Could someone please explain why this compiles
Prelude> 1 :: Num a => a
and this doesn't
Prelude> 1.0 :: Num a => a
Second example would work with Fractional, but Num is superclass of Fractional. Just like it's superclass of Integral.
If we have
x :: Num a => a
the user of x can pick a as wanted. E.g.
x :: Int
What would this evaluate to if x = 1.5 ?
For this reason a floating point literal can't be given the polytype Num a => a, since its value won't (in general) fit all Numeric types.
Integral literals instead fit every numeric type, so they are allowed.
The superclass relationship between type classes does not establish relationships between types. It tells us only that if type t is a member of type class C, and B is a super class of C, then t will also be member of B.
It doesn't say that each value v::t can be used at any type that is also member of B. But this is what you are stating with:
3.14159 :: Num a => a
It's important to distinguish between polymorphism in OO languages and polymorphism in Haskell. OO polymorphism is covariant, while Haskell's parametric polymorphism is contravariant.
What this means is: in an OO language, if you have
class A {...}
class B: A {...}
i.e. A is a superclass of B, then any value of type B is also a value of type A. (Note that any particular value is actually not polymorphic but has a concrete type!) Thus, if you had
class Num {...}
class Fractional: Num {...}
then a Fractional value could indeed be used as a Num value. That's roughly what covariant means: any subclass value is also a superclass value; the values hierarchy goes the same direction as the type hierarchy.
In Haskell, classes are different. There is no such thing as a “value of type Num”, only values of concrete types a. That type may be in the Num class.
Unlike in OO languages, a value like 1 :: Num a => a is polymorphic: it can take on whatever type the environment demands, provided the type is in the Num class. (Actually that syntax is just shorthand for 1 :: ∀ a . Num a => a, to be read as “for all types a, you can have a value 1 of type a.) For example,
Prelude> let x = 1 :: Num a => a
Prelude> x :: Int
1
Prelude> x :: Double
1.0
You can also give x a more specific constraint of Fractional, since that's a subclass of Num. That just restricts what type the polymorphic value can be instantiated to:
Prelude> let x = 1 :: Fractional a => a
Prelude> x :: Int
<interactive>:6:1:
No instance for (Fractional Int) arising from a use of ‘x’
...
Prelude> x :: Double
1.0
because Int is not a fractional type.
Thus, Haskell's polymorphism is contravariant: polymorphic values restricted to a superclass can also be restricted to a subclass instead, but not the other way around. In particular, you can obviously have
Prelude> let y = 1.0 :: Fractional a => a
(y is the same as x'), but you can not generalise this to y' = 1.0 :: Num a => a. Which is a good thing as Ingo remarked since otherwise it would be possible to do
Prelude> 3.14159 :: Int
????
Related
I am confused with classes in Haskell as follows.
I can define a function that takes an Integral argument, and successfully supply it with a Num argument:
gi :: Integral a => a -> a
gi i = i
gin = gi (3 :: Num a => a)
I can define a function that takes a Num argument, and successfully supply it with an Integral argument:
fn :: Num a => a -> a
fn n = n
fni = fn (3 :: Integral a => a)
I can define an Integral value and assign a Num to it
i :: Integral a => a
i = (3 :: Num a => a)
But if I try to define a Num value, then I get a parse error if I assign an Integral value to it
- this doesn't work
n :: Num a => a
n = (3 :: Integral a => a)
Maybe I am being confused by my OO background. But why do function variables appear to let you go 'both ways' i.e. can provide a value of a subclass when a superclass is 'expected' and can provide a value of a superclass when a subclass is expected, whereas in value assignment you can provide a superclass to a subclass value but can't assign a subclass to a superclass value?
For comparison, in OO programming you can typically assign a child value to a parent type, but not vice-versa. In Haskell, the opposite appears to be the case in the second pair of examples.
The first two examples don't actually have anything to do with the relationship between Num and Integral.
Take a look at the type of gin and fni. Let's do it together:
> :t gin
gin :: Integer
> :t fni
fni :: Integer
What's going on? This is called "type defaulting".
Technically speaking, any numeric literal like 3 or 5 or 42 in Haskell has type Num a => a. So if you wanted it to just be an integer number dammit, you'd have to always write 42 :: Integer instead of just 42. This is mighty inconvenient.
So to work around that, Haskell has certain rules that in certain special cases prescribe concrete types to be substituted when the type comes out generic. And in case of both Num and Integral the default type is Integer.
So when the compiler sees 3, and it's used as a parameter for gi, the compiler defaults to Integer. That's it. Your additional constraint of Num a has no further effect, because Integer is, in fact, already an instance of Num.
With the last two examples, on the other hand, the difference is that you explicitly specified the type signature. You didn't just leave it to the compiler to decide, no! You specifically said that n :: Num a => a. So the compiler can't decide that n :: Integer anymore. It has to be generic.
And since it's generic, and constrained to be Num, an Integral type doesn't work, because, as you have correctly noted, Num is not a subclass of Integral.
You can verify this by giving fni a type signature:
-- no longer works
fni :: Num a => a
fni = fn (3 :: Integral a => a)
Wait, but shouldn't n still work? After all, in OO this would work just fine. Take C#:
class Num {}
class Integral : Num {}
class Integer : Integral {}
Num a = (Integer)3
// ^ this is valid (modulo pseudocode), because `Integer` is a subclass of `Num`
Ah, but this is not a generic type! In the above example, a is a value of a concrete type Num, whereas in your Haskell code a is itself a type, but constrained to be Num. This is more like a C# interface than a C# class.
And generic types (whether in Haskell or not) actually work the other way around! Take a value like this:
x :: a
x = ...
What this type signature says is that "Whoever has a need of x, come and take it! But first name a type a. Then the value x will be of that type. Whichever type you name, that's what x will be"
Or, in plainer terms, it's the caller of a function (or consumer of a value) that chooses generic types, not the implementer.
And so, if you say that n :: Num a => a, it means that value n must be able to "morph" into any type a whatsoever, as long as that type has a Num instance. Whoever will use n in their computation - that person will choose what a is. You, the implementer of n, don't get to choose that.
And since you don't get to choose what a is, you don't get to narrow it down to be not just any Num, but an Integral. Because, you know, there are some Nums that are not Integrals, and so what are you going to do if whoever uses n chooses one of those non-Integral types to be a?
In case of i this works fine, because every Integral must also be Num, and so whatever the consumer of i chooses for a, you know for sure that it's going to be Num.
I am going through the Haskell Book (http://haskellbook.com/) and got stuck at the following exercise:
f :: Float
f = 1.0
-- Question: Can you replace `f`'s signature by `f :: Num a => a`
At first, I thought the answer would be Yes. Float provides an instance for Num, so substituting a Num a => a by a Float value should be fine (I am thinking co-variance here).
This does not compile however:
Could not deduce (Fractional a) arising from the literal ‘1.0’
from the context: Num a
bound by the type signature for:
f :: forall a. Num a => a
at ...
Possible fix:
add (Fractional a) to the context of
the type signature for:
f :: forall a. Num a => a
• In the expression: 1.0
In an equation for ‘f’: f = 1.0
If I do this however, no problem:
f :: Fractional a => a
f = 1.0
Why can't I use a less specific constraint here like Num a => a?
UPD:
Actually, you can sum this up to:
1.0::(Num a => a)
vs
1.0::(Fractional a => a)
Why is the second working but not the first one? I thought Fractional was a subset of Num (meaning Fractional is compatible with Num)
UPD 2:
Thanks for your comments, but I am still confused. Why this works:
f :: Num a => a -> a
f a = a
f 1.0
while this not:
f :: Num a => a
f = 1.0
UPD 3:
I have just noticed something:
f :: Num a => a
f = (1::Int)
does not work either.
UPD 4
I have been reading all the answers/comments and as far as I understand:
f :: Num a => a
is the Scala equivalent of
def f[A: Num]: A
which would explain why many mentioned that a is defined by the caller. The only reason why we can write this:
f :: Num a => a
f = 1
is because 1 is typed as a Num a => a. Could someone please confirm this assumption? In any case, thank you all for your help.
If I have f :: Num a => a this means that I can use f wherever I need any numeric type. So, all of f :: Int, f :: Double must type check.
In your case, we can't have 1.0 :: Int for the same reason we can't have 543.24 :: Int, that is, Int does not represent fractional values. However, 1.0 does fit any fractional type (as 543.24 does).
Fractional indeed can be thought as a subset of Num. But if I have a value inside all the fractionals f :: forall a . Fractional a => a, I don't necessarily have a value in all the numeric types f :: forall a . Num a => a.
Note that, in a sense, the constraints are on the left side of =>, which makes them behave contra-variantly. I.e. cars are a subset of vehichles, but I can't conclude that a wheel that can be used in any car will be able to be used with any vehicle. Rather, the opposite: a wheel that can be used in any vehicle will be able to be used with any car.
So, you can roughly regard f :: forall a . Num a => a (values fitting in any numeric type, like 3 and 5) as a subtype of f :: forall a . Fractional a => a (values fitting in any fractional type, like 3,5, and 32.43).
Let's start with the monomorphic case:
f :: Float
f = 1.0
Here, you've said that f is a Float; not an Int, not a Double, not any other type. 1.0, on the other hand, is a polymorphic constant; it has type Fractional a => a, and so can be used to provide a value of any type that has a Fractional instance. You can restrict it to Float, or Double, etc. Since f has to be a Float, that's what 1.0 is restricted to.
If you try to change the signature
f :: Num a => a
f = 1.0
now you have a problem. You've now promised that f can be used to provide a value of any type that has a Num instance. That includes Int and Integer, but neither of those types has a Fractional instance. So, the compiler refuses to let you do this. 1.0 simply can't produce an Int should the need arise.
Likewise,
1.0::(Num a => a)
is a lie, because this isn't a restriction; it's an attempt at expanding the set of types that 1.0 can produce. Something with type Num a => a should be able to give me an Int, but 1.0 cannot do that.
1.0::(Fractional a => a)
works because you are just restating something that is already true. It's neither restricting 1.0 to a smaller set of types or trying to expand it.
Now we get something a little more interesting, because you are specifying a function type, not just a value type.
f :: Num a => a -> a
f a = a
This just says that f can take as its argument any value that is no more polymorphic than Num a => a. a can be any type that implements Num, or it can be a polymorphic value that is a subset of the types that represent Num.
You chose
f 1.0
which means a gets unified with Fractional a => a. Type inference then decides that the return type is also Fractional a => a, and returning the same value you passed in is allowed.
We've already covered why this
f :: Num a => a
f = 1.0
isn't allowed above, and
f :: Num a => a
f = (1::Int)
fails for the same reason. Int is simply too specific; it is not the same as Num a => a.
For example, (+) :: Num a => a -> a -> a requires two arguments of the same type. So, I might try to write
1 :: Integer + f
or
1 :: Float + f
. In both cases, I need f to be able to provide a value with the same type as the other argument, to satisfy the type of (+): if I want an Integer, I should be able to get an Integer, and if I want a Float, I should be able to get a Float. But if you could specify a value with something less specific than Num a => a, you wouldn't be able to keep that promise. 1.0 :: Fractional a => a can't provide an Integer, and 1 :: Int can't provide anything except an Int, not even an Integer.
Think of a polymorphic type as a function from a concrete type to a value; you can literally do this if you enable the TypeApplications extension.
Prelude> :set -XTypeApplications
Prelude> :{
Prelude| f :: Num a => a
Prelude| f = 1
Prelude| :}
Prelude> :t f
f :: Num a => a
Prelude> :t f #Int
f #Int :: Int
Prelude> f #Int
1
Prelude> :t f #Float
f #Float :: Float
Prelude> f #Float
1.0
Prelude> :t f #Rational
f #Rational :: Rational
Prelude> f #Rational
1 % 1
The reason these all work is because you promised that you could pass any type with a Num instance to f, and it could return a value of that type. But if you had been allowed to say f = 1.0, there is no way that f #Int could, in fact, return an Int, because 1.0 simply is not capable of producing an Int.
When values with polymorphic types like Num a => a are involved, there are two sides. The source of the value, and the use of the value. One side gets the flexibility to choose any specific type compatible with the polymorphic type (like using Float for Num a => a). The other side is restricted to using code that will work regardless of which specific type is involved - it can only make use of features that will work for every type compatible with the polymorphic type.
There is no free lunch; both sides cannot have the same freedom to pick any compatible type they like.
This is just as true for object-oriented subclass polymorphism, however OO polymorphism rules give the flexibility to the source side, and put all the restrictions on the use side (except for generics, which works like parametric polymorphism), while Haskell's parametric polymorphism gives the flexibility to the use side, and puts all the restrictions on the source side.
For example in an OO language with a general number class Num, and Int and Double as subclasses of this, then a method returning something of type Num would work the way you're expecting. You can return 1.0 :: Double, but the caller can't use any methods on the value that are provided specifically by Double (like say one that splits off the fractional part) because the caller must be programmed to work the same whether you return an Int or a Double (or even any brand new subclass of Num that is private to your code, that the caller cannot possibly know about).
Polymorphism in Haskell is based on type parameters rather than subtyping, which switches things around. The place in the code where f :: Num a => a is used has the freedom to demand any particular choice for a (subject to the Num constraint), and the code for f that is the source of the value must be programmed to work regardless of the use-site's choice. The use-site is even free to demand values of a type that is private to the code using f, that the implementer of f cannot possibly know about. (I can literally open up a new file, make any bizarre new type I like and give it an instance for Num, and any of the standard library functions written years ago that are polymorphic in Num will work with my type)
So this works:
f :: Float
f = 1.0
Because there are no type variables, so both source and use-site simply have to treat this as a Float. But this does not:
f :: Num a => a
f = 1.0
Because the place where f is used can demand any valid choice for a, and this code must be able to work for that choice. (It would not work when Int is chosen, for example, so the compiler must reject this definition of f). But this does work:
f :: Fractional a => a
f = 1.0
Because now the use site is only free to demand any type that is in Fractional, which excludes the ones (like Int) that floating point literals like 1.0 cannot support.
Note that this is exactly how "generics" work in object oriented languages, so if you're familiar with any language supporting generics, just treat Haskell types the way you do generic ones in OO languages.
One further thing that may be confusing you is that in this:
f :: Float
f = 1.0
The literal 1.0 isn't actually definitively a Float. Haskell literals are much more flexible than those of most other languages. Whereas e.g. Java says that 1.0 is definitely a value of type double (with some automatic conversion rules if you use a double where certain other types are expected), in Haskell that 1.0 is actually itself a thing with a polymorphic type. 1.0 has type Fractional a => a.
So the reason the f :: Fractional a => a definition worked is obvious, it's actually the f :: Float definition that needs some explanation. It's making use of exactly the rules I described above in the first section of my post. Your code f = 1.0 is a use-site of the value represented by 1.0, so it can demand any type it likes (subject to Fractional). In particular, it can demand that the literal 1.0 supply a value of type Float.
This again reinforces why the f :: Num a => a definition can't work. Here the type for f is promising to f's callers that they can demand any type they like (subject to Num). But it's going to fulfill that demand by just passing the demand down the chain to the literal 1.0, which has the most general type of Fractional a => a. So if the use-site of f demands a type that is in Num but outside Fractional, f would then try to demand that 1.0 supply that same non-Fractional type, which it can't.
Names do not have types. Values have types, and the type is an intrinsic part of the value. 1 :: Int and 1 :: Integer are two different values. Haskell does not implicitly convert values between types, though it can be trivial to define functions that take values of one type and return values of another. (For example, f :: Int -> Integer with f x = x will "convert" its Int arugment to an Integer.
A declaration like
f :: Num a => a
does not say that f has type Num a => a, it says that you can assign values of type Num a => a to f.
You can think of a polymorphic value like 1 :: Num a => a being all 1-like values in every type with a Num instance, including 1 :: Int, 1 :: Integer, 1 :: Rational, etc.
An assignment like f = 1 succeeds because the literal 1 has the expected type Num a => a.
An assignment like f = 1.0 fails because the literal 1.0 has a different type, Fractional a => a, and that type is too specific. It does not include all 1-like values that Num a => a may be called on to produce.
Suppose you declared g :: Fractional a => a. You can say g = 1.0, because the types match. You cannot say g = (1.0 :: Float), because the types do not match; Float has a Fractional instance, but it is just one of a possibly infinite set of types that could have Fractional instances.
You can say g = 1, because Fractional a => a is more specific than Num a => a, and has Fractional has Num as its superclass. The assignment "selects" the subset of 1 :: Num a => a that overlaps with (and for all intents and purposes is) 1 :: Fractional a => a and assigns that to g. Put another way, just 1 :: Num a => a can produce a value for any single type that has a Num instance, it can produce a value for any subset of types implied by a subclass of Num.
I am puzzled by this as well.
What gets stuck in my mind is something like:
// in pseudo-typescript
type Num = Int | Float;
let f: Num;
f = (1.0 as Float); // why this doesn't work
Fact is, Num a => a is just not a simple sum of numeric types.
It represents something that can morph into various kind of numeric types.
Thanks to chepner's explanation, now I can persuade myself like this:
if I have a Num a => a then I can get a Int from it, I can also get a Float from it, as well as a Double....
If I am able to install 1.1 into a Num a => a, then there is no way I can safely derive a Int from 1.1.
The expression 1 is able to bind to Num a => a is due to the fact that 1 itself is a polymorphic constant with type signature Num a => a.
I tested the numeric coercion by using GHCI:
>> let c = 1 :: Integer
>> 1 / 2
0.5
>> c / 2
<interactive>:15:1: error:
• No instance for (Fractional Integer) arising from a use of ‘/’
• In the expression: c / 2
In an equation for ‘it’: it = c / 2
>> :t (/)
(/) :: Fractional a => a -> a -> a -- (/) needs Fractional type
>> (fromInteger c) / 2
0.5
>>:t fromInteger
fromInteger :: Num a => Integer -> a -- Just convert the Integer to Num not to Fractional
I can use fromInteger function to convert a Integer type to Num (fromInteger has the type fromInteger :: Num a => Integer -> a), but I cannot understand that how can the type Num be converted to Fractional implicitly?
I know that if an instance has type Fractional it must have type Num (class Num a => Fractional a where), but does it necessary that if an instance has type Num it can be used as an instance with Fractional type?
#mnoronha Thanks for your detailed reply. There is only one question confuse me. I know the reason that type a cannot be used in function (/) is that type a is with type Integer which is not an instance of type class Fractional (the function (/) requires that the type of arguments must be instance of Fractional). What I don't understand is that even by calling fromInteger to convert the type integer to atype which be an instance of Num, it does not mean a type be an instance of Fractional (because Fractional type class is more constrained than Num type class, so a type may not implement some functions required by Fractional type class). If a type does not fully fit the condition Fractional type class requires, how can it be use in the function (/) which asks the arguments type be instance of Fractional. Sorry for not native speaker and really thanks for your patience!
I tested that if a type only fits the parent type class, it cannot be used in a function which requires more constrained type class.
{-# LANGUAGE OverloadedStrings #-}
module Main where
class ParentAPI a where
printPar :: int -> a -> String
class (ParentAPI a) => SubAPI a where
printSub :: a -> String
data ParentDT = ParentDT Int
instance ParentAPI ParentDT where
printPar i p = "par"
testF :: (SubAPI a) => a -> String
testF a = printSub a
main = do
let m = testF $ ParentDT 10000
return ()
====
test-typeclass.hs:19:11: error:
• No instance for (SubAPI ParentDT) arising from a use of ‘testF’
• In the expression: testF $ ParentDT 10000
In an equation for ‘m’: m = testF $ ParentDT 10000
In the expression:
do { let m = testF $ ParentDT 10000;
return () }
I have found a doc explaining the numeric overloading ambiguity very clearly and may help others with the same confusion.
https://www.haskell.org/tutorial/numbers.html
First, note that both Fractional and Num are not types, but type classes. You can read more about them in the documentation or elsewhere, but the basic idea is that they define behaviors for types. Num is the most inclusive numeric typeclass, defining behaviors functions like (+), negate, which are common to pretty much all "numeric types." Fractional is a more constrained type class that describes "fractional numbers, supporting real division."
If we look at the type class definition for Fractional, we see that it is actually defined as a subclass of Num. That is, for a type a to be an have an instance Fractional, it must first be a member of the typeclass Num:
class Num a => Fractional a where
Let's consider some type that is constrained by Fractional. We know it implements the basic behaviors common to all members of Num. However, we can't expect it to implement behaviors from other type classes unless multiple constraints are specified (ex. (Num a, Ord a) => a. Take, for example, the function div :: Integral a => a -> a -> a (integral division). If we try to apply the function with an argument that is constrained by the typeclass Fractional (ex. 1.2 :: Fractional t => t), we encounter an error. Type classes restrict the sort of values a function deals with, allowing us to write more specific and useful functions for types that share behaviors.
Now let's look at the more general typeclass, Num. If we have a type variable a that is only constrained by Num a => a, we know that it will implement the (few) basic behaviors included in the Num type class definition, but we'd need more context to know more. What does this mean practically? We know from our Fractional class declaration that functions defined in the Fractional type class are applied to Num types. However, these Num types are a subset of all possible Num types.
The importance of all this, ultimately, has to do with the ground types (where type class constraints are most commonly seen in functions). a represents a type, with the notation Num a => a telling us that a is a type that includes an instance of the type class Num. a could be any of the types that include the instance (ex. Int, Natural). Thus, if we give a value a general type Num a => a, we know it can implement functions for every type where there is a type class defined. For example:
ghci>> let a = 3 :: (Num a => a)
ghci>> a / 2
1.5
Whereas if we'd defined a as a specific type or in terms of a more constrained type class, we would have not been able to expect the same results:
ghci>> let a = 3 :: Integral a => a
ghci>> a / 2
-- Error: ambiguous type variable
or
ghci>> let a = 3 :: Integer
ghci>> a / 2
-- Error: No instance for (Fractional Integer) arising from a use of ‘/’
(Edit responding to followup question)
This is definitely not the most concrete explanation, so readers feel free to suggest something more rigorous.
Suppose we have a function a that is just a type class constrained version of the id function:
a :: Num a => a -> a
a = id
Let's look at type signatures for some applications of the function:
ghci>> :t (a 3)
(a 3) :: Num a => a
ghci>> :t (a 3.2)
(a 3.2) :: Fractional a => a
While our function had the general type signature, as a result of its application the the type of the application is more restricted.
Now, let's look at the function fromIntegral :: (Num b, Integral a) => a -> b. Here, the return type is the general Num b, and this will be true regardless of input. I think the best way to think of this difference is in terms of precision. fromIntegral takes a more constrained type and makes it less constrained, so we know we'll always expect the result will be constrained by the type class from the signature. However, if we give an input constraint, the actual input could be more restricted than the constraint and the resulting type would reflect that.
The reason why this works comes down to the way universal quantification works. To help explain this I am going to add in explicit forall to the type signatures (which you can do yourself if you enable -XExplicitForAll or any other forall related extension), but if you just removed them (forall a. ... becomes just ...), everything will work fine.
The thing to remember is that when a function involves a type constrained by a typeclass, then what that means is that you can input/output ANY type within that typeclass, so it's actually better to have a less constrained typeclass.
So:
fromInteger :: forall a. Num a => Integer -> a
fromInteger 5 :: forall a. Num a => a
Means that you have a value that is of EVERY Num type. So not only can you use it in a function taking it in a Fractional, you could use it in a function that only takes in MyWeirdTypeclass a => ... as long as there is one single type that implements both Num and MyWeirdTypeclass. Hence why you can get the following just fine:
fromInteger 5 / 2 :: forall a. Fractional a => a
Now of course once you decide to divide by 2, it now wants the output type to be Fractional, and thus 5 and 2 will be interpreted as some Fractional type, so we won't run into issues where we try to divide Int values, as trying to make the above have type Int will fail to type check.
This is really powerful and awesome, but very much unfamiliar, as generally other languages either don't support this, or only support it for input arguments (e.g print in most languages can take in any printable type).
Now you may be curious when the whole superclass / subclass stuff comes into play, so when you are defining a function that takes in something of type Num a => a, then because a user can pass in ANY Num type, you are correct that in this situation you cannot use functions defined on some subclass of Num, only things that work on ALL Num values, like *:
double :: forall a. Num a => a -> a
double n = n * 2 -- in here `n` really has type `exists a. Num a => a`
So the following does not type check, and it wouldn't type check in any language, because you don't know that the argument is a Fractional.
halve :: Num a => a -> a
halve n = n / 2 -- in here `n` really has type `exists a. Num a => a`
What we have up above with fromInteger 5 / 2 is more equivalent to the following, higher rank function, note that the forall within parenthesis is required, and you need to use -XRankNTypes:
halve :: forall b. Fractional b => (forall a. Num a => a) -> b
halve n = n / 2 -- in here `n` has type `forall a. Num a => a`
Since this time you are taking in EVERY Num type (just like the fromInteger 5 you were dealing with before), not just ANY Num type. Now the downside of this function (and one reason why no one wants it) is that you really do have to pass in something of EVERY Num type:
halve (2 :: Int) -- does not work
halve (3 :: Integer) -- does not work
halve (1 :: Double) -- does not work
halve (4 :: Num a => a) -- works!
halve (fromInteger 5) -- also works!
I hope that clears things up a little. All you need for the fromInteger 5 / 2 to work is that there exists ONE single type that is both a Num and a Fractional, or in other words just a Fractional, since Fractional implies Num. Type defaulting doesn't help much with clearing up this confusion, as what you may not realize is that GHC is just arbitrarily picking Double, it could have picked any Fractional.
If I force Haskell to infer the type of a number I'll get Num a => a. For example:
Prelude> :t 1
1 :: Num a => a
But what does a => a mean?
1 :: Num a => a means that 1 has some type a, where a is an instance of the Num typeclass. Note that Num is not a type, but a typeclass, which describes common properties of various types. The Num typeclass, for example, describes types that are numeric, and so support basic arithmetic. The native machine integer type Int is an instance of Num, as is the arbitrary-sized Integer, the floating point type Double, and even the rational number type Rational.
a => a doesn't mean anything. The complete phrase is Num a => a. This means "a" is an instance of the Num type class.
You could also read it as (Num a) => a. It provides a context to say that a is numerical.
However, it is not the type of a; it just says that a should be in the Num type class .
Type classes are a little like interfaces in object-oriented programming, in that they define certain behaviours, without defining a in detail.
Note that there is a difference between -> and => . The first is used for the function signature; the second is used to show type classes.
The typing 1 :: Num a => a means "1 is of type a, for all types a in typeclass Num". More succintly, it means "1 is of any numeric type".
Because of this typing, you can pass 1 to any function requiring any numeric type, such as Int, Double, etc.
Extending your example a bit, we also have [1,2,3] :: Num a => [a]. This means "[1,2,3] is a list of values of type a, for all types a in typeclass Num". In other words, "[1,2,3] is a list of values of any numeric type".
Expression 1::Num a => a can be broken down in 3 parts for reading/understanding purposes.
Let's build step by step :
1 :: says 1 has the type of
c :: Char says c has the type of Char
Everything before => is a "class constraint",so inserting Num a between :: and =>
i.e. 1 :: Num a => a says "1 has the type of a, but not just any type but with a class constraint where a is a member of Num class.
1 has the type Num a => a
1.0 has the type Fractional a => a
Why does 1+1.0 have the type Fractional a => a
This seems strange to me because 1 is not fractional. Only 1.0 is fractional. So how did 1 turn into fractional and get combined with 1.0 to form a fractional?
Since only Num has the + operator, it would seem more natural to me if 1.0 turned into Num, got combined with 1 to produce a final Num (although that would be strange too because we would lose information going from 1.0 to 1).
Every Fractional is a Num, but not every Num is a Fractional. So if we have a Num like 1, it could either be a Fractional (because some Nums are Fractionals) or it could not be. However 1.0 can only be a Fractional, it definitely can't be some other Num like Integer.
So when the compiler sees that you add 1 to a Fractional, it realizes that 1 must be a Fractional as well in this case - otherwise you'd not be allowed to add it to a Fractional.
Here's an example of a similar example that only involves user-defined type classes instead of Nums. Maybe this makes things clearer for you:
class Foo a where
foo :: a
class Foo a => Bar a where
bar :: a
combine :: a -> a -> a
Through the above type classes we now have the following methods:
foo :: Foo a => a
bar :: Bar a => a
combine :: Bar a => a -> a -> a
So now let's try to combine foo and bar like this:
combine foo bar
This is roughly equivalent to you trying to add 1 (of type Num a => a) and 1.0 (of type Fractional a => a) in your example. And just like your example, this works fine and has the type Bar a => a.
Type classes are very much not like OO classes, this can't be overemphasized.
In particular, “if 1.0 turned into Num” doesn't make any sense. Num is a type class, not a type, so nothing can ever “turn into a Num”. In fact, nothing ever turns into something else at all in Haskell – everything has a concrete type, that is fixed.
Now you ask, how do polymorphic functions work then? Well, it's called parametric polymorphism for a reason: what seems to be an “arbitrary type a” is really a type parameter. Like a function parameter, these aren't variables in the sense that they can ever change their value after the fact, but they are variable in the sense that the caller of the function is allowed to choose any particular “type value” for a – provided it fulfills the type-class constraint.
So in a sense, the literal 1 is a function: it accepts a type argument a, and returns a value 1 :: a. What it requires is that a is in class Num.
Then we have (+) and 1.0, both of which also need that same a argument. (+) again requires Num, nothing new; but 1.0 requires Fractional a. So all in all, 1 + 1.0 is a function that accepts “three copies of” the type argument a, and requires
Num a
Num a
Fractional a – which also requires Num a because Num is superclass of Fractional.
It would be pretty awkward if we actually had to write the type out as
(1 + 1.0) :: (Num a, Num a, Fractional a, Num a) => a
so it is allowed to leave out the redundant constraints, only leaving Fractional a, which implies all the rest. What we can't do is only leave one of the Num constraints, because that would not imply Fractional.
Number 1 (which technically stands for fromInteger applied to the Integer value 1) belongs to all the types in class Num.
All the types in class Fractional belong to class Num as well.
Ergo,
Number 1 belongs to all the types in class Fractional.