[Haskell-beginners] fromIntegral
Russ Abbott
russ.abbott at gmail.com
Sat Oct 2 01:24:38 EDT 2010
Thanks. I'm still wondering what [ ] refers to. I can load the following
file without error.
null' xs = xs == [ ]
test =
let x = [ ]
in tail [1] == x && tail ['a'] == x
(I know I can define null' differently. I'm defining it this way so that I
can ask this question.)
When I execute test I get True.
> test
True
So my question is: what is x after compilation? Is it really a thing of type
(Eq a) => [a] ?
If so, how should I think of such a thing being stored so that it can be
found equal to both tail [1] and tail ['a']? Furthermore, this seems to
show that (==) is not transitive since one can't even compile
tail [1] == tail ['a']
much less find them to be equal. Yet they are each == to x.
-- Russ
On Fri, Oct 1, 2010 at 9:08 AM, Daniel Fischer <daniel.is.fischer at web.de>wrote:
> On Friday 01 October 2010 17:08:08, Russ Abbott wrote:
> > Thanks, Filipe
> >
> > I clearly over-stated my case. I'd like to break it into a number of
> > different question. Please see below.
> >
> > On Thu, Sep 30, 2010 at 10:25 PM, Felipe Lessa
> <felipe.lessa at gmail.com>wrote:
> > > I'll try to clarify some concepts. Please correct me if I am
> > > wrong, and please forgive me if I misunderstood you.
> > >
> > > On Fri, Oct 1, 2010 at 12:54 AM, Russ Abbott <russ.abbott at gmail.com>
> > >
> > > wrote:
> > > > In explaining fromIntegral, I want to say that it is really a
> > > > collection
> > >
> > > of
> > >
> > > > overloaded functions:
> > > >
> > > > Integer -> Double
> > > > Int -> Float
> > > > ...
> > > >
> > > > When GHC compiles a line of code with fromIntegral it in, it must
> > > > decide
> > >
> > > at
> > >
> > > > compile time which of these overloaded functions to compile to.
> > > > This is
> > >
> > > in
> > >
> > > > contrast to saying that fromIngetral is a function of the type
> > > > (Integral
> > >
> > > a,
> > >
> > > > Num b) => a -> b. In reality there is no (single) function of the
> > > > type (Integral a, Num b) => a -> b because (among other things)
> > > > every function must map between actual types, but (Integral a, Num
> > > > b) => a -> b does not say which actual types are mapped between.
> > > >
> > > > Is the preceding a reasonable thing to say?
> > >
> > > First of all, I do think that polymorphic functions are plain ol'
> > > functions. For example
> > >
> > > id :: a -> a
> > > id x = x
> > >
> > > is a function. Moreover, in GHC 'id' has only one
> > > representation, taking a thunk and returning a thunk, so even at
> > > the machine code level this is only one function.
> >
> > Agree. I over stated my case. The same can be said for
> > length :: [a] -> Int
> > It doesn't matter what the type of element in the list is. length runs
> > the same way no matter what. So this is pure polymorphism.
> >
> > > Now, maybe 'fromIntegral' has something more than polymorphism?
> > > Well, it has typeclasses. But we can represent those as
> > > data types, so we could write
> > >
> > > fromIntegralD :: Integral a -> Num b -> a -> b
> > > fromIntegralD intrDictA numDictB =
> > > fromIntegral numDictB . toInteger intrDictA
> >
> > I'm afraid I don't understand this. Moreover, I couldn't get the
> > preceding to load without error.
> >
>
> No wonder, Integral and Num are type classes and not datatypes (unless you
> have defined such datatypes in scope).
>
> The point is, you can represent type classes as dictionaries, e.g.
>
> data Num a = NumDict
> { plus :: a -> a -> a
> , minus :: a -> a -> a
> , ...
> , fromIntegerD :: Integer -> a
> }
>
> data Integral a = IntegralDict
> { quotD :: a -> a -> a
> , ...
> , toIntegerD a
> }
>
> Then a type-class polymorphic function like fromIntegral becomes a function
> with some dictionaries as additional arguments.
>
> foo :: (Class1 a, Class2 b) => a -> b
>
> becomes
>
> fooDict :: Class1Dict a -> Class2Dict b -> a -> b
>
> To do that explicitly is of course somewhat cumbersome as one has to always
> carry the dictionaries around and one can have more than one dictionary per
> type (e.g.
>
> intNum1 :: Num Int
> intNum1 = NumDict
> { plus = (+)
> , ...
> , fromIntegerD = fromInteger
> }
>
> intNum2 :: Num Int
> intNum2 = NumDict
> { plus = quot
> , -- more nonsense
> , fromInteger = const 1
> }
> ).
>
> Internally, GHC implements type classes via dictionaries and passes them as
> extra arguments to overloaded functions, as you can see by inspecting the
> Core output (-ddump-simpl).
>
> > > Better yet, the compiler could write this code for us internally.
>
> And GHC does.
>
> > > Now, using thunks we can get a single machine code for
> > > 'fromIntegralD' as well.
>
> But that's not terribly efficient, so with -O, GHC tries to eliminate
> dictionaries and use the specialised functions (like
> plusInteger :: Integer -> Integer -> Integer).
>
> > >
> > > In sum, I think all functions are really just that, functions.
> > >
> > > --
> > >
> > > You may call functions that have typeclass constraints
> > > "overloaded functions", but they still are functions.
> > >
> > > Functions that are polymorphic but do not have constraints are
> > > not really overloaded because of parametricity, which means that
> > > they can't change the way they work based on the specific choices
> > > of types you make.
> >
> > I don't understand the preceding paragraph. Would you mind elaborating.
> >
>
> For a function like
>
> length :: [a] -> Int
>
> , because it doesn't know anything about the type a at which it will be
> called, it cannot do anything with the contents of the list (well, it could
> call seq on them, but it would do that for every type), it can only inspect
> the spine of the list.
>
> The code is completely independent of what type of data the pointers to the
> contents point to, so `length [True,False]' and `length [()]' can and do
> call the exact same machine code.
>
> > > > If so, can I say the same sort of thing about constants like 1 and
> > > > []? In particular there is no single value []. Instead [] is a
> > > > symbol which at compile time must be compiled to the empty list of
> > > > some particular type, e.g., [Int]. There is no such Haskell value
> > > > as [] :: [a] since [a] (as type) is not an actual type. I want to
> > > > say the same thing about 1, i.e., that there is no such Haskell
> > > > value as 1 :: (Num t) => t. When the symbol
> > >
> > > 1
> > >
> > > > appears in a program, the compiler must decide during compilation
> > > > whether
> > >
> > > it
> > >
> > > > is intended to be 1::Int or 1::Integer or 1::Double, etc.
> > >
> > > Well, [a] *is* an actual type, a polymorphic one.
> >
> > Here is the example that raised that issue for me. Let's say I define
> > null' as follows.
> >
> > null' xs = xs == [ ]
> >
> > If I don't include a declaration in the file, Haskell (reasonably)
> > concludes the following.
> >
> > > :t null'
> >
> > null' :: (Eq a) => [a] -> Bool
> >
> > If I write the following at the top level,
>
> You seem to mean the ghci prompt here, not the top level of a module.
>
> > everything is fine.
> >
> > > null' [ ]
> >
> > True
> >
> > But if I include the following in the file that defines null', I get an
> > error message.
> >
> > test = null' [ ]
> >
> > Ambiguous type variable `a' in the constraint:
> > `Eq a' arising from a use of `null'' at null.hs:6:17-24
> > Probable fix: add a type signature that fixes these type
> > variable(s)
> >
> > Why is that?
>
> null' has an Eq constraint, so to evaluate test, an Eq dictionary is
> needed, but there's no way to determine which one should be used.
>
> At a lower level, the type of null' is
>
> null' :: EqDict a -> [a] -> Bool
>
> The (hidden) first argument is missing and GHC doesn't know which one to
> pass.
>
> At the ghci-prompt, ghci's extended default rules let it selet the Eq
> dictionary of () and all's fine.
>
> In a source file, GHC restricts itself to the default rules specified in
> the language report, which state that for defaulting to take place, at
> least one of the constraints must be a numeric class. There's none here, so
> no defaulting and the type variable of the constraint remains ambiguous.
>
> > And how can it be fixed? I know I can fix it as follows.
> >
> > test = null' ([ ] :: [Integer])
> >
> > > :reload
> > >
> > > test
> >
> > True
>
> In that situation, I think giving a type signature is the only way¹.
>
> test = null' ([] :: Num a => [a])
>
> also works.
>
> ¹ -XExtendedDefaultRules might work too.
> >
> > That's what suggested to me that [ ] had to be compiled into a concrete
> > value.
>
> Try
>
> null'' [] = True
> null'' _ = False
>
> test'' = null'' []
>
> No type class constraints, no problems.
>
> >
> >
> > It seemed to me that similar reasoning applied to things like 1. How is
> > the following explained?
> >
> > Prelude> 111111111111111111111111111111111111111111
> > 111111111111111111111111111111111111111111
> > it :: (Num t) => t
> > Prelude> maxBound :: Int
> > 2147483647
> > it :: Int
> > Prelude> 111111111111111111111111111111111111111111 - (1::Int)
> > -954437178
> > it :: Int
> >
> > Does it make sense to say that the long string of 1's is really of type
> > (Num t) => t?
>
> Integer literals stand for (fromInteger Integer-value-of-literal), so the
> literal itself can have any type belonging to Num. If you force it to have
> a particular type, the corresponding fromInteger function is determined and
> can be applied if the value is needed.
>
> >
> > If so, what does the compiler think it's doing when it processes(?) it
> > as an Int so that it can subtract 1 :: Int from it? It didn't treat it
> > as maxBound :: Int. And yet it didn't generate an error message.
>
> For efficiency, fromInteger wraps, for a b-bit Integral type, the result of
> fromInteger n is n `mod` 2^b.
>
> >
> > Thanks
> >
> > -- Russ
>
>
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