Commit 3214ec5a authored by Simon Peyton Jones's avatar Simon Peyton Jones

Comments only

parent 350ed083
......@@ -1794,6 +1794,19 @@ seqCos (co:cos) = seqCo co `seq` seqCos cos
%* *
Note [Computing a coercion kind and role]
To compute a coercion's kind is straightforward: see coercionKind.
But to compute a coercion's role, in the case for NthCo we need
its kind as well. So if we have two separate functions (one for kinds
and one for roles) we can get exponentially bad behaviour, sinc each
NthCo node makes a seaprate call to coercionKind, which traverses the
sub-tree again. This was part of the problem in Trac #9233.
Solution: compute both together; hence coercionKindRole. We keep a
separate coercionKind function because it's a bit more efficient if
the kind is all you wan.
coercionType :: Coercion -> Type
coercionType co = case coercionKindRole co of
......@@ -1843,9 +1856,8 @@ coercionKind co = go co
coercionKinds :: [Coercion] -> Pair [Type]
coercionKinds tys = sequenceA $ map coercionKind tys
-- | Get a coercion's kind and role. More efficient than getting
-- each individually, but less efficient than calling just
-- 'coercionKind' if that's all you need.
-- | Get a coercion's kind and role.
-- Why both at once? See Note [Computing a coercion kind and role]
coercionKindRole :: Coercion -> (Pair Type, Role)
coercionKindRole = go
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment