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T12785b.hs:29:63: error:
• Could not deduce: Payload ('S n) (Payload n s1) ~ s
arising from a use of ‘SBranchX’
from the context: m ~ 'S n
bound by a pattern with constructor:
Branch :: forall a (n :: Peano).
a -> HTree n (HTree ('S n) a) -> HTree ('S n) a,
in an equation for ‘nest’
at T12785b.hs:29:7-51
‘s’ is a rigid type variable bound by
a pattern with constructor:
Hide :: forall a (n :: Peano) (f :: a -> *) (s :: HTree n a).
STree n f s -> Hidden n f,
in an equation for ‘nest’
at T12785b.hs:29:7-12
• In the second argument of ‘($)’, namely ‘a `SBranchX` tr’
In the expression: Hide $ a `SBranchX` tr
In an equation for ‘nest’:
nest (Hide a `Branch` (nest . hmap nest -> Hide tr))
= Hide $ a `SBranchX` tr
• Relevant bindings include
tr :: STree n (STree ('S n) (STree ('S ('S n)) f)) s1
(bound at T12785b.hs:29:49)
a :: STree ('S m) f s (bound at T12785b.hs:29:12)
nest :: HTree m (Hidden ('S m) f) -> Hidden m (STree ('S m) f)
(bound at T12785b.hs:27:1)
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