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Type Constraints in Data Declaration Haskell
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I am trying to make a normal form game solver for game theory, and I'm trying to make it as generic as possible for good practice and for my own convenience. I would like to use the same functions to solve both zero-sum and non-zero-sum games, so I am using the following data type:
data Payoffs = (Num a, Eq a, Ord a) => ZS a
| (Num a, Eq a, Ord a) => NZS (a,a)
However, this is not correct syntax. Is there any way to constrain a so that it must satisfy those type constraints?
Short answer (and probably not the one you need):
To make your code work as is, you need a forall quantifier (for which you need to enable ExistentialQuantification):
{-# LANGUAGE ExistentialQuantification #-}
data Payoffs =
forall a. (Num a, Eq a, Ord a) => ZS a
| forall a. (Num a, Eq a, Ord a) => NZS (a,a)
If you have a type variable in the data constructor (i.e. ZS a), then you have two choices: either that variable has to appear in the type constructor (i.e. data Payoffs a =), or you need to say "I don't care what type it is, as long as it supports these classes" - which is achieved via the forall quantifier.
But this looks kinda useless to me, which suggests that you may be misunderstanding what it means. If you write the above code, every value of your Payoffs type will be able to wrap a value of any type, as long as that type supports Num, Eq, and Ord. One subtle consequence of this is that, if you have two values of Payoffs lying around, they will not necessarily wrap the same type. For example:
let x = ZS (42 :: Int) -- wraps an Int
let y = NZS (2.71 :: Double, 3.14) -- wraps two Doubles
This means that, upon unpacking them, you won't be able to, for example, add them together, because, even though they both implement Num, the compiler doesn't have any proof that they're actually the same type.
What I suspect you actually need is a parametrized type, like this:
data Payoffs a = ZS a | NZS (a, a)
But then, of course, you lose the constraints: anybody can go and create ZS String or something. You can use the GADT syntax (with the GADTs extension) to bring them back:
{-# LANGUAGE GADTs #-}
data Payoffs a where
ZS :: (Num a, Ord a, Eq a) => a -> Payoffs a
NZS :: (Num a, Ord a, Eq a) => (a, a) -> Payoffs a
This notation is equivalent to ZS a | NZS (a, a), except you get to define each constructor with the same syntax as any function - including constraints. A type defined like this won't allow for creating values of type Payoffs a unless a satisfies the constraints.
At the same time, if you have a value of a type like this lying around, you know what type it wraps inside. And this allows you to tell if two Payoffs values wrap the same type or different. And then, if you know that they're the same, you can do things with them using the supported classes, for example:
addPayoffs :: Payoffs a -> Payoffs a -> Payoffs a
addPayoffs (ZS a) (ZS b) = ZS (a + b)
addPayoffs (ZS a) (NZS (x,y)) = NZS (a+x, a+y)
... etc.
Related
How is a function defined for different types in Haskell?
Given
func :: Integral a => a -> a
func x = x
func' :: (RealFrac a , Integral b) => a -> b
func' x = truncate x
How could they be combined into one function with the signature
func :: (SomeClassForBoth a, Integral b) => a -> b
With a typeclass.
class TowardsZero a where towardsZero :: Integral b => a -> b
instance TowardsZero Int where towardsZero = fromIntegral
instance TowardsZero Double where towardsZero = truncate
-- and so on
Possibly a class with an associated type family constraint is closer to what you wrote (though perhaps not closer to what you had in mind):
{-# LANGUAGE TypeFamilies #-}
import GHC.Exts
class TowardsZero a where
type RetCon a b :: Constraint
towardsZero :: RetCon a b => a -> b
instance TowardsZero Int where
type RetCon Int b = Int ~ b
towardsZero = id
instance TowardsZero Double where
type RetCon Double b = Integral b
towardsZero = truncate
-- and so on
This is known as ad hoc polymorphism, where you execute different code depending on the type. The way this is done in Haskell is using typeclasses. The most direct way is to define a new class
class Truncable a where
trunc :: Integral b => a -> b
And then you can define several concrete instances.
instance Truncable Integer where trunc = fromInteger
instance Truncable Double where trunc = truncate
This is unsatisfying because it requires an instance for each concrete type, when there are really only two families of identical-looking instances. Unfortunately, this is one of the cases where it is hard to reduce boilerplate, for technical reasons (being able to define "instance families" like this interferes with the open-world assumption of typeclasses, among other difficulties with type inference). As a hint of the complexity, note that your definition assumes that there is no type that is both RealFrac and Integral, but this is not guaranteed -- which implementation should we pick in this case?
There is another issue with this typeclass solution, which is that the Integral version doesn't have the type
trunc :: Integral a => a -> a
as you specified, but rather
trunc :: (Integral a, Integral b) => a -> b
Semantically this is not a problem, as I don't believe it is possible to end up with some polymorphic code where you don't know whether the type you are working with is Integral, but you do need to know that when it is, the result type is the same as the incoming type. That is, I claim that whenever you would need the former rather than the latter signature, you already know enough to replace trunc by id in your source. (It's a gut feeling though, and I would love to be proven wrong, seems like a fun puzzle)
There may be performance implications, however, since you might unnecessarily call fromIntegral to convert a type to itself, and I think the way around this is to use {-# RULES #-} definitions, which is a dark scary bag of complexity that I've never really dug into, so I don't know how hard or easy this is.
I don't recommend this, but you can hack at it with a GADT:
data T a where
T1 :: a -> T a
T2 :: RealFrac a => a -> T b
func :: Integral a => T a -> a
func (T1 x) = x
func (T2 x) = truncate x
The T type says, "Either you already know the type of the value I'm wrapping up, or it's some unknown instance of RealFrac". The T2 constructor existentially quantifies a and packs up a RealFrac dictionary, which we use in the second clause of func to convert from (unknown) a to b. Then, in func, I'm applying an Integral constraint to the a which may or may not be inside the T.
I'm having trouble wrapping my mind around the relationship (and interactions) between Haskell's forall and => (and for that matter the . that often connects them).
For example
λ> :t (+)
λ> :t id
give
(+) :: forall a. Num a => a -> a -> a
id :: forall a. a -> a
and while I understand how these work in these specific cases, I'm not comfortable parsing the expressions (signatures?) forall a. Num a => or forall a. themselves into something meaningful, or that I can generally understand in more complex contexts.
What do forall a. Num a => and forall a. mean? Specifically, what is the roles played in each by forall, => and a?
(As another perspective, without invoking the "implicit dictionary passing" implementation of type classes):
forall a. in Haskell means "for every type a".1 It's introducing a type variable, and declaring that the rest of the type expression has to be valid whatever choice is made for a.
You usually don't see it in basic Haskell (without turning on any extensions in GHC), because it's not necessary; you just use type variables in your type signature, and GHC automatically assumes there are foralls introducing those variables at the start of the expression.
For example:
zip :: forall a. ( forall b. ( [a] -> [b] -> [(a, b)] ))
zip :: forall a. forall b. [a] -> [b] -> [(a, b)]
zip :: forall a b. [a] -> [b] -> [(a, b)]
zip :: [a] -> [b] -> [(a, b)]
The above are all the same; they just tell us that zip can be a way of zipping a list of a together with a list of b to make a list of (a, b) pairs, whatever choice we feel like making for a and b.
forall mainly comes into play with extensions, because then you can introduce type variables with scopes other than the default ones assumed by GHC if you don't explicitly write them.
Now, the constraints => type syntax can be read roughly as "these constraints imply this type", or "provided these constraints hold, you can use this type". It's used all the time, even in vanilla Haskell with no extensions, so it's important to understand what it means and how it works and not just copy and paste and hope.
The => arrow allows us to state a set of constraints on the variables in the rest of the type expression; it lets us put limitations on what choices can be made to introduce the type variable. You should read it first by ignoring everything left of the => arrow, and reading the the right part on its own. This gives you the "shape" of the type. The stuff to the left of the => arrow tells you what kind of types you can use the rest of the type with.
An example:
(+) :: Num a => a -> a -> a
This means that (+) is exactly the same kind of thing as anything with a simpler type like a -> a -> a, except the Num a => is telling us that we're not free to choose any type a. We can only choose a type for a when we know that it is a member of the Num type class (another slightly more precise way of saying "a is a member of Num is "the constraint Num a holds").
Note that GHC is still assuming that there's an implicit forall a to introduce the type variable a here, so it really looks like:
(+) :: forall a. Num a => a -> a -> a
In which case you can read this off moderately easily as an English sentence once you know what forall a. and Num a => means: "For every type a, provided Num a holds, plus has the type a -> a -> a".
1 If you're familiar with formal logic at all, it's just an ASCII-friendly way of writing ∀a, a "universally quantified variable".
As the forall matter appears to be settled, I'll attempt to explain the => a bit. The things to the left of the => are arguments, much like ones to the left of a ->. But you don't apply these arguments manually, and they can only have specific types.
f :: Num a => a -> a
is a function that takes two arguments:
A Num a dictionary.
An a.
When you apply f, you just provide the a. GHC has to provide the Num a. If it's applied to a specific concrete type like Int, GHC knows Num Int and can supply it at the call site. Otherwise, it checks that Num a is provided by some outer context and uses that one. The great thing about Haskell's typeclass system is that it ensures that any two Num a dictionaries, however they are found, will be identical. So it doesn't matter where the dictionary comes from—it is sure to be the right one.
Further discussion
A lot of these things we're talking about aren't exactly part of Haskell so much as they're part of the way GHC interprets Haskell by translation to GHC core, AKA System FC, an extension of the very well-studied System F, AKA the Girard-Reynolds calculus. System FC is an explicitly typed polymorphic lambda calculus with algebraic datatypes, etc., but no type inference, no instance resolution, etc. After GHC checks the types in your Haskell code, it translates that code to System FC by a thoroughly mechanical process. It can do this confidently because the type checker "decorates" the code with all the information the desugarer needs to plumb all the dictionaries around. If you have a Haskell function that looks like
foo :: forall a . Num a => a -> a -> a
foo x y= x + y
then that will translate to something that looks like
foo :: forall a . Num a -> a -> a -> a
foo = /\ (a :: *) -> \ (d :: Num a) -> \ (x :: a) -> \ (y :: a) -> (+) #a d x y
The /\ is a type lambda—it's just line a normal lambda except it takes a type variable. The # represents application of a type to a function that takes one. The + is really just a record selector. It chooses the right field from the dictionary it's passed.
I suppose it helps if we add the implied parentheses:
(+) :: ∀ a . ( Num a => (a -> (a -> a)) )
id :: ∀ a . ( a -> a )
The ∀ always goes together with a .. It's basically special syntax meaning “anything between ∀ and . are type variables that I want to introduce into the following scope”†
=> denotes what Idris calls an implicit function: Num a is a dictionary for the instance Num a, and such a dictionary is implicitly needed whenever you're adding numbers. But whether a is a type variable here that was previously introduced by some ∀, or a fixed type, doesn't really matter. You could also have
(+) :: Num Int => Int -> Int -> Int
That's just superfluous, because the compiler knows that Int is a Num instance and hence automatically (implicitly!) chooses the right dictionary.
Really, there's no particular relationship between ∀ and =>, they just happen to be used often together.
†Actually this is a type-level lambda. The type expression ∀ a . b behaves analogously to the value level expression \a -> b.
I know this question has been asked and answered lots of times but I still don't really understand why putting constraints on a data type is a bad thing.
For example, let's take Data.Map k a. All of the useful functions involving a Map need an Ord k constraint. So there is an implicit constraint on the definition of Data.Map. Why is it better to keep it implicit instead of letting the compiler and programmers know that Data.Map needs an orderable key.
Also, specifying a final type in a type declaration is something common, and one can see it as a way of "super" constraining a data type.
For example, I can write
data User = User { name :: String }
and that's acceptable. However is that not a constrained version of
data User' s = User' { name :: s }
After all 99% of the functions I'll write for the User type don't need a String and the few which will would probably only need s to be IsString and Show.
So, why is the lax version of User considered bad:
data (IsString s, Show s, ...) => User'' { name :: s }
while both User and User' are considered good?
I'm asking this, because lots of the time, I feel I'm unnecessarily narrowing my data (or even function) definitions, just to not have to propagate constraints.
Update
As far as I understand, data type constraints only apply to the constructor and don't propagate. So my question is then, why do data type constraints not work as expected (and propagate)? It's an extension anyway, so why not have a new extension doing data properly, if it was considered useful by the community?
TL;DR:
Use GADTs to provide implicit data contexts.
Don't use any kind of data constraint if you could do with Functor instances etc.
Map's too old to change to a GADT anyway.
Scroll to the bottom if you want to see the User implementation with GADTs
Let's use a case study of a Bag where all we care about is how many times something is in it. (Like an unordered sequence. We nearly always need an Eq constraint to do anything useful with it.
I'll use the inefficient list implementation so as not to muddy the waters over the Data.Map issue.
GADTs - the solution to the data constraint "problem"
The easy way to do what you're after is to use a GADT:
Notice below how the Eq constraint not only forces you to use types with an Eq instance when making GADTBags, it provides that instance implicitly wherever the GADTBag constructor appears. That's why count doesn't need an Eq context, whereas countV2 does - it doesn't use the constructor:
{-# LANGUAGE GADTs #-}
data GADTBag a where
GADTBag :: Eq a => [a] -> GADTBag a
unGADTBag (GADTBag xs) = xs
instance Show a => Show (GADTBag a) where
showsPrec i (GADTBag xs) = showParen (i>9) (("GADTBag " ++ show xs) ++)
count :: a -> GADTBag a -> Int -- no Eq here
count a (GADTBag xs) = length.filter (==a) $ xs -- but == here
countV2 a = length.filter (==a).unGADTBag
size :: GADTBag a -> Int
size (GADTBag xs) = length xs
ghci> count 'l' (GADTBag "Hello")
2
ghci> :t countV2
countV2 :: Eq a => a -> GADTBag a -> Int
Now we didn't need the Eq constraint when we found the total size of the bag, but it didn't clutter up our definition anyway. (We could have used size = length . unGADTBag just as well.)
Now lets make a functor:
instance Functor GADTBag where
fmap f (GADTBag xs) = GADTBag (map f xs)
oops!
DataConstraints_so.lhs:49:30:
Could not deduce (Eq b) arising from a use of `GADTBag'
from the context (Eq a)
That's unfixable (with the standard Functor class) because I can't restrict the type of fmap, but need to for the new list.
Data Constraint version
Can we do as you asked? Well, yes, except that you have to keep repeating the Eq constraint wherever you use the constructor:
{-# LANGUAGE DatatypeContexts #-}
data Eq a => EqBag a = EqBag {unEqBag :: [a]}
deriving Show
count' a (EqBag xs) = length.filter (==a) $ xs
size' (EqBag xs) = length xs -- Note: doesn't use (==) at all
Let's go to ghci to find out some less pretty things:
ghci> :so DataConstraints
DataConstraints_so.lhs:1:19: Warning:
-XDatatypeContexts is deprecated: It was widely considered a misfeature,
and has been removed from the Haskell language.
[1 of 1] Compiling Main ( DataConstraints_so.lhs, interpreted )
Ok, modules loaded: Main.
ghci> :t count
count :: a -> GADTBag a -> Int
ghci> :t count'
count' :: Eq a => a -> EqBag a -> Int
ghci> :t size
size :: GADTBag a -> Int
ghci> :t size'
size' :: Eq a => EqBag a -> Int
ghci>
So our EqBag count' function requires an Eq constraint, which I think is perfectly reasonable, but our size' function also requires one, which is less pretty. This is because the type of the EqBag constructor is EqBag :: Eq a => [a] -> EqBag a, and this constraint must be added every time.
We can't make a functor here either:
instance Functor EqBag where
fmap f (EqBag xs) = EqBag (map f xs)
for exactly the same reason as with the GADTBag
Constraintless bags
data ListBag a = ListBag {unListBag :: [a]}
deriving Show
count'' a = length . filter (==a) . unListBag
size'' = length . unListBag
instance Functor ListBag where
fmap f (ListBag xs) = ListBag (map f xs)
Now the types of count'' and show'' are exactly as we expect, and we can use standard constructor classes like Functor:
ghci> :t count''
count'' :: Eq a => a -> ListBag a -> Int
ghci> :t size''
size'' :: ListBag a -> Int
ghci> fmap (Data.Char.ord) (ListBag "hello")
ListBag {unListBag = [104,101,108,108,111]}
ghci>
Comparison and conclusions
The GADTs version automagically propogates the Eq constraint everywhere the constructor is used. The type checker can rely on there being an Eq instance, because you can't use the constructor for a non-Eq type.
The DatatypeContexts version forces the programmer to manually propogate the Eq constraint, which is fine by me if you want it, but is deprecated because it doesn't give you anything more than the GADT one does and was seen by many as pointless and annoying.
The unconstrained version is good because it doesn't prevent you from making Functor, Monad etc instances. The constraints are written exactly when they're needed, no more or less. Data.Map uses the unconstrained version partly because unconstrained is generally seen as most flexible, but also partly because it predates GADTs by some margin, and there needs to be a compelling reason to potentially break existing code.
What about your excellent User example?
I think that's a great example of a one-purpose data type that benefits from a constraint on the type, and I'd advise you to use a GADT to implement it.
(That said, sometimes I have a one-purpose data type and end up making it unconstrainedly polymorphic just because I love to use Functor (and Applicative), and would rather use fmap than mapBag because I feel it's clearer.)
{-# LANGUAGE GADTs #-}
import Data.String
data User s where
User :: (IsString s, Show s) => s -> User s
name :: User s -> s
name (User s) = s
instance Show (User s) where -- cool, no Show context
showsPrec i (User s) = showParen (i>9) (("User " ++ show s) ++)
instance (IsString s, Show s) => IsString (User s) where
fromString = User . fromString
Notice since fromString does construct a value of type User a, we need the context explicitly. After all, we composed with the constructor User :: (IsString s, Show s) => s -> User s. The User constructor removes the need for an explicit context when we pattern match (destruct), becuase it already enforced the constraint when we used it as a constructor.
We didn't need the Show context in the Show instance because we used (User s) on the left hand side in a pattern match.
Constraints
The problem is that constraints are not a property of the data type, but of the algorithm/function that operates on them. Different functions might need different and unique constraints.
A Box example
As an example, let's assume we want to create a container called Box which contains only 2 values.
data Box a = Box a a
We want it to:
be showable
allow the sorting of the two elements via sort
Does it make sense to apply the constraint of both Ord and Show on the data type? No, because the data type in itself could be only shown or only sorted and therefore the constraints are related to its use, not it's definition.
instance (Show a) => Show (Box a) where
show (Box a b) = concat ["'", show a, ", ", show b, "'"]
instance (Ord a) => Ord (Box a) where
compare (Box a b) (Box c d) =
let ca = compare a c
cb = compare b d
in if ca /= EQ then ca else cb
The Data.Map case
Data.Map's Ord constraints on the type is really needed only when we have > 1 elements in the container. Otherwise the container is usable even without an Ord key. For example, this algorithm:
transf :: Map NonOrd Int -> Map NonOrd Int
transf x =
if Map.null x
then Map.singleton NonOrdA 1
else x
Live demo
works just fine without the Ord constraint and always produce a non empty map.
Using DataTypeContexts reduces the number of programs you can write. If most of those illegal programs are nonsense, you might say it's worth the runtime cost associated with ghc passing in a type class dictionary that isn't used. For example, if we had
data Ord k => MapDTC k a
then #jefffrey's transf is rejected. But we should probably have transf _ = return (NonOrdA, 1) instead.
In some sense the context is documentation that says "every Map must have ordered keys". If you look at all of the functions in Data.Map you'll get a similar conclusion "every useful Map has ordered keys". While you can create maps with unordered keys using
mapKeysMonotonic :: (k1 -> k2) -> Map k1 a -> Map k2 a
singleton :: k2 a -> Map k2 a
But the moment you try to do anything useful with them, you'll wind up with No instance for Ord k2 somewhat later.
here's my question:
this works perfectly:
type Asdf = [Integer]
type ListOfAsdf = [Asdf]
Now I want to do the same but with the Integral class restriction:
type Asdf2 a = (Integral a) => [a]
type ListOfAsdf2 = (Integral a) => [Asdf2 a]
I got this error:
Illegal polymorphic or qualified type: Asdf2 a
Perhaps you intended to use -XImpredicativeTypes
In the type synonym declaration for `ListOfAsdf2'
I have tried a lot of things but I am still not able to create a type with a class restriction as described above.
Thanks in advance!!! =)
Dak
Ranting Against the Anti-Existentionallists
I always dislike the anti-existential type talk in Haskell as I often find existentials useful. For example, in some quick check tests I have code similar to (ironically untested code follows):
data TestOp = forall a. Testable a => T String a
tests :: [TestOp]
tests = [T "propOne:" someProp1
,T "propTwo:" someProp2
]
runTests = mapM runTest tests
runTest (T s a) = putStr s >> quickCheck a
And even in a corner of some production code I found it handy to make a list of types I'd need random values of:
type R a = Gen -> (a,Gen)
data RGen = forall a. (Serialize a, Random a) => RGen (R a)
list = [(b1, str1, random :: RGen (random :: R Type1))
,(b2, str2, random :: RGen (random :: R Type2))
]
Answering Your Question
{-# LANGUAGE ExistentialQuantification #-}
data SomeWrapper = forall a. Integral a => SW a
If you need a context, the easiest way would be to use a data declaration:
data (Integral a) => IntegralData a = ID [a]
type ListOfIntegralData a = [IntegralData a]
*Main> :t [ ID [1234,1234]]
[ID [1234,1234]] :: Integral a => [IntegralData a]
This has the (sole) effect of making sure an Integral context is added to every function that uses the IntegralData data type.
sumID :: Integral a => IntegralData a -> a
sumID (ID xs) = sum xs
The main reason a type synonym isn't working for you is that type synonyms are designed as
just that - something that replaces a type, not a type signature.
But if you want to go existential the best way is with a GADT, because it handles all the quantification issues for you:
{-# LANGUAGE GADTs #-}
data IntegralGADT where
IG :: Integral a => [a] -> IntegralGADT
type ListOfIG = [ IntegralGADT ]
Because this is essentially an existential type, you can mix them up:
*Main> :t [IG [1,1,1::Int], IG [234,234::Integer]]
[IG [1,1,1::Int],IG [234,234::Integer]] :: [ IntegralGADT ]
Which you might find quite handy, depending on your application.
The main advantage of a GADT over a data declaration is that when you pattern match, you implicitly get the Integral context:
showPointZero :: IntegralGADT -> String
showPointZero (IG xs) = show $ (map fromIntegral xs :: [Double])
*Main> showPointZero (IG [1,2,3])
"[1.0,2.0,3.0]"
But existential quantification is sometimes used for the wrong reasons,
(eg wanting to mix all your data up in one list because that's what you're
used to from dynamically typed languages, and you haven't got used to
static typing and its advantages yet).
Here I think it's more trouble than it's worth, unless you need to mix different
Integral types together without converting them. I can't see a reason
why this would help, because you'll have to convert them when you use them.
For example, you can't define
unIG (IG xs) = xs
because it doesn't even type check. Rule of thumb: you can't do stuff that mentions the type a on the right hand side.
However, this is OK because we convert the type a:
unIG :: Num b => IntegralGADT -> [b]
unIG (IG xs) = map fromIntegral xs
Here existential quantification has forced you convert your data when I think your original plan was to not have to!
You may as well convert everything to Integer instead of this.
If you want things simple, keep them simple. The data declaration is the simplest way of ensuring you don't put data in your data type unless it's already a member of some type class.
I'm new to Haskell and, in general, to functional programming, and I'm a bit uncomfortable with its syntax.
In the following code what does the => denote? And also (Num a, Ord a)?
loop :: (Num a, Ord a) => a -> (t -> t) -> t -> t
This is a typeclass constraint; (Num a, Ord a) => ... means that loop works with any type a that is an instance of the Num and Ord typeclasses, corresponding to numeric types and ordered types respectively. Basically, you can think of loop as having the type on the right hand side of the =>, except that a is required to be an instance of Num and Ord.
You can think of typeclasses as basically similar to OOP interfaces (but they're not the same thing!) — they encapsulate a set of definitions which any instance must support, and generic code can be written using these definitions. For instance, Num includes numeric operations like addition and multiplication, while Ord includes less than, greater than, and so on.
For more information on typeclasses, see this introduction from Learn You a Haskell.
=> separates two parts of a type signature:
On the left, typeclass constraints
On the right, the actual type
So you can think of (Num a, Ord a) => a -> (t -> t) -> t -> t as meaning "the type is a -> (t -> t) -> t -> t and also there must be a Num instance for a and an Ord instance for a".
For more on typeclasses see http://www.learnyouahaskell.com/types-and-typeclasses
One way to think about it is that Ord a and Num a are additional inputs to the function. They are a special kind of input though: dictionaries. When you use this function with a particular type a, there must also be dictionaries available for the Ord and Num operations on the type a as well.
Any function that makes use of a function with dictionary inputs must also have the same dictionary inputs.
foo :: (Num a, Ord a) => a -> t
foo x = loop x someFunc someT
However, you do not have to explicitly pass these dictionaries around. Haskell will take care of that for you, assuming there is a dictionary available. You can create a dictionary with a typeclass instance.
instance Num MyType with
x + y = ...
x - y = ...
...
This creates a dictionary for the Num operations on MyType, therefore MyType can be used anywhere that Num a is a required input (assuming it satisfies the other requirements, of course).
On the left hand side of the => you declare constraints for the types that are used on the right.
In the example you give, it means that a is constrained to being an instance of both the Ord type class and the Num type class.