[GHC] #15683: coerce fails for Coercible type families
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Fri Sep 28 12:29:46 UTC 2018
#15683: coerce fails for Coercible type families
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Reporter: Iceland_jack | Owner: (none)
Type: bug | Status: new
Priority: normal | Milestone: 8.6.1
Component: Compiler | Version: 8.6.1
Resolution: | Keywords: TypeFamilies
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 simonpj):
This is by design. You are expecting `T Int` and `T Bool` to be
representationally equal because
if you look at every field of every constructor of `T`, and reduce every
type-family application
therein, on the particular argument types `Int` and `Bool`, then the
answers are reprsentationally
equal. Quite true! But this could be laborious if `T` had many
constructors
{{{
data T a = T1 (X a )
| T2 [Y a] (Z [a])
..etc...
}}}
If you add `type instance X Char = Char`, then certainly `T Int` is not
coercible to `T Char`,
so that check would have to be done for every individual instantation --
and recursively so.
Instead, the type system analyses the data type declaration for `T`
computes a single summary -- the "role signature" of `T` --that decides,
one and for all,
when `(T t1)` is representationally equal to
`T t2`.
In this case, because of the possibility of `type instance T Char = Char`,
the decision is
that `T t1` is coercible to `T t2` only if `t1` is (nominally) equal to
`t2`.
[https://www.microsoft.com/en-us/research/publication/safe-coercions/ The
paper] explains all this in some details. I don't see a decently feasibly
way to fix this, I'm afraid.
--
Ticket URL: <http://ghc.haskell.org/trac/ghc/ticket/15683#comment:3>
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