Say I've got a Distributive
instance written for some complex custom type, Foo
.
Is it possible to write Foo
's Representable
instance using only the properties available from its Distributive
instance? And, if not, then why is Distributive
a superclass of Representable
?
How to write a Representable instance using only Distributive properties?
323 views Asked by dbanas At
1
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The superclass relationship between
Distributive
andRepresentable
...... means that if
f
is aRepresentable
, then it also must be aDistributive
, and not the other way around. When used for subclassing,=>
should be read as "is a prerequisite of", rather than "implies". (This is, in fact, opposite to how it is when=>
is used for constraints in type signatures. Purescript uses<=
for subclassing for this very reason.)For most other pairs of superclass and subclass, the story would end here.
Distributive
andRepresentable
, however, have a special relationship, in thatDistributive
functors are actually representable, as stated by the documentation of bothDistributive
...... and
Representable
:The hierarchy is set up the way it is, with
Distributive
as the superclass, becauseDistributive
is meant to have a simpler interface which is expressible in Haskell 98, unlikeRepresentable
(which uses a type family) andAdjunction
(which is a multi-parameter type class). From a more conceptual point of view, while the distributive laws imply that everyDistributive
is representable, they are not enough for figuring out what the representation is. Getting our hands on the representation requires specifying it, either directly (as inRepresentable
) or indirectly (as inAdjunction
).