In Haskell, when we talk type declaration.
I've seen both -> and =>.
As an example: I can make my own type declaration.
addMe :: Int -> Int -> Int
addMe x y = x + y
And it works just fine.
But if we take a look at :t sqrt we get:
sqrt :: Floating a => a -> a
At what point do we use => and when do we use ->?
When do we use "fat arrow" and when do we use "thin arrow"?
-> is for explicit functions. I.e. when f is something that can be written in an expression of the form f x, the signature must have one of these arrows in it†. Specifically, the type of x (the argument) must appear to the left of a -> arrow.
It's best to not think of => as a function arrow at all, at least at first‡. It's an implication arrow in the logical sense: if a is a type with the property Floating a, then it follows that the signature of sqrt is a -> a.
For your addMe example, which is a function with two arguments, the signature must always have the form x -> y -> z. Possibly there can also be a q => in front of that; that doesn't influence the function-ishness, but may have some saying in what particular types are allowed. Generally, such constraints are not needed if the types are already fixed and concrete. Like, you could in principle impose a constraint on Int:
addMe :: Num Int => Int -> Int -> Int
addMe x y = x + y
...but that doesn't really accomplish anything, because everybody knows that the particular type Int is an instance of the Num class. Where you need such constraints is when the type is not fixed but a type variable (i.e. lowercase), i.e. if the function is polymorphic. You can't just write
addMe' :: a -> a -> a
addMe' x y = x + y
because that signature would suggest the function works for any type a whatsoever, but it can't work for all types (how would you add, for example, two strings? ok perhaps not the best example, but how would you multiply two strings?)
Hence you need the constraint
addMe' :: Num a => a -> a -> a
addMe' x y = x + y
This means, you don't care what exact type a is, but you do require it to be a numerical type. Anybody can use the function with their own type MyNumType, but they need to ensure that Num MyNumType is fulfilled: then it follows that addMe' can have signature MyNumType -> MyNumType -> MyNumType.
The way to ensure this is to either use a standard type which you know to be numerical, for instance addMe' 5.9 3.7 :: Double would work, or give an instance declaration for your custom type and the Num class. Only do the latter if you're sure it's a good idea; usually the standard num types are all you'll need.
†Note that the arrow may not be visible in the signature: it's possible to have a type synonym for a function type, for example when type IntEndofunc = Int -> Int, then f :: IntEndofunc; f x = x+x is ok. But you can think of the typedef as essentially just a syntactic wrapper; it's still the same type and does have the arrow in it.
‡It so happens that logical implication and function application can be seen as two aspects of the same mathematical concept. Furthermore, GHC actually implements class constraints as function arguments, so-called dictionaries. But all this happens behind the scenes, so if anything they're implicit functions. In standard Haskell, you will never see the LHS of a => type as the type of some actual argument the function is applied to.
The "thin arrow" is used for function types (t1 -> t2 being the type of a function that takes a value of type t1 and produces a value of type t2).
The "fat arrow" is used for type constraints. It separates the list of type constraints on a polymorphic function from the rest of the type. So given Floating a => a -> a, we have the function type a -> a, the type of a function that can take arguments of any type a and produces a result of that same type, with the added constraint Floating a, meaning that the function can in fact only be used with types that implement the Floating type class.
the -> is the constructor of functions and the => is used to constraints, a sort of "interface" in Haskell called typeclass.
A little example:
sum :: Int -> Int -> Int
sum x y = x + y
that function only allows Int types, but if you want a huge int or a small int, you probably want Integer, and how to tell it to use both?
sum2 :: Integral a => a -> a -> a
sum2 x y = x + y
now if you try to do:
sum2 3 1.5
it will give you an error
also, you may want to know if two data are equals, you want:
equals :: Eq a => a -> a -> Bool
equals x y = x == y
now if you do:
3 == 4
that's ok
but if you create:
data T = A | B
equals A B
it will give to you:
error:
• No instance for (Eq T) arising from a use of ‘equals’
• In the expression: equals A B
In an equation for ‘it’: it = equals A B
if you want for that to work, you must just do:
data T = A | B deriving Eq
equals A B
False
Related
https://youtu.be/brE_dyedGm0?t=1362
data T a where
T1 :: Bool -> T Bool
T2 :: T a
f x y = case x of
T1 x -> True
T2 -> y
Simon is saying that f could be typed as T a -> a -> a, but I would think the return value MUST be a Bool since that is an explicit result in a branch of the case expression. This is in regards to Haskell GADTs. Why is this the case?
This is kind of the whole point of GADTs. Matching on a constructor can cause additional type information to come into scope.
Let's look at what happens when GHC checks the following definition:
f :: T a -> a -> a
f x y = case x of
T1 _ -> True
T2 -> y
Let's look at the T2 case first, just to get it out of the way. y and the result have the same type, and T2 is polymorphic, so you can declare its type is also T a. All good.
Then comes the trickier case. Note that I removed the binding of the name x inside the case as the shadowing might be confusing, the inner value isn't used, and it doesn't change anything in the explanation. When you match on a GADT constructor like T1, one that explicitly sets a type variable, it introduces additional constraints inside that branch which add that type information. In this case, matching on T1 introduces a (a ~ Bool) constraint. This type equality says that a and Bool match each other. Therefore the literal True with type Bool matches the a written in the type signature. y isn't used, so the branch is consistent with T a -> a -> a.
So it all matches up. T a -> a -> a is a valid type for that definition. But as Simon is saying, it's ambiguous. T a -> Bool -> Bool is also a valid type for that definition. Neither one is more general than the other, so the definition doesn't have a principle type. So the definition is rejected unless a type is provided, because inference cannot pick a single most-correct type for it.
A value of type T a with a different from Bool can never have the form T1 x (since that has only type T Bool).
Hence, in such case, the T1 x branch in the case becomes inaccessible and can be ignored during type checking/inference.
More concretely: GADTs allow the type checker to assume type-level equations during pattern matching, and exploit such equations later on. When checking
f :: T a -> a -> a
f x y = case x of
T1 x -> True
T2 -> y
the type checker performs the following reasoning:
f :: T a -> a -> a
f x y = case x of
T1 x -> -- assume: a ~ Bool
True -- has type Bool, hence it has also type a
T2 -> -- assume: a~a (pointless)
y -- has type a
Thanks to GADTs, both branches of the case have type a, hence the whole case expression has type a and the function definition type checks.
More generally, when x :: T A and the GADT constructor was defined as K :: ... -> T B then, when type checking we can make the following assumption:
case x of
K ... -> -- assume: A ~ B
Note that A and B can be types involving type variables (as in a~Bool above), so that allows one obtain useful information about them and exploit it later on.
I'm learning Haskell. This is the very first program that I've ever wrote and I just thought of making a simple function that returns the product of the given arguments.
mult :: a -> a -> a
mult x y = x * y
When I write it this way, I get an "inferred type" error.
ERROR "uno.hs":5 - Inferred type is not general enough
*** Expression : mult
*** Expected type : a -> a -> a
*** Inferred type : Integer -> Integer -> Integer
It wants me to write it this way:
mult :: Int -> Int -> Int
mult x y = x * y
...which works perfectly fine, but then I wonder... can't this function work on Floats as well?
It can.
mult :: Float -> Float -> Float
mult x y = x * y
--then
mult 4 5 == 20.0
Finally, I wonder:
Can I make a polymorphic mult function, that takes any numeric type?
Why does the parser infers Integrer from a function that also works with other types?
I found no answers so far, yet this seems like a simple question. I'll be grateful for help me.
Short answer: mult :: Num a => a -> a -> a and it picked Integer because it felt like it was a good choice.
Long answer:
mult can't be a -> a -> a because that would mean it works on every type a, including String, Char, Bool, etc. The Num a => part is called a constraint, and it restricts what a can be by requiring that it is an instance of the Num type class. It's like saying "this function works on any type, as long as that type has the properties (has an instance) of this specific type class". The properties for the Num type class are defined in the standard library, linked below.
It's worth noting that the type of (*) itself is Num a => a -> a -> a, and that your function mult could also have been defined as mult = (*).
The reason it told you to use Integer -> Integer -> Integer is because the default for Num is Integer. You should never rely on this being the case, and should instead write out the full, type class constrained type signature.
When you specified the type signature Float -> Float -> Float, the compiler specializes the function to only work on Floats, as you would expect. This resolves the "over-promising" problem arising from a -> a -> a by choosing a specific type for a that the compiler knows has a Num instance.
Read more about type classes here, and the specific definition of Num is documented here and the code is linked on the side of the page. (Not enough rep to post the link directly)
In Haskell, I just know that
:type ((+)(1))
((+)(1)) :: Num a => a -> a
((+)(1） 2
3
But how about
:type abs(sqrt)
abs(sqrt) :: (Floating a, Num (a -> a)) => a -> a
Actually, I try many times but fail to use the function 'abs(sqrt)'. Then I have a few questions. What is the type(class?) '(Floating a, Num (a -> a))'? Is it possible to use the function 'abs(sqrt)'? How?
A type class is a way to generalize functions so that they can be polymorphic and others can implement those functions for their own types. Take as an example the type class Show, which in a simplified form looks like
class Show a where
show :: a -> String
This says that any type that implements the Show typeclass can be converted to a String (there's some more complication for more realistic constraints, but the point of having Show is to be able to convert values to Strings).
In this case, the function show has the full type Show a => a -> String.
If we examine the function sqrt, its type is
> :type sqrt
sqrt :: Floating a => a -> a
And for abs:
> :type abs
abs :: Num b => b -> b
If you ask GHCi what the types are it will use the type variable a in both cases, but I've used b in the type signature for abs to make it clear that these are different type variables of the same name, and it will help avoid confusion in the next step.
These type signatures mean that sqrt takes a value whose type implements the Floating typeclass (use :info Floating to see all the members) and returns a value of that same type, and that the abs function takes a value whose type implements the Num typeclass and returns a value of that same type.
The expression abs(show) is equivalently parsed as abs sqrt, meaning that sqrt is the first and only argument passed to abs. However, we just said that abs takes a value of a Num type, but sqrt is a function, not a number. Why does Haskell accept this instead of complaining? The reason can be seen a little more clearly when we perform substitution with the type signatures. Since sqrt's type is Floating a => a -> a, this must match the argument b in abs's type signature, so by substituting b with Floating a => a -> a we get that abs sqrt :: (Floating a, Num (a -> a)) => a -> a.
Haskell actually allows the function type to implement the Num typeclass, you could do it yourself although it would likely be nonsensical. However, just because something wouldn't seem to make sense to GHC, so long as the types can be cleanly solved it will allow it.
You can't really use this function, it just doesn't really make sense. There is no built-in instance of Num (a -> a) for any a, so you'd have to define your own. You can, however, compose the functions abs and sqrt using the composition operator .:
> :type abs . sqrt
abs . sqrt :: Floating c => c -> c
And this does make sense. This function is equivalent to
myfunc x = abs (sqrt x)
Note here that x is first applied to sqrt, and then the result of that computation is passed to abs, rather than passing the function sqrt to abs.
When you see Num (a -> a) it generally means you made a mistake somewhere.
Perhaps you really wanted: abs . sqrt which has type Floating c => c -> c - i.e. it's a function of a Floating type (e.g. Float, Double) to the same Floating type.
It is probably not possible to use this function.
What's likely happening here is that the type is saying that abs(sqrt) has the constraints that a must be of type class Floating and (a -> a) must be of type class Num. In other words, the sqrt function needs to be able to be treated as if it was a number.
Unfortunately, sqrt is not of type class Num so there won't be any input that will work here (not that it would make sense anyway). However, some versions of GHCi allow you to get the type of as if it were possible.
Have a look at Haskell type length + 1 for a similar type problem.
As ErikR has said, perhaps you meant to write abs . sqrt instead.
I'm just starting with Haskell, and I thought I'd start by making a random image generator. I looked around a bit and found JuicyPixels, which offers a neat function called generateImage. The example that they give doesn't seem to work out of the box.
Their example:
imageCreator :: String -> IO ()
imageCreator path = writePng path $ generateImage pixelRenderer 250 300
where pixelRenderer x y = PixelRGB8 x y 128
when I try this, I get that generateImage expects an Int -> Int -> PixelRGB8 whereas pixelRenderer is of type Pixel8 -> Pixel8 -> PixelRGB8. PixelRGB8 is of type Pixel8 -> Pixel8 -> Pixel8 -> PixelRGB8, so it makes sense that pixelRenderer is doing some type inference to determine that x and y are of type Pixel8. If I define a type signature that asserts that they are of type Int (so the function gets accepted by generateImage, PixelRGB8 complains that it needs Pixel8s, not Ints.
Pixel8 is just a type alias for Word8. After some hair pulling, I discovered that the way to convert an Int to a Word8 is by using fromIntegral.
The type signature for fromIntegral is (Integral a, Num b) => a -> b. It seems to me that the function doesn't actually know what you want to convert it to, so it converts to the very generic Num class. So theoretically, the output of this is a variable of any type that fits the type class Num (correct me if I'm mistaken here--as I understand it, classes are kind of like "interfaces" where types are more like classes/primitives in OOP). If I assign a variable
let n = fromIntegral 5
:t n -- n :: Num b => b
So I'm wondering... what is 'b'? I can use this variable as anything, and it will implicitly cast to any numeric type, as it seems. Not only will it implicitly cast to a Word8, it will implicitly cast to a Pixel8, meaning fromPixel effectively gets turned from (as I understood it) (Integral a, Num b) => a -> b to (Integral a) => a -> Pixel8 depending on context.
Can someone please clarify exactly what's happening here? Why can I use a generic Num as any type that fits Num, both mechanically and "ethically"? I don't understand how the implicit conversion is implemented (if I were to create my own class, I feel like I would need to add explicit conversion functions). I also don't really know why this works; here I can use a pretty unsafe type and convert it implicitly to anything else. (for example, fromIntegral 50000 gets translated to 80 if I implicitly convert it to a Word8)
A common implementation of type classes such as Num is dictionary-passing. Roughly, when the compiler sees something like
f :: Num a => a -> a
f x = x + 2
it transforms it into something like
f :: (Integer -> a, a -> a -> a) -> a -> a
-- ^-- the "dictionary"
f (dictFromInteger, dictPlus) x = dictPlus x (dictFromInteger 2)
The latter basically says: "pass me an implementation for these methods of class Num for your type a, and I will use them to produce a function a -> a for you".
Values such as your n :: Num b => b are no different. They are compiled into things such as
n :: (Integer -> b) -> b
n dictFromInteger = dictFromInteger 5 -- roughly
As you can see, this turns innocent-looking integer literals into functions, which can (and does) impact performance. However, in many circumstances the compiler can realize that the full polymorphic version is not actually needed, and remove all the dictionaries.
For instance, if you write f 3 but f expects Int, the "polymorphic" 3 can be converted at compile time. So type inference can aid the optimization phase (and user-written type annotation can greatly help here). Further, some other optimizations can be triggered manually, e.g. using the GHC SPECIALIZE pragma. Finally, the dreaded monomorphism restriction tries hard to force non-functions to remain non-functions after translation, at the cost of some loss of polymorphism. However, the MR is now being regarded as harmful, since it can cause puzzling type errors in some contexts.
Still new to Haskell, I have hit a wall with the following:
I am trying to define some type classes to generalize a bunch of functions that use gaussian elimination to solve linear systems of equations.
Given a linear system
M x = k
the type a of the elements m(i,j) \elem M can be different from the type b of x and k. To be able to solve the system, a should be an instance of Num and b should have multiplication/addition operators with b, like in the following:
class MixedRing b where
(.+.) :: b -> b -> b
(.*.) :: (Num a) => b -> a -> b
(./.) :: (Num a) => b -> a -> b
Now, even in the most trivial implementation of these operators, I'll get Could not deduce a ~ Int. a is a rigid type variable errors (Let's forget about ./. which requires Fractional)
data Wrap = W { get :: Int }
instance MixedRing Wrap where
(.+.) w1 w2 = W $ (get w1) + (get w2)
(.*.) w s = W $ ((get w) * s)
I have read several tutorials on type classes but I can find no pointer to what actually goes wrong.
Let us have a look at the type of the implementation that you would have to provide for (.*.) to make Wrap an instance of MixedRing. Substituting Wrap for b in the type of the method yields
(.*.) :: Num a => Wrap -> a -> Wrap
As Wrap is isomorphic to Int and to not have to think about wrapping and unwrapping with Wrap and get, let us reduce our goal to finding an implementation of
(.*.) :: Num a => Int -> a -> Int
(You see that this doesn't make the challenge any easier or harder, don't you?)
Now, observe that such an implementation will need to be able to operate on all types a that happen to be in the type class Num. (This is what a type variable in such a type denotes: universal quantification.) Note: this is not the same (actually, it's the opposite) of saying that your implementation can itself choose what a to operate on); yet that is what you seem to suggest in your question: that your implementation should be allowed to pick Int as a choice for a.
Now, as you want to implement this particular (.*.) in terms of the (*) for values of type Int, we need something of the form
n .*. s = n * f s
with
f :: Num a => a -> Int
I cannot think of a function that converts from an arbitary Num-type a to Int in a meaningful way. I'd therefore say that there is no meaningful way to make Int (and, hence, Wrap) an instance of MixedRing; that is, not such that the instance behaves as you would probably expect it to do.
How about something like:
class (Num a) => MixedRing a b where
(.+.) :: b -> b -> b
(.*.) :: b -> a -> b
(./.) :: b -> a -> b
You'll need the MultiParamTypeClasses extension.
By the way, it seems to me that the mathematical structure you're trying to model is really module, not a ring. With the type variables given above, one says that b is an a-module.
Your implementation is not polymorphic enough.
The rule is, if you write a in the class definition, you can't use a concrete type in the instance. Because the instance must conform to the class and the class promised to accept any a that is Num.
To put it differently: Exactly the class variable is it that must be instantiated with a concrete type in an instance definition.
Have you tried:
data Wrap a = W { get :: a }
Note that once Wrap a is an instance, you can still use it with functions that accept only Wrap Int.