Keyed container types with type-checked proofs of key presence.
Table of contents
Have you ever known that a key could be found in a certain map? Were you tempted to
error to handle the "impossible" case, when you knew that
lookup should give
Just v? (and did shifting requirements ever make the impossible
become possible after all?)
Data.Map.Justified provides a zero-cost
newtype wrapper around
that enables you to separate the proof that a key is present from the operations using the key.
Once you prove that a key is present, you can use it
Maybe-free in any number of other
operations -- sometimes even operations on other maps!
None of the functions in this module can cause a run-time error, and very few
of the operations return a
Data.Map.Justified.Tutorial module for usage examples, or browse the API of the
most recent release on Hackage.
withMap test_table $ \table -> do case member 1 table of Nothing -> putStrLn "Sorry, I couldn't prove that the key is present." Just key -> do -- We have proven that the key is present, and can now use it Maybe-free... putStrLn ("Found key: " ++ show key) putStrLn ("Value for key: " ++ lookup key table) -- ...even in certain other maps! let table' = reinsert key "howdy" table putStrLn ("Value for key in updated map: " ++ lookup key table')
Found key: Key 1 Value for key: hello Value for key in updated map: howdy
Suppose you have a key-value mapping using
Map k v. Anybody making
Map k v to look up or modify a value must take into account the possibility
that the given key is not present.
Data.Map, there are two strategies for dealing with absent keys:
Cause a runtime error (e.g.
(!)when the key is absent)
The first option introduces partial functions, so is not very palatable. But what is wrong with the second option?
To understand the problem with returning a
Maybe value, let's ask what the
Maybe v in
lookup :: k -> Map k v -> Maybe v
really does for us. By returning
Maybe v value,
lookup key table is saying "Your program must account
for the possibility that
key cannot be found in
table. I will ensure that you
account for this possibility by forcing you to handle the
Data.Map is requiring the user to prove they have handled the
possibility that a key is absent whenever they use the
Laziness (the bad kind)
Every programmer has probably had the experience of knowing, somehow, that a certain
key is going to be present in a map. In this case, the
Maybe v feels like a burden:
I already know that this key is in the map, why should I have to handle the
In this situation, it is tempting to reach for the partial function
or a pattern match like
Nothing -> error "The impossible happened!". But as parts of
the program are changed over time, you may find the impossible has become possible after
all (or perhaps youll see the dreaded and unhelpful
*** Exception: Maybe.fromJust: Nothing)
It is tempting to reach for partial functions or "impossible" runtime errors here, because
the programmer has proven that the key is a member of the map in some other way. They
lookup should return a
Just v --- but the compiler doesnt know this!
The idea behind
Data.Map.Justified is to encode the programmers knowledge that a key
is present within the type system, where it can be checked at compile-time. Once a key
is known to be present,
lookup will never fail. Your justification
How it works
Evidence that a key can indeed be found in a map is carried by a phantom type parameter
shared by both the
Data.Map.Justified.Key types. If you are
able to get your hands on a value of type
Key ph k, then you must have already proven that
the key is present in any value of type
Map ph k v.
Key ph k type is simply a
newtype wrapper around
k, but the phantom type
Key ph k to represent both a key of type
k and a proof that the key is present in
all maps of type
Map ph k v.
There are several ways to prove that a key belongs to a map, but the simplest is to just use
member function. In
has the type
member :: Ord k => k -> Map k v -> Bool
and reports whether or not the key can be found in the map. In
member has the type
member :: Ord k => k -> Map ph k v -> Maybe (Key ph k)
Instead of a boolean,
member either says
the key is not present
Nothing), or gives back the same key, augmented with evidence that they key
is present. This key-plus-evidence can then be used to do any number of
operations on the map.
Data.Map.Justified uses the same rank-2 polymorphism trick used in the
Control.Monad.ST monad to
ensure that the
ph phantom type can not be extracted; in effect, the proof that a key is
present can't leak to contexts where the proof would no longer be valid.
You can interpret the
ph phantom type as a concrete set of keys; under this interpretation,
a value of type
Key ph k is a key of type
k, belonging to the subset described by
Map ph k v is a map whose keys are exactly the subset of
k described by
From this perspective, the maps behave as if they were total, leading to their
Why all the continuations?
Many of the functions in
justified-containers make use of continuations, but why? As a case-study,
consider the basic function
withMap that promotes a standard
Data.Map.Map to a `Data.Map.Justified.Map':
import Data.Map.Justified import qualified Data.Map as M withMap :: M.Map k v -> (forall ph. Map ph k v -> t) -> t
(forall ph. Map ph k v -> t) -> t part is the continuation.
The idea is that we know there is some set of keys
ph belonging to this particular map, but
at compile-time we may not know exactly what it is. But it does exist, after all, so we should be
able to write
withMap :: M.Map k v -> exists ph. Map ph k v
inserting function could look like
inserting :: k -> v -> Map ph k v -> exists ph'. Map ph' k v
which can be read as "after inserting a key/value pair, we get a (possibly) different set of keys
But in this case, we actually know a bit more: first, the inserted key will be found in the new map. And
second, every key in
ph can also be found in
can encode that knowledge by giving an explicit inclusion of
ph', encoded as a function of
Key ph k -> Key ph' k. So we could re-write
inserting with the type
inserting :: k -> v -> Map ph k v -> exists ph'. (Key ph' k, Key ph k -> Key ph' k, Map ph' k v) -- \_______/ \___________________/ \_________/ -- the new key______| | | -- the inclusion__| | -- the new map_____|
Likewise, when deleting a key from a map with keys
ph, we get a new map with keys
with a guarantee that
ph' is a subset of
ph. Compared to
inserting, the inclusion goes the
other way: there is an inclusion of
ph, encoded as a function of type
Key ph' k -> Key ph l.
Altogether, we could give
deleting the type
deleting :: k -> Map ph k v -> exists ph'. (Key ph' k -> Key ph k, Map ph' k v) -- \___________________/ \_________/ -- | | -- the reversed inclusion__| | -- the new map_____|
A similar pattern works for other map operations like
But what about the continuations?
In the last section, we argued that
deleting should have a type like
deleting :: k -> Map ph k v -> exists ph'. (Key ph' k -> Key ph k, Map ph' k v) -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But if you check the documentation, you'll see the type
deleting :: k -> Map ph k v -> (forall ph'. (Key ph' k -> Key ph k, Map ph' k v) -> t) -> t -- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
What happened to the underlined part of the type?
The problem is that Haskell doesn't support existential types directly: the
exists part of
the type we wrote out is just wishful thinking. Instead, we have to go about things a little
indirectly: we'll encode existentially-quantified types, via rank-2 universally-quantified types.
The idea can be understood via the Curry-Howard correspondence:
In classical logic, we have an equivalence between the propositions
∀T. ((∀X.P(X) => T) => T). It turns out that this equivalence remains valid
in constructive logic, so we can transport it via the Curry-Howard correspondence to get
an isomorphism between types:
exists ph. Map ph k v ~ (forall ph. Map ph k v -> t) -> t
In other words, instead of returning the existentially-quantified type directly we say "tell me what you wanted to do with that existentially-quantified type, and I'll do it for you".