What does '#' mean in Haskell? - haskell

I've tried googling but come up short. I am furthering my Haskell knowledge by reading some articles and I came across one that uses a syntax I've never seen before.
An example would be:
reconstruct node#(Node a b c l r) parent#(Node b d le ri)
I've never seen these #'s before. I tried searching online for an answer but came up short. Is this simply a way to embed tags to help make things clearer, or do they have an actual impact on the code?

It is used in pattern matching. Now node variable will refer to the entire Node data type for the argument Node a b c l r. So instead of passing to the function as Node a b c l r, you can use node instead to pass it up.
A much simpler example to demonstrate it:
data SomeType = Leaf Int Int Int | Nil deriving Show
someFunction :: SomeType -> SomeType
someFunction leaf#(Leaf _ _ _) = leaf
someFunction Nil = Leaf 0 0 0
The someFunction can also be written as:
someFunction :: SomeType -> SomeType
someFunction (Leaf x y z) = Leaf x y z
someFunction Nil = Leaf 0 0 0
See how simpler was the first version ?

Using #t as a type indicator
Besides the argument pattern matching usage described in the answer of #Sibi, in Haskell the "at" character ('#', also known as an arobase character) can be used in some contexts to force a typing decision. This is mentioned in the comments by #Josh.F.
This is not part of the default language features, and is known as the Type Application Haskell language extension. In summary, the extension allows you to give explicit type arguments to a polymorphic function such as read. In a classic .hs source file, the relevant pragma must be included:
{-# LANGUAGE TypeApplications #-}
Example:
$ ghci
GHCi, version 8.2.2: http://www.haskell.org/ghc/ :? for help
λ>
λ> let x = (read #Integer "33")
<interactive>:4:10: error:
Pattern syntax in expression context: read#Integer
Did you mean to enable TypeApplications?
λ>
λ> :set -XTypeApplications
λ>
λ> let x = (read #Integer "33")
λ>
λ> :type x
x :: Integer
λ>
λ> x
33
λ>
Further details
For the read polymorphic function, the type indicator introduced by # relates to the type of the result returned by read. But this is not generally true.
Generally speaking, you have to consider the type variables that appear in the type signature of the function at hand. For example, let's have a look at the fmap library function.
fmap :: Functor ft => (a -> b) -> ft a -> ft b
So here, we have 3 type variables, in order of appearance: ft, a, b. If we specialize fmap like this:
myFmap = fmap #type1 #type2 #type3
then type1 will relate to ft, type2 will relate to a and type3 will relate to b. Also, there is a special dummy type indicator #_ which means: “here, any type goes”.
For example, we can force the output type of fmap to be Integer and the functor to be the plain list [], leaving the input type a unspecified:
λ>
λ> myFmap = fmap #[] #_ #Integer
λ>
λ> :type myFmap
myFmap :: (_ -> Integer) -> [_] -> [Integer]
λ>
As for the read function, its type is:
read :: Read a => String -> a
So there is only room for one type indicator, and it relates to the type of the result returned by read, as displayed above.

Related

Besides as-pattern, what else can # mean in Haskell?

I am studying Haskell currently and try to understand a project that uses Haskell to implement cryptographic algorithms. After reading Learn You a Haskell for Great Good online, I begin to understand the code in that project. Then I found I am stuck at the following code with the "#" symbol:
-- | Generate an #n#-dimensional secret key over #rq#.
genKey :: forall rq rnd n . (MonadRandom rnd, Random rq, Reflects n Int)
=> rnd (PRFKey n rq)
genKey = fmap Key $ randomMtx 1 $ value #n
Here the randomMtx is defined as follows:
-- | A random matrix having a given number of rows and columns.
randomMtx :: (MonadRandom rnd, Random a) => Int -> Int -> rnd (Matrix a)
randomMtx r c = M.fromList r c <$> replicateM (r*c) getRandom
And PRFKey is defined below:
-- | A PRF secret key of dimension #n# over ring #a#.
newtype PRFKey n a = Key { key :: Matrix a }
All information sources I can find say that # is the as-pattern, but this piece of code is apparently not that case. I have checked the online tutorial, blogs and even the Haskell 2010 language report at https://www.haskell.org/definition/haskell2010.pdf. There is simply no answer to this question.
More code snippets can be found in this project using # in this way too:
-- | Generate public parameters (\( \mathbf{A}_0 \) and \(
-- \mathbf{A}_1 \)) for #n#-dimensional secret keys over a ring #rq#
-- for gadget indicated by #gad#.
genParams :: forall gad rq rnd n .
(MonadRandom rnd, Random rq, Reflects n Int, Gadget gad rq)
=> rnd (PRFParams n gad rq)
genParams = let len = length $ gadget #gad #rq
n = value #n
in Params <$> (randomMtx n (n*len)) <*> (randomMtx n (n*len))
I deeply appreciate any help on this.
That #n is an advanced feature of modern Haskell, which is usually not covered by tutorials like LYAH, nor can be found the the Report.
It's called a type application and is a GHC language extension. To understand it, consider this simple polymorphic function
dup :: forall a . a -> (a, a)
dup x = (x, x)
Intuitively calling dup works as follows:
the caller chooses a type a
the caller chooses a value x of the previously chosen type a
dup then answers with a value of type (a,a)
In a sense, dup takes two arguments: the type a and the value x :: a. However, GHC is usually able to infer the type a (e.g. from x, or from the context where we are using dup), so we usually pass only one argument to dup, namely x. For instance, we have
dup True :: (Bool, Bool)
dup "hello" :: (String, String)
...
Now, what if we want to pass a explicitly? Well, in that case we can turn on the TypeApplications extension, and write
dup #Bool True :: (Bool, Bool)
dup #String "hello" :: (String, String)
...
Note the #... arguments carrying types (not values). Those are something that exists at compile time, only -- at runtime the argument does not exist.
Why do we want that? Well, sometimes there is no x around, and we want to prod the compiler to choose the right a. E.g.
dup #Bool :: Bool -> (Bool, Bool)
dup #String :: String -> (String, String)
...
Type applications are often useful in combination with some other extensions which make type inference unfeasible for GHC, like ambiguous types or type families. I won't discuss those, but you can simply understand that sometimes you really need to help the compiler, especially when using powerful type-level features.
Now, about your specific case. I don't have all the details, I don't know the library, but it's very likely that your n represents a kind of natural-number value at the type level. Here we are diving in rather advanced extensions, like the above-mentioned ones plus DataKinds, maybe GADTs, and some typeclass machinery. While I can't explain everything, hopefully I can provide some basic insight. Intuitively,
foo :: forall n . some type using n
takes as argument #n, a kind-of compile-time natural, which is not passed at runtime. Instead,
foo :: forall n . C n => some type using n
takes #n (compile-time), together with a proof that n satisfies constraint C n. The latter is a run-time argument, which might expose the actual value of n. Indeed, in your case, I guess you have something vaguely resembling
value :: forall n . Reflects n Int => Int
which essentially allows the code to bring the type-level natural to the term-level, essentially accessing the "type" as a "value". (The above type is considered an "ambiguous" one, by the way -- you really need #n to disambiguate.)
Finally: why should one want to pass n at the type level if we then later on convert that to the term level? Wouldn't be easier to simply write out functions like
foo :: Int -> ...
foo n ... = ... use n
instead of the more cumbersome
foo :: forall n . Reflects n Int => ...
foo ... = ... use (value #n)
The honest answer is: yes, it would be easier. However, having n at the type level allows the compiler to perform more static checks. For instance, you might want a type to represent "integers modulo n", and allow adding those. Having
data Mod = Mod Int -- Int modulo some n
foo :: Int -> Mod -> Mod -> Mod
foo n (Mod x) (Mod y) = Mod ((x+y) `mod` n)
works, but there is no check that x and y are of the same modulus. We might add apples and oranges, if we are not careful. We could instead write
data Mod n = Mod Int -- Int modulo n
foo :: Int -> Mod n -> Mod n -> Mod n
foo n (Mod x) (Mod y) = Mod ((x+y) `mod` n)
which is better, but still allows to call foo 5 x y even when n is not 5. Not good. Instead,
data Mod n = Mod Int -- Int modulo n
-- a lot of type machinery omitted here
foo :: forall n . SomeConstraint n => Mod n -> Mod n -> Mod n
foo (Mod x) (Mod y) = Mod ((x+y) `mod` (value #n))
prevents things to go wrong. The compiler statically checks everything. The code is harder to use, yes, but in a sense making it harder to use is the whole point: we want to make it impossible for the user to try adding something of the wrong modulus.
Concluding: these are very advanced extensions. If you're a beginner, you will need to slowly progress towards these techniques. Don't be discouraged if you can't grasp them after only a short study, it does take some time. Make a small step at a time, solve some exercises for each feature to understand the point of it. And you'll always have StackOverflow when you are stuck :-)

Getting all function arguments in haskel as list

Is there a way in haskell to get all function arguments as a list.
Let's supose we have the following program, where we want to add the two smaller numbers and then subtract the largest. Suppose, we can't change the function definition of foo :: Int -> Int -> Int -> Int. Is there a way to get all function arguments as a list, other than constructing a new list and add all arguments as an element of said list? More importantly, is there a general way of doing this independent of the number of arguments?
Example:
module Foo where
import Data.List
foo :: Int -> Int -> Int -> Int
foo a b c = result!!0 + result!!1 - result!!2 where result = sort ([a, b, c])
is there a general way of doing this independent of the number of arguments?
Not really; at least it's not worth it. First off, this entire idea isn't very useful because lists are homogeneous: all elements must have the same type, so it only works for the rather unusual special case of functions which only take arguments of a single type.
Even then, the problem is that “number of arguments” isn't really a sensible concept in Haskell, because as Willem Van Onsem commented, all functions really only have one argument (further arguments are actually only given to the result of the first application, which has again function type).
That said, at least for a single argument- and final-result type, it is quite easy to pack any number of arguments into a list:
{-# LANGUAGE FlexibleInstances #-}
class UsingList f where
usingList :: ([Int] -> Int) -> f
instance UsingList Int where
usingList f = f []
instance UsingList r => UsingList (Int -> r) where
usingList f a = usingList (f . (a:))
foo :: Int -> Int -> Int -> Int
foo = usingList $ (\[α,β,γ] -> α + β - γ) . sort
It's also possible to make this work for any type of the arguments, using type families or a multi-param type class. What's not so simple though is to write it once and for all with variable type of the final result. The reason being, that would also have to handle a function as the type of final result. But then, that could also be intepreted as “we still need to add one more argument to the list”!
With all respect, I would disagree with #leftaroundabout's answer above. Something being
unusual is not a reason to shun it as unworthy.
It is correct that you would not be able to define a polymorphic variadic list constructor
without type annotations. However, we're not usually dealing with Haskell 98, where type
annotations were never required. With Dependent Haskell just around the corner, some
familiarity with non-trivial type annotations is becoming vital.
So, let's take a shot at this, disregarding worthiness considerations.
One way to define a function that does not seem to admit a single type is to make it a method of a
suitably constructed class. Many a trick involving type classes were devised by cunning
Haskellers, starting at least as early as 15 years ago. Even if we don't understand their
type wizardry in all its depth, we may still try our hand with a similar approach.
Let us first try to obtain a method for summing any number of Integers. That means repeatedly
applying a function like (+), with a uniform type such as a -> a -> a. Here's one way to do
it:
class Eval a where
eval :: Integer -> a
instance (Eval a) => Eval (Integer -> a) where
eval i = \y -> eval (i + y)
instance Eval Integer where
eval i = i
And this is the extract from repl:
λ eval 1 2 3 :: Integer
6
Notice that we can't do without explicit type annotation, because the very idea of our approach is
that an expression eval x1 ... xn may either be a function that waits for yet another argument,
or a final value.
One generalization now is to actually make a list of values. The science tells us that
we may derive any monoid from a list. Indeed, insofar as sum is a monoid, we may turn arguments to
a list, then sum it and obtain the same result as above.
Here's how we can go about turning arguments of our method to a list:
class Eval a where
eval2 :: [Integer] -> Integer -> a
instance (Eval a) => Eval (Integer -> a) where
eval2 is i = \j -> eval2 (i:is) j
instance Eval [Integer] where
eval2 is i = i:is
This is how it would work:
λ eval2 [] 1 2 3 4 5 :: [Integer]
[5,4,3,2,1]
Unfortunately, we have to make eval binary, rather than unary, because it now has to compose two
different things: a (possibly empty) list of values and the next value to put in. Notice how it's
similar to the usual foldr:
λ foldr (:) [] [1,2,3,4,5]
[1,2,3,4,5]
The next generalization we'd like to have is allowing arbitrary types inside the list. It's a bit
tricky, as we have to make Eval a 2-parameter type class:
class Eval a i where
eval2 :: [i] -> i -> a
instance (Eval a i) => Eval (i -> a) i where
eval2 is i = \j -> eval2 (i:is) j
instance Eval [i] i where
eval2 is i = i:is
It works as the previous with Integers, but it can also carry any other type, even a function:
(I'm sorry for the messy example. I had to show a function somehow.)
λ ($ 10) <$> (eval2 [] (+1) (subtract 2) (*3) (^4) :: [Integer -> Integer])
[10000,30,8,11]
So far so good: we can convert any number of arguments into a list. However, it will be hard to
compose this function with the one that would do useful work with the resulting list, because
composition only admits unary functions − with some trickery, binary ones, but in no way the
variadic. Seems like we'll have to define our own way to compose functions. That's how I see it:
class Ap a i r where
apply :: ([i] -> r) -> [i] -> i -> a
apply', ($...) :: ([i] -> r) -> i -> a
($...) = apply'
instance Ap a i r => Ap (i -> a) i r where
apply f xs x = \y -> apply f (x:xs) y
apply' f x = \y -> apply f [x] y
instance Ap r i r where
apply f xs x = f $ x:xs
apply' f x = f [x]
Now we can write our desired function as an application of a list-admitting function to any number
of arguments:
foo' :: (Num r, Ord r, Ap a r r) => r -> a
foo' = (g $...)
where f = (\result -> (result !! 0) + (result !! 1) - (result !! 2))
g = f . sort
You'll still have to type annotate it at every call site, like this:
λ foo' 4 5 10 :: Integer
-1
− But so far, that's the best I can do.
The more I study Haskell, the more I am certain that nothing is impossible.

How to "iterate" over a function whose type changes among iteration but the formal definition is the same

I have just started learning Haskell and I come across the following problem. I try to "iterate" the function \x->[x]. I expect to get the result [[8]] by
foldr1 (.) (replicate 2 (\x->[x])) $ (8 :: Int)
This does not work, and gives the following error message:
Occurs check: cannot construct the infinite type: a ~ [a]
Expected type: [a -> a]
Actual type: [a -> [a]]
I can understand why it doesn't work. It is because that foldr1 has type signature foldr1 :: Foldable t => (a -> a -> a) -> a -> t a -> a, and takes a -> a -> a as the type signature of its first parameter, not a -> a -> b
Neither does this, for the same reason:
((!! 2) $ iterate (\x->[x]) .) id) (8 :: Int)
However, this works:
(\x->[x]) $ (\x->[x]) $ (8 :: Int)
and I understand that the first (\x->[x]) and the second one are of different type (namely [Int]->[[Int]] and Int->[Int]), although formally they look the same.
Now say that I need to change the 2 to a large number, say 100.
My question is, is there a way to construct such a list? Do I have to resort to meta-programming techniques such as Template Haskell? If I have to resort to meta-programming, how can I do it?
As a side node, I have also tried to construct the string representation of such a list and read it. Although the string is much easier to construct, I don't know how to read such a string. For example,
read "[[[[[8]]]]]" :: ??
I don't know how to construct the ?? part when the number of nested layers is not known a priori. The only way I can think of is resorting to meta-programming.
The question above may not seem interesting enough, and I have a "real-life" case. Consider the following function:
natSucc x = [Left x,Right [x]]
This is the succ function used in the formal definition of natural numbers. Again, I cannot simply foldr1-replicate or !!-iterate it.
Any help will be appreciated. Suggestions on code styles are also welcome.
Edit:
After viewing the 3 answers given so far (again, thank you all very much for your time and efforts) I realized this is a more general problem that is not limited to lists. A similar type of problem can be composed for each valid type of functor (what if I want to get Just Just Just 8, although that may not make much sense on its own?).
You'll certainly agree that 2 :: Int and 4 :: Int have the same type. Because Haskell is not dependently typed†, that means foldr1 (.) (replicate 2 (\x->[x])) (8 :: Int) and foldr1 (.) (replicate 4 (\x->[x])) (8 :: Int) must have the same type, in contradiction with your idea that the former should give [[8]] :: [[Int]] and the latter [[[[8]]]] :: [[[[Int]]]]. In particular, it should be possible to put both of these expressions in a single list (Haskell lists need to have the same type for all their elements). But this just doesn't work.
The point is that you don't really want a Haskell list type: you want to be able to have different-depth branches in a single structure. Well, you can have that, and it doesn't require any clever type system hacks – we just need to be clear that this is not a list, but a tree. Something like this:
data Tree a = Leaf a | Rose [Tree a]
Then you can do
Prelude> foldr1 (.) (replicate 2 (\x->Rose [x])) $ Leaf (8 :: Int)
Rose [Rose [Leaf 8]]
Prelude> foldr1 (.) (replicate 4 (\x->Rose [x])) $ Leaf (8 :: Int)
Rose [Rose [Rose [Rose [Leaf 8]]]]
†Actually, modern GHC Haskell has quite a bunch of dependently-typed features (see DaniDiaz' answer), but these are still quite clearly separated from the value-level language.
I'd like to propose a very simple alternative which doesn't require any extensions or trickery: don't use different types.
Here is a type which can hold lists with any number of nestings, provided you say how many up front:
data NestList a = Zero a | Succ (NestList [a]) deriving Show
instance Functor NestList where
fmap f (Zero a) = Zero (f a)
fmap f (Succ as) = Succ (fmap (map f) as)
A value of this type is a church numeral indicating how many layers of nesting there are, followed by a value with that many layers of nesting; for example,
Succ (Succ (Zero [['a']])) :: NestList Char
It's now easy-cheesy to write your \x -> [x] iteration; since we want one more layer of nesting, we add one Succ.
> iterate (\x -> Succ (fmap (:[]) x)) (Zero 8) !! 5
Succ (Succ (Succ (Succ (Succ (Zero [[[[[8]]]]])))))
Your proposal for how to implement natural numbers can be modified similarly to use a simple recursive type. But the standard way is even cleaner: just take the above NestList and drop all the arguments.
data Nat = Zero | Succ Nat
This problem indeed requires somewhat advanced type-level programming.
I followed #chi's suggestion in the comments, and searched for a library that provided inductive type-level naturals with their corresponding singletons. I found the fin library, which is used in the answer.
The usual extensions for type-level trickery:
{-# language DataKinds, PolyKinds, KindSignatures, ScopedTypeVariables, TypeFamilies #-}
Here's a type family that maps a type-level natural and an element type to the type of the corresponding nested list:
import Data.Type.Nat
type family Nested (n::Nat) a where
Nested Z a = [a]
Nested (S n) a = [Nested n a]
For example, we can test from ghci that
*Main> :kind! Nested Nat3 Int
Nested Nat3 Int :: *
= [[[[Int]]]]
(Nat3 is a convenient alias defined in Data.Type.Nat.)
And here's a newtype that wraps the function we want to construct. It uses the type family to express the level of nesting
newtype Iterate (n::Nat) a = Iterate { runIterate :: (a -> [a]) -> a -> Nested n a }
The fin library provides a really nifty induction1 function that lets us compute a result by induction on Nat. We can use it to compute the Iterate that corresponds to every Nat. The Nat is passed implicitly, as a constraint:
iterate' :: forall n a. SNatI n => Iterate (n::Nat) a
iterate' =
let step :: forall m. SNatI m => Iterate m a -> Iterate (S m) a
step (Iterate recN) = Iterate (\f a -> [recN f a])
in induction1 (Iterate id) step
Testing the function in ghci (using -XTypeApplications to supply the Nat):
*Main> runIterate (iterate' #Nat3) pure True
[[[[True]]]]

Omitting constructor arguments in Haskell case statements

Omitting function arguments is a nice tool for concise Haskell code.
h :: String -> Int
h = (4 +) . length
What about omitting data constructor arguments in case statements. The following code might be considered a little grungy, where s and i are the final arguments in A and B but are repeated as the final arguments in the body of each case match.
f :: Foo -> Int
f = \case
A s -> 4 + length s
B i -> 2 + id i
Is there a way to omit such arguments in case pattern matching? For constructors with a large number of arguments, this would radically shorten code width. E.g. the following pseudo code.
g :: Foo -> Int
g = \case
{- match `A` constructor -> function application to A's arguments -}
A -> (4 +) . length
{- match `B` constructor -> function application to B's arguments -}
B -> (2 +) . id
The GHC extension RecordWildCards lets you concisely bring all the fields of a constructor into scope (of course, this requires you to give names to those fields).
{-# LANGUAGE LambdaCase, RecordWildCards #-}
data Foo = Foo {field1, field2 :: Int} | Bar {field1 :: Int}
baz = \case
Foo{..} -> 4 + field2
Bar{..} -> 2 + field1
-- plus it also "sucks in" fields from a scope
mkBar400 = let field1 = 400 in Bar{..}
`
You can always refactor case statements on constructors into a single function so that from then on you only pass your concise function definitions as arguments to these specific functions. Allow me to illustrate.
Consider the Maybe a datatype:
data Maybe a = Nothing | Just a
Should you now need to define a function f :: Maybe a -> b (for some fixed b and perhaps also a), instead of writing it like
f Nothing = this
f (Just x) = that x
you could start by first defining a function
maybe f _ Nothing = f
maybe _ g (Just x) = g x
and then f can by defined as maybe this that. Pretty much as what happens with all the familiar recursion patterns.
This way you're effectively refactoring out case statements. The code gets arguably cleaner and it does not require language extensions.

Reflection on inputs to a function in Haskell?

I have a reflective situation where I want to display the types of inputs/outputs of a function. I could just add it to a separate data structure, but then I have duplication and would have to make sure they stay in sync manually.
For example, a function:
myFunc :: (String, MyType, Double) -> (Int, SomeType TypeP, MyOtherType a)
and so now I want to have something like (can be somewhat flexible, esp. when params involved):
input = ["String", "MyType", "Double"]
output = ["Int", "SomeType TypeP", "MyOtherType a"]
defined automatically. It doesn't have to be Strings directly. Is there a simple way to do this?
You don't really need a custom parser. You can just reflect on the TypeRep value you receive. For instance the following would work:
module ModuleReflect where
import Data.Typeable
import Control.Arrow
foo :: (Int, Bool, String) -> String -> String
foo = undefined
-- | We first want to in the case that no result is returned from splitTyConApp
-- to just return the input type, this. If we find an arrow constructor (->)
-- we want to get the start of the list and then recurse on the tail if any.
-- if we get anything else, like [] Char then we want to return the original type
-- [Char]
values :: Typeable a => a -> [TypeRep]
values x = split (typeOf x)
where split t = case first tyConString (splitTyConApp t) of
(_ , []) -> [t]
("->", [x]) -> [x]
("->", x) -> let current = init x
next = last x
in current ++ split next
(_ , _) -> [t]
inputs :: Typeable a => a -> [TypeRep]
inputs x = case values x of
(x:xs) | not (null xs) -> x : init xs
_ -> []
output :: Typeable a => a -> TypeRep
output x = last $ values x
and a sample session
Ok, modules loaded: ModuleReflect.
*ModuleReflect Data.Function> values foo
[(Int,Bool,[Char]),[Char],[Char]]
*ModuleReflect Data.Function> output foo
[Char]
*ModuleReflect Data.Function> inputs foo
[(Int,Bool,[Char]),[Char]]
This is just some quick barely tested code, I'm sure it can be cleaned up. And the reason I don't use typeRepArgs is that in the case of other constructors they would be broken up to, e.g [] Char returns Char instead of [Char].
This version does not treat elements of tuples as seperate results, but it can be easily changed to add that.
However as mentioned before this has a limitation, It'll only work on monomorphic types. If you want to be able to find this for any type, then you should probably use the GHC api. You can load the module, ask it to Typecheck and then inspect the Typecheck AST for the function and thus find it's type. Or you can use Hint to ask for the type of a function. and parse that.
Have a look at Data.Typeable: It defined a functiontypeOf :: Typeable a => a -> TypeRep, TypeRep is instance of Show. For example:
$ ghci
GHCi, version 6.12.1: http://www.haskell.org/ghc/ :? for help
:m Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
Prelude> :m +Data.Typeable
Prelude Data.Typeable> :i TypeRep
data TypeRep
= Data.Typeable.TypeRep !Data.Typeable.Key TyCon [TypeRep]
-- Defined in Data.Typeable
instance [overlap ok] Eq TypeRep -- Defined in Data.Typeable
instance [overlap ok] Show TypeRep -- Defined in Data.Typeable
instance [overlap ok] Typeable TypeRep -- Defined in Data.Typeable
Prelude Data.Typeable> typeOf (+)
Integer -> Integer -> Integer
Prelude Data.Typeable> typeOf (\(a,(_:x),1) -> (a :: (),x :: [()]))
((),[()],Integer) -> ((),[()])
Now, you only need a custom parser that translated this output into something suitable for you. This is left as an exercise to the reader.
PS: There seems to be one limitation: All types must be known before, for instance typeOf id will fail.

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