variable, might be easier than the general case.
--
Ticket URL: <http://ghc.haskell.org/trac/ghc/ticket/14333#comment:5>
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