On a ghci prompt everywhere (mkT (\x -> 2 * x)) (8.7, 21, "word") evaluates to (8.7, 42, "word").
I expected the 8.7 to be doubled as well. Why am I wrong?
This is the result of mkT monomorphizing its argument in this particular case, but it turns out there's no broader way to address the issue. mkT isn't doing anything wrong.
It's worth looking first at why everywhere (* 2) doesn't type-check.
ghci> :t everywhere
everywhere
:: (forall a. Data a => a -> a) -> forall a. Data a => a -> a
ghci> :t (* 2)
(* 2) :: Num a => a -> a
ghci> :t everywhere (* 2)
<interactive>:1:13: error:
• Could not deduce (Num a) arising from a use of ‘*’
from the context: Data a
bound by a type expected by the context:
forall a. Data a => a -> a
at <interactive>:1:12-16
Possible fix:
add (Num a) to the context of
a type expected by the context:
forall a. Data a => a -> a
• In the expression: (*)
In the first argument of ‘everywhere’, namely ‘(* 2)’
In the expression: everywhere (* 2)
everywhere has a higher-rank type - the first forall a. is inside the parentheses. I kind of dislike documenting the type that way - it uses a as a type variable in two completely separate ways. But there are two different scopes, and that matters. What it's saying is that any function passed to it must be polymorphic over all instances of Data.
But the type of (* 2) doesn't match up there. It won't work with any instance of Data. It requires more - it requires that it be provided an instance of Num. So the error message dutifully reports that it can't deduce (Num a) from the context Data a. So this isn't going to work. The pieces don't fit together.
This is where mkT comes into play:
ghci> :t mkT
mkT :: (Typeable a, Typeable b) => (b -> b) -> a -> a
Its type is a bit funny. It looks almost like it does nothing at all, but Typeable is a funny class. mkT actually compares a and b for type equality, using those Typeable constraints. If they're the same, it applies the function you provided. Otherwise, it just acts as the identity function.
What it does when it's applied to a function is where things are going wrong for you:
ghci> :t mkT (* 2)
mkT (* 2) :: Typeable a => a -> a
It's still polymorphic in a, but the b it used to have has vanished. It had to pick a specific type b to work against, and it did that by defaulting to Integer. (See ghc's extended defaulting rules for details on how that works in ghci.) So...
ghci> mkT (* 2) 3.5
3.5
ghci> mkT (* 2) 7
14
ghci> mkT (* 2) (7 :: Int)
7
At the type level, mkT has to monomorphize its argument. That's the only way it can make use of the Typeable constraint when used in a context where a relevant variable no longer appears in its type.
(To tie the loop back to everywhere, the reason mkT (* 2) works as an argument to everywhere is because Data is a subclass of Typeable. The Data constraint implies that the Typeable requirement will be satisfied.)
So what can you do about this? Well, it's impossible to write it truly generically because of Haskell's open world assumption. Anywhere in the program, any type might be declared an instance of Num with arbitrary implementations of (*) and fromInteger. In order to work with everywhere, there would need to be some mechanism to go from knowing something is an instance of Data to looking up its Num instance. This just isn't possible at run time. Types have been erased. There may be some residues like Typeable dictionaries being carried around, but they don't provide any means to look up other instance dictionaries. And while you might be able to envision a language where that sort of lookup is possible, it actually would be very harmful to allow it in Haskell. It would invalidate the ability to reason about types parametrically, which would be a giant loss.
The best you can do is write transformation functions that work on multiple types:
ghci> let f = mkT (* (2 :: Int)) . mkT (* (2 :: Double)) . mkT (* (2 :: Integer))
ghci> f 5
10
ghci> f 2.7
5.4
ghci> f (9 :: Int)
18
ghci> f "hello"
"hello"
It's verbose and you can probably write something better by hand if you so desire. But it at least works, at least to some extent. And it doesn't require breaking foundational assumptions in the language design, which is always a bonus.
Here is a simplification of your case that doesn't use any Data stuff.
module MyModule where
dbl x = 2 * x
myId :: (a->a) -> a -> a
myId f = f
myDbl = myId dbl
Don't type this to the ghci prompt, rather, create a .hs file and load it.
Now check what type myDbl has.
Prelude > :l MyModule
[1 of 1] Compiling MyModule ( MyModule.hs, interpreted )
Ok, one module loaded.
*MyModule > :t MyModule.myDbl
MyModule.myDbl :: Integer -> Integer
Surprise! Why is it compiling at all? And why the weird types?
Because of the defaulting rules. (Basically, "if you don't know what to do with Num a, just use Integer"). Since myId cannot deal with dbl :: Num a => a -> a, Haskell allows it to take the Integer version.
Disable defaulting by adding default () at the top, and this module no longer compiles.
mkT is no different from myId in this respect.
I have a function
mySucc :: (Enum a, Bounded a, Eq a, Show a) => a -> Maybe a
mySucc int
| int == maxBound = Nothing
| otherwise = Just $ succ int
When I want to print the output of this function in ghci, Haskell seems to be confused as to which instance of Show to use. Why is that? Shouldn't Haskell automatically resolve a's type during runtime and use it's Show?
My limited understanding of type class is that, if you mention a type (in my case a) and say that it belongs to a type class (Show), Haskell should automatically resolve the type. Isn't that how it resolves Bounded, Enum and Eq? Please correct me if my understanding is wrong.
Shouldn't Haskell automatically resolve a's type during runtime and use it's Show?
Generally speaking, types don't exist at runtime. The compiler typechecks your code, resolves any polymorphism, and then erases the types. But, this is kind of orthogonal to your main question.
My limited understanding of type class is that, if you mention a type (in my case a) and say that it belongs to a type class (Show), Haskell should automatically resolve the type
No. The compiler will automatically resolve the instance. What that means is, you don't need to explicitly pass a showing-method into your function. For example, instead of the function
showTwice :: Show a => a -> String
showTwice x = show x ++ show x
you could have a function that doesn't use any typeclass
showTwice' :: (a -> String) -> a -> String
showTwice' show' x = show' x ++ show' x
which can be used much the same way as showTwice if you give it the standard show as the first argument. But that argument would need to be manually passed around at each call site. That's what you can avoid by using the type class instead, but this still requires the type to be known first.
(Your mySucc doesn't actually use show in any way at all, so you might as well omit the Show a constraint completely.)
When your call to mySucc appears in a larger expression, chances are the type will in fact also be inferred automatically. For example, mySucc (length "bla") will use a ~ Int, because the result of length is fixed to Int; or mySucc 'y' will use a ~ Char. However, if all the subexpressions are polymorphic (and in Haskell, even number literals are polymorphic), then the compiler won't have any indication what type you actually want. In that case you can always specify it explicitly, either in the argument
> mySucc (3 :: Int)
Just 4
or in the result
> mySucc 255 :: Maybe Word8
Nothing
Are you writing mySucc 1? In this case, you get a error because 1 literal is a polymorphic value of type Num a => a.
Try calling mySucc 1 :: Maybe Int and it will work.
This question already has an answer here:
What is the monomorphism restriction?
(1 answer)
Closed 6 years ago.
Example 1
Following definition without the type declaration will throw an error:
f :: Eq t => (t,t) -> Bool -- omiting this line will result in an error
f = \(x,y) -> x==y
(I know this function can be written shorter, but this is not the point here.)
Example 2
On the other hand using the same lambda function in a function using map does work without producing an error:
g l = map (\(x,y) -> x==y) l
(Just as an illustration: g [(3,4),(5,5),(7,6)] will produce [False,True,False]
Example 3
Also following code is perfectly fine and it seems to do exactly the same as the original f from above. Here the type inference seems to work.
f' (x,y) = x==y
Question
So my question is: Why do we need a type declaration in the first case, but not in the second and in the third?
If you use:
{-# LANGUAGE NoMonomorphismRestriction #-}
f = \(x,y) -> x==y
you won't get the error.
Update
The Haskell Wiki page on Monomorphism Restriction (link) offers some details on why these definitions are treated differently:
f1 x = show x
f2 = \x -> show x
The difference between the first and second version is that the first version binds x via a "function binding" (see section 4.4.3 of the Haskell 2010 Report), and is therefore unrestricted, but the second version does not. The reason why one is allowed and the other is not is that it's considered clear that sharing f1 will not share any computation, and less clear that sharing f2 will have the same effect. If this seems arbitrary, that's because it is. It is difficult to design an objective rule which disallows subjective unexpected behaviour. Some people are going to fall foul of the rule even though they're doing quite reasonable things.
As #ErikR notes in the comments, this is due to the Monomorphism restriction. We also see this in the error message:
No instance for (Eq a0) arising from a use of '=='
The type variable 'a0' is ambiguous
Possible cause: the monomorphism restriction applied to the following:
f :: (a0, a0) -> Bool
(bound at ...
Note: there are several potential instances:
instance Eq a => Eq (GHC.Real.Ratio a) -- Defined in 'GHC.Real'
instance Eq () -- Defined in 'GHC.Classes'
instane (Eq a, Eq b) => Eq (a,b) -- Defined in 'GHC.Classes'
..plus 22 others
The monomorphism restriction implies that the compiler tries to instantiate an ambiguous type into a non-ambiguous type. (Source: What is the monomorphism restriction?).
So, Haskell wants to put a single instance, but it can't - it finds several, and doesn't know which to choose.
This explains why adding the type solves the problem: now the compiler knows what to choose.
The "monomorphism restriction" is a counter-intuitive rule in Haskell type inference. If you forget to provide a type signature, sometimes this rule will fill the free type variables with specific types using "type defaulting" rules.
When I play with checking types of functions in Haskell with :t, for example like those in my previous question, I tend to get results such as:
Eq a => a -> [a] -> Bool
(Ord a, Num a, Ord a1, Num a1) => a -> a1 -> a
(Num t2, Num t1, Num t, Enum t2, Enum t1, Enum t) => [(t, t1, t2)]
It seems that this is not such a trivial question - how does the Haskell interpreter pick literals to symbolize typeclasses? When would it choose a rather than t? When would it choose a1 rather than b? Is it important from the programmer's point of view?
The names of the type variables aren't significant. The type:
Eq element => element -> [element] -> Bool
Is exactly the same as:
Eq a => a -> [a] -> Bool
Some names are simply easier to read/remember.
Now, how can an inferencer choose the best names for types?
Disclaimer: I'm absolutely not a GHC developer. However I'm working on a type-inferencer for Haskell in my bachelor thesis.
During inferencing the names chosen for the variables aren't probably that readable. In fact they are almost surely something along the lines of _N with N a number or aN with N a number.
This is due to the fact that you often have to "refresh" type variables in order to complete inferencing, so you need a fast way to create new names. And using numbered variables is pretty straightforward for this purpose.
The names displayed when inference is completed can be "pretty printed". The inferencer can rename the variables to use a, b, c and so on instead of _1, _2 etc.
The trick is that most operations have explicit type signatures. Some definitions require to quantify some type variables (class, data and instance for example).
All these names that the user explicitly provides can be used to display the type in a better way.
When inferencing you can somehow keep track of where the fresh type variables came from, in order to be able to rename them with something more sensible when displaying them to the user.
An other option is to refresh variables by adding a number to them. For example a fresh type of return could be Monad m0 => a0 -> m0 a0 (Here we know to use m and a simply because the class definition for Monad uses those names). When inferencing is finished you can get rid of the numbers and obtain the pretty names.
In general the inferencer will try to use names that were explicitly provided through signatures. If such a name was already used it might decide to add a number instead of using a different name (e.g. use b1 instead of c if b was already bound).
There are probably some other ad hoc rules. For example the fact that tuple elements have like t, t1, t2, t3 etc. is probably something done with a custom rule. In fact t doesn't appear in the signature for (,,) for example.
How does GHCi pick names for type variables? explains how many of these variable names come about. As Ganesh Sittampalam pointed out in a comment, something strange seems to be happening with arithmetic sequences. Both the Haskell 98 report and the Haskell 2010 report indicate that
[e1..] = enumFrom e1
GHCi, however, gives the following:
Prelude> :t [undefined..]
[undefined..] :: Enum t => [t]
Prelude> :t enumFrom undefined
enumFrom undefined :: Enum a => [a]
This makes it clear that the weird behavior has nothing to do with the Enum class itself, but rather comes in from some stage in translating the syntactic sequence to the enumFrom form. I wondered if maybe GHC wasn't really using that translation, but it really is:
{-# LANGUAGE NoMonomorphismRestriction #-}
module X (aoeu,htns) where
aoeu = [undefined..]
htns = enumFrom undefined
compiled using ghc -ddump-simpl enumlit.hs gives
X.htns :: forall a_aiD. GHC.Enum.Enum a_aiD => [a_aiD]
[GblId, Arity=1]
X.htns =
\ (# a_aiG) ($dEnum_aiH :: GHC.Enum.Enum a_aiG) ->
GHC.Enum.enumFrom # a_aiG $dEnum_aiH (GHC.Err.undefined # a_aiG)
X.aoeu :: forall t_aiS. GHC.Enum.Enum t_aiS => [t_aiS]
[GblId, Arity=1]
X.aoeu =
\ (# t_aiV) ($dEnum_aiW :: GHC.Enum.Enum t_aiV) ->
GHC.Enum.enumFrom # t_aiV $dEnum_aiW (GHC.Err.undefined # t_aiV)
so the only difference between these two representations is the assigned type variable name. I don't know enough about how GHC works to know where that t comes from, but at least I've narrowed it down!
Ørjan Johansen has noted in a comment that something similar seems to happen with function definitions and lambda abstractions.
Prelude> :t \x -> x
\x -> x :: t -> t
but
Prelude> :t map (\x->x) $ undefined
map (\x->x) $ undefined :: [b]
In the latter case, the type b comes from an explicit type signature given to map.
Are you familiar with the concepts of alpha equivalence and alpha substitution? This captures the notion that, for example, both of the following are completely equivalent and interconvertible (in certain circumstances) even though they differ:
\x -> (x, x)
\y -> (y, y)
The same concept can be extended to the level of types and type variables (see "System F" for further reading). Haskell in fact has a notion of "lambdas at the type level" for binding type variables, but it's hard to see because they're implicit by default. However, you can make them explicit by using the ExplicitForAll extension, and play around with explicitly binding your type variables:
ghci> :set -XExplicitForAll
ghci> let f x = x; f :: forall a. a -> a
In the second line, I use the forall keyword to introduce a new type variable, which is then used in a type.
In other words, it doesn't matter whether you choose a or t in your example, as long as the type expressions satisfy alpha-equivalence. Choosing type variable names so as to maximize human convenience is an entirely different topic, and probably far more complicated!
When metaprogramming, it may be useful (or necessary) to pass along to Haskell's type system information about types that's known to your program but not inferable in Hindley-Milner. Is there a library (or language extension, etc) that provides facilities for doing this—that is, programmatic type annotations—in Haskell?
Consider a situation where you're working with a heterogenous list (implemented using the Data.Dynamic library or existential quantification, say) and you want to filter the list down to a bog-standard, homogeneously typed Haskell list. You can write a function like
import Data.Dynamic
import Data.Typeable
dynListToList :: (Typeable a) => [Dynamic] -> [a]
dynListToList = (map fromJust) . (filter isJust) . (map fromDynamic)
and call it with a manual type annotation. For example,
foo :: [Int]
foo = dynListToList [ toDyn (1 :: Int)
, toDyn (2 :: Int)
, toDyn ("foo" :: String) ]
Here foo is the list [1, 2] :: [Int]; that works fine and you're back on solid ground where Haskell's type system can do its thing.
Now imagine you want to do much the same thing but (a) at the time you write the code you don't know what the type of the list produced by a call to dynListToList needs to be, yet (b) your program does contain the information necessary to figure this out, only (c) it's not in a form accessible to the type system.
For example, say you've randomly selected an item from your heterogenous list and you want to filter the list down by that type. Using the type-checking facilities supplied by Data.Typeable, your program has all the information it needs to do this, but as far as I can tell—this is the essence of the question—there's no way to pass it along to the type system. Here's some pseudo-Haskell that shows what I mean:
import Data.Dynamic
import Data.Typeable
randList :: (Typeable a) => [Dynamic] -> IO [a]
randList dl = do
tr <- randItem $ map dynTypeRep dl
return (dynListToList dl :: [<tr>]) -- This thing should have the type
-- represented by `tr`
(Assume randItem selects a random item from a list.)
Without a type annotation on the argument of return, the compiler will tell you that it has an "ambiguous type" and ask you to provide one. But you can't provide a manual type annotation because the type is not known at write-time (and can vary); the type is known at run-time, however—albeit in a form the type system can't use (here, the type needed is represented by the value tr, a TypeRep—see Data.Typeable for details).
The pseudo-code :: [<tr>] is the magic I want to happen. Is there any way to provide the type system with type information programatically; that is, with type information contained in a value in your program?
Basically I'm looking for a function with (pseudo-) type ??? -> TypeRep -> a that takes a value of a type unknown to Haskell's type system and a TypeRep and says, "Trust me, compiler, I know what I'm doing. This thing has the value represented by this TypeRep." (Note that this is not what unsafeCoerce does.)
Or is there something completely different that gets me the same place? For example, I can imagine a language extension that permits assignment to type variables, like a souped-up version of the extension enabling scoped type variables.
(If this is impossible or highly impractical,—e.g., it requires packing a complete GHCi-like interpreter into the executable—please try to explain why.)
No, you can't do this. The long and short of it is that you're trying to write a dependently-typed function, and Haskell isn't a dependently typed language; you can't lift your TypeRep value to a true type, and so there's no way to write down the type of your desired function. To explain this in a little more detail, I'm first going to show why the way you've phrased the type of randList doesn't really make sense. Then, I'm going to explain why you can't do what you want. Finally, I'll briefly mention a couple thoughts on what to actually do.
Existentials
Your type signature for randList can't mean what you want it to mean. Remembering that all type variables in Haskell are universally quantified, it reads
randList :: forall a. Typeable a => [Dynamic] -> IO [a]
Thus, I'm entitled to call it as, say, randList dyns :: IO [Int] anywhere I want; I must be able to provide a return value for all a, not simply for some a. Thinking of this as a game, it's one where the caller can pick a, not the function itself. What you want to say (this isn't valid Haskell syntax, although you can translate it into valid Haskell by using an existential data type1) is something more like
randList :: [Dynamic] -> (exists a. Typeable a => IO [a])
This promises that the elements of the list are of some type a, which is an instance of Typeable, but not necessarily any such type. But even with this, you'll have two problems. First, even if you could construct such a list, what could you do with it? And second, it turns out that you can't even construct it in the first place.
Since all that you know about the elements of the existential list is that they're instances of Typeable, what can you do with them? Looking at the documentation, we see that there are only two functions2 which take instances of Typeable:
typeOf :: Typeable a => a -> TypeRep, from the type class itself (indeed, the only method therein); and
cast :: (Typeable a, Typeable b) => a -> Maybe b (which is implemented with unsafeCoerce, and couldn't be written otherwise).
Thus, all that you know about the type of the elements in the list is that you can call typeOf and cast on them. Since we'll never be able to usefully do anything else with them, our existential might just as well be (again, not valid Haskell)
randList :: [Dynamic] -> IO [(TypeRep, forall b. Typeable b => Maybe b)]
This is what we get if we apply typeOf and cast to every element of our list, store the results, and throw away the now-useless existentially typed original value. Clearly, the TypeRep part of this list isn't useful. And the second half of the list isn't either. Since we're back to a universally-quantified type, the caller of randList is once again entitled to request that they get a Maybe Int, a Maybe Bool, or a Maybe b for any (typeable) b of their choosing. (In fact, they have slightly more power than before, since they can instantiate different elements of the list to different types.) But they can't figure out what type they're converting from unless they already know it—you've still lost the type information you were trying to keep.
And even setting aside the fact that they're not useful, you simply can't construct the desired existential type here. The error arises when you try to return the existentially-typed list (return $ dynListToList dl). At what specific type are you calling dynListToList? Recall that dynListToList :: forall a. Typeable a => [Dynamic] -> [a]; thus, randList is responsible for picking which a dynListToList is going to use. But it doesn't know which a to pick; again, that's the source of the question! So the type that you're trying to return is underspecified, and thus ambiguous.3
Dependent types
OK, so what would make this existential useful (and possible)? Well, we actually have slightly more information: not only do we know there's some a, we have its TypeRep. So maybe we can package that up:
randList :: [Dynamic] -> (exists a. Typeable a => IO (TypeRep,[a]))
This isn't quite good enough, though; the TypeRep and the [a] aren't linked at all. And that's exactly what you're trying to express: some way to link the TypeRep and the a.
Basically, your goal is to write something like
toType :: TypeRep -> *
Here, * is the kind of all types; if you haven't seen kinds before, they are to types what types are to values. * classifies types, * -> * classifies one-argument type constructors, etc. (For instance, Int :: *, Maybe :: * -> *, Either :: * -> * -> *, and Maybe Int :: *.)
With this, you could write (once again, this code isn't valid Haskell; in fact, it really bears only a passing resemblance to Haskell, as there's no way you could write it or anything like it within Haskell's type system):
randList :: [Dynamic] -> (exists (tr :: TypeRep).
Typeable (toType tr) => IO (tr, [toType tr]))
randList dl = do
tr <- randItem $ map dynTypeRep dl
return (tr, dynListToList dl :: [toType tr])
-- In fact, in an ideal world, the `:: [toType tr]` signature would be
-- inferable.
Now, you're promising the right thing: not that there exists some type which classifies the elements of the list, but that there exists some TypeRep such that its corresponding type classifies the elements of the list. If only you could do this, you would be set. But writing toType :: TypeRep -> * is completely impossible in Haskell: doing this requires a dependently-typed language, since toType tr is a type which depends on a value.
What does this mean? In Haskell, it's perfectly acceptable for values to depend on other values; this is what a function is. The value head "abc", for instance, depends on the value "abc". Similarly, we have type constructors, so it's acceptable for types to depend on other types; consider Maybe Int, and how it depends on Int. We can even have values which depend on types! Consider id :: a -> a. This is really a family of functions: id_Int :: Int -> Int, id_Bool :: Bool -> Bool, etc. Which one we have depends on the type of a. (So really, id = \(a :: *) (x :: a) -> x; although we can't write this in Haskell, there are languages where we can.)
Crucially, however, we can never have a type that depends on a value. We might want such a thing: imagine Vec 7 Int, the type of length-7 lists of integers. Here, Vec :: Nat -> * -> *: a type whose first argument must be a value of type Nat. But we can't write this sort of thing in Haskell.4 Languages which support this are called dependently-typed (and will let us write id as we did above); examples include Coq and Agda. (Such languages often double as proof assistants, and are generally used for research work as opposed to writing actual code. Dependent types are hard, and making them useful for everyday programming is an active area of research.)
Thus, in Haskell, we can check everything about our types first, throw away all that information, and then compile something that refers only to values. In fact, this is exactly what GHC does; since we can never check types at run-time in Haskell, GHC erases all the types at compile-time without changing the program's run-time behavior. This is why unsafeCoerce is easy to implement (operationally) and completely unsafe: at run-time, it's a no-op, but it lies to the type system. Consequently, something like toType is completely impossible to implement in the Haskell type system.
In fact, as you noticed, you can't even write down the desired type and use unsafeCoerce. For some problems, you can get away with this; we can write down the type for the function, but only implement it with by cheating. That's exactly how fromDynamic works. But as we saw above, there's not even a good type to give to this problem from within Haskell. The imaginary toType function allows you to give the program a type, but you can't even write down toType's type!
What now?
So, you can't do this. What should you do? My guess is that your overall architecture isn't ideal for Haskell, although I haven't seen it; Typeable and Dynamic don't actually show up that much in Haskell programs. (Perhaps you're "speaking Haskell with a Python accent", as they say.) If you only have a finite set of data types to deal with, you might be able to bundle things into a plain old algebraic data type instead:
data MyType = MTInt Int | MTBool Bool | MTString String
Then you can write isMTInt, and just use filter isMTInt, or filter (isSameMTAs randomMT).
Although I don't know what it is, there's probably a way you could unsafeCoerce your way through this problem. But frankly, that's not a good idea unless you really, really, really, really, really, really know what you're doing. And even then, it's probably not. If you need unsafeCoerce, you'll know, it won't just be a convenience thing.
I really agree with Daniel Wagner's comment: you're probably going to want to rethink your approach from scratch. Again, though, since I haven't seen your architecture, I can't say what that will mean. Maybe there's another Stack Overflow question in there, if you can distill out a concrete difficulty.
1 That looks like the following:
{-# LANGUAGE ExistentialQuantification #-}
data TypeableList = forall a. Typeable a => TypeableList [a]
randList :: [Dynamic] -> IO TypeableList
However, since none of this code compiles anyway, I think writing it out with exists is clearer.
2 Technically, there are some other functions which look relevant, such as toDyn :: Typeable a => a -> Dynamic and fromDyn :: Typeable a => Dynamic -> a -> a. However, Dynamic is more or less an existential wrapper around Typeables, relying on typeOf and TypeReps to know when to unsafeCoerce (GHC uses some implementation-specific types and unsafeCoerce, but you could do it this way, with the possible exception of dynApply/dynApp), so toDyn doesn't do anything new. And fromDyn doesn't really expect its argument of type a; it's just a wrapper around cast. These functions, and the other similar ones, don't provide any extra power that isn't available with just typeOf and cast. (For instance, going back to a Dynamic isn't very useful for your problem!)
3 To see the error in action, you can try to compile the following complete Haskell program:
{-# LANGUAGE ExistentialQuantification #-}
import Data.Dynamic
import Data.Typeable
import Data.Maybe
randItem :: [a] -> IO a
randItem = return . head -- Good enough for a short and non-compiling example
dynListToList :: Typeable a => [Dynamic] -> [a]
dynListToList = mapMaybe fromDynamic
data TypeableList = forall a. Typeable a => TypeableList [a]
randList :: [Dynamic] -> IO TypeableList
randList dl = do
tr <- randItem $ map dynTypeRep dl
return . TypeableList $ dynListToList dl -- Error! Ambiguous type variable.
Sure enough, if you try to compile this, you get the error:
SO12273982.hs:17:27:
Ambiguous type variable `a0' in the constraint:
(Typeable a0) arising from a use of `dynListToList'
Probable fix: add a type signature that fixes these type variable(s)
In the second argument of `($)', namely `dynListToList dl'
In a stmt of a 'do' block: return . TypeableList $ dynListToList dl
In the expression:
do { tr <- randItem $ map dynTypeRep dl;
return . TypeableList $ dynListToList dl }
But as is the entire point of the question, you can't "add a type signature that fixes these type variable(s)", because you don't know what type you want.
4 Mostly. GHC 7.4 has support for lifting types to kinds and for kind polymorphism; see section 7.8, "Kind polymorphism and promotion", in the GHC 7.4 user manual. This doesn't make Haskell dependently typed—something like TypeRep -> * example is still out5—but you will be able to write Vec by using very expressive types that look like values.
5 Technically, you could now write down something which looks like it has the desired type: type family ToType :: TypeRep -> *. However, this takes a type of the promoted kind TypeRep, and not a value of the type TypeRep; and besides, you still wouldn't be able to implement it. (At least I don't think so, and I can't see how you would—but I am not an expert in this.) But at this point, we're pretty far afield.
What you're observing is that the type TypeRep doesn't actually carry any type-level information along with it; only term-level information. This is a shame, but we can do better when we know all the type constructors we care about. For example, suppose we only care about Ints, lists, and function types.
{-# LANGUAGE GADTs, TypeOperators #-}
import Control.Monad
data a :=: b where Refl :: a :=: a
data Dynamic where Dynamic :: TypeRep a -> a -> Dynamic
data TypeRep a where
Int :: TypeRep Int
List :: TypeRep a -> TypeRep [a]
Arrow :: TypeRep a -> TypeRep b -> TypeRep (a -> b)
class Typeable a where typeOf :: TypeRep a
instance Typeable Int where typeOf = Int
instance Typeable a => Typeable [a] where typeOf = List typeOf
instance (Typeable a, Typeable b) => Typeable (a -> b) where
typeOf = Arrow typeOf typeOf
congArrow :: from :=: from' -> to :=: to' -> (from -> to) :=: (from' -> to')
congArrow Refl Refl = Refl
congList :: a :=: b -> [a] :=: [b]
congList Refl = Refl
eq :: TypeRep a -> TypeRep b -> Maybe (a :=: b)
eq Int Int = Just Refl
eq (Arrow from to) (Arrow from' to') = liftM2 congArrow (eq from from') (eq to to')
eq (List t) (List t') = liftM congList (eq t t')
eq _ _ = Nothing
eqTypeable :: (Typeable a, Typeable b) => Maybe (a :=: b)
eqTypeable = eq typeOf typeOf
toDynamic :: Typeable a => a -> Dynamic
toDynamic a = Dynamic typeOf a
-- look ma, no unsafeCoerce!
fromDynamic_ :: TypeRep a -> Dynamic -> Maybe a
fromDynamic_ rep (Dynamic rep' a) = case eq rep rep' of
Just Refl -> Just a
Nothing -> Nothing
fromDynamic :: Typeable a => Dynamic -> Maybe a
fromDynamic = fromDynamic_ typeOf
All of the above is pretty standard. For more on the design strategy, you'll want to read about GADTs and singleton types. Now, the function you want to write follows; the type is going to look a bit daft, but bear with me.
-- extract only the elements of the list whose type match the head
firstOnly :: [Dynamic] -> Dynamic
firstOnly [] = Dynamic (List Int) []
firstOnly (Dynamic rep v:xs) = Dynamic (List rep) (v:go xs) where
go [] = []
go (Dynamic rep' v:xs) = case eq rep rep' of
Just Refl -> v : go xs
Nothing -> go xs
Here we've picked a random element (I rolled a die, and it came up 1) and extracted only the elements that have a matching type from the list of dynamic values. Now, we could have done the same thing with regular boring old Dynamic from the standard libraries; however, what we couldn't have done is used the TypeRep in a meaningful way. I now demonstrate that we can do so: we'll pattern match on the TypeRep, and then use the enclosed value at the specific type the TypeRep tells us it is.
use :: Dynamic -> [Int]
use (Dynamic (List (Arrow Int Int)) fs) = zipWith ($) fs [1..]
use (Dynamic (List Int) vs) = vs
use (Dynamic Int v) = [v]
use (Dynamic (Arrow (List Int) (List (List Int))) f) = concat (f [0..5])
use _ = []
Note that on the right-hand sides of these equations, we are using the wrapped value at different, concrete types; the pattern match on the TypeRep is actually introducing type-level information.
You want a function that chooses a different type of values to return based on runtime data. Okay, great. But the whole purpose of a type is to tell you what operations can be performed on a value. When you don't know what type will be returned from a function, what do you do with the values it returns? What operations can you perform on them? There are two options:
You want to read the type, and perform some behaviour based on which type it is. In this case you can only cater for a finite list of types known in advance, essentially by testing "is it this type? then we do this operation...". This is easily possible in the current Dynamic framework: just return the Dynamic objects, using dynTypeRep to filter them, and leave the application of fromDynamic to whoever wants to consume your result. Moreover, it could well be possible without Dynamic, if you don't mind setting the finite list of types in your producer code, rather than your consumer code: just use an ADT with a constructor for each type, data Thing = Thing1 Int | Thing2 String | Thing3 (Thing,Thing). This latter option is by far the best if it is possible.
You want to perform some operation that works across a family of types, potentially some of which you don't know about yet, e.g. by using type class operations. This is trickier, and it's tricky conceptually too, because your program is not allowed to change behaviour based on whether or not some type class instance exists – it's an important property of the type class system that the introduction of a new instance can either make a program type check or stop it from type checking, but it can't change the behaviour of a program. Hence you can't throw an error if your input list contains inappropriate types, so I'm really not sure that there's anything you can do that doesn't essentially involve falling back to the first solution at some point.