How is (,) defined internally? - haskell

It's trivial to redefine the function
(,) :: a -> b -> (a,b)
(,) a b = (a,b)
The weird (to me) thing is that this function is defined for arbitrary length tuples. So, for example, there is actually a function:
(,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,) :: a -> b -> c -> ... -> (a,b,c,...)
How is this done? Why isn't it done for all the standard functions on tuples, like zip?
Hoogle gives me no results, and I don't see how template Haskell could do this, so I assume it must be some sort of magic inside the compiler. But that seems very un-Haskelly to me.

How is this done?
Compiler support. Haskell language report mandates (,) to be supported for at least up to 15 arguments (6.1.4), but GHC goes a bit further and generates them for a lot more (last time we've tested this, it could handle hundreds or even thousands). zip and other tuple functions have to be defined for up to 7-tuples. I don't know if GHC generates those for larger amounts.

My understanding is that (,) isn't a regular function, it's a constructor with special syntax, hard-wired into the langage. It's similar to syntax [1, 2, 3], which you can't define yourself because it's hard-wired.

It is specified in the language definition and hard-wired into the compiler. You do not define tuples in Haskell, The definition of Haskell includes tuples.
There is accepted (,,,) syntax for tuples, which are anonymous product types, which are quite fundamental. These also play perfectly well with type inference, since each component is present and can be inferred.
There is no accepted syntax for anonymous sum types, and they may not play well with type inference.
Haskell offers user-definable infix syntax but not user-definable brackets (outside of the broad quasi-quoting possibilities).

Related

Pre-defined infix operator list in Haskell?

Basically, I need to define infix operator for function composition, flipped g . f manner.
(#) :: (a -> b) -> (b -> c) -> a -> c
(#) = flip (.)
infixl 9 #
Here, temporarily, I just have chosen the symbol: #, but I'm not certain this choice is not problematic in terms of collision to other definitions.
The weird thing is somehow I could not find a Pre-defined infix operator list in Haskell.
I want the composition operator as concise as possible, hopefully, a single character like ., and if possible, I want to make it &. Is it OK?
Is there any guidance or best pracice tutorial? Please advise.
Related Q&A:
What characters are permitted for Haskell operators?
You don't see a list of Haskell built-in operators for the same reason you don't see a list of all built-in functions. They're everywhere. Some are in Prelude, some are in Control.Monad, etc., etc. Operators aren't special in Haskell; they're ordinary functions with neat syntax. And Haskellers are generally pretty operator-happy in general. Spend any time inside your favorite lens library and you'll find plenty of amusing-looking operators.
In terms of (#), it might be best to avoid. I don't know of any built-in operators called that, but # can be a bit special when it comes to parsing. Specifically, a compiler extension enables # at the end of ordinary identifiers, and GHC defines a lot of built-in (primitive) types following this practice. For instance, Int is the usual (boxed) integer type, whereas Int# is a primitive integer. Most people don't need to interface with this directly, but it is a use that # has in Haskell that would be confused slightly by the addition of your operator.
Your (#) operator is called (>>>) in Haskell, and it works on all Category instances, including functions. Its companion is (<<<), the generalization of (.) to all Category instances. If three characters is too long for you, I've seen it called (|>) in some other languages, but Haskell already uses that operator for something else. If you're not using Data.Sequence, you could use that operator. But personally, I'd just go with (>>>). Any Haskeller will recognize it pretty quickly.

Why do we need Control.Lens.Reified?

Why do we need Control.Lens.Reified? Is there some reason I can't place a Lens directly into a container? What does reify mean anyway?
We need reified lenses because Haskell's type system is predicative. I don't know the technical details of exactly what that means, but it prohibits types like
[Lens s t a b]
For some purposes, it's acceptable to use
Functor f => [(a -> f b) -> s -> f t]
instead, but when you reach into that, you don't get a Lens; you get a LensLike specialized to some functor or another. The ReifiedBlah newtypes let you hang on to the full polymorphism.
Operationally, [ReifiedLens s t a b] is a list of functions each of which takes a Functor f dictionary, while forall f . Functor f => [LensLike f s t a b] is a function that takes a Functor f dictionary and returns a list.
As for what "reify" means, well, the dictionary will say something, and that seems to translate into a rather stunning variety of specific meanings in Haskell. So no comment on that.
The problem is that, in Haskell, type abstraction and application are completely implicit; the compiler is supposed to insert them where needed. Various attempts at designing 'impredicative' extensions, where the compiler would make clever guesses where to put them, have failed; so the safest thing ends up being relying on the Haskell 98 rules:
Type abstractions occur only at the top level of a function definition.
Type applications occur immediately whenever a variable with a polymorphic type is used in an expression.
So if I define a simple lens:[1]
lensHead f [] = pure []
lensHead f (x:xn) = (:xn) <$> f x
and use it in an expression:
[lensHead]
lensHead gets automatically applied to some set of type parameters; at which point it's no longer a lens, because it's not polymorphic in the functor anymore. The take-away is: an expression always has some monomorphic type; so it's not a lens. (You'll note that the lens functions take arguments of type Getter and Setter, which are monomorphic types, for similar reasons to this. But a [Getter s a] isn't a list of lenses, because they've been specialized to only getters.)
What does reify mean? The dictionary definition is 'make real'. 'Reifying' is used in philosophy to refer to the act of regarding or treating something as real (rather than ideal or abstract). In programming, it tends to refer to taking something that normally can't be treated as a data structure and representing it as one. For example, in really old Lisps, there didn't use to be first-class functions; instead, you had to use S-Expressions to pass 'functions' around, and eval them when you needed to call the function. The S-Expressions represented the functions in a way you could manipulate in the program, which is referred to as reification.
In Haskell, we don't typically need such elaborate reification strategies as Lisp S-Expressions, partly because the language is designed to avoid needing them; but since
newtype ReifiedLens s t a b = ReifiedLens (Lens s t a b)
has the same effect of taking a polymorphic value and turning it into a true first-class value, it's referred to as reification.
Why does this work, if expressions always have monomorphic types? Well, because the Rank2Types extension adds a third rule:
Type abstractions occur at the top-level of the arguments to certain functions, with so-called rank 2 types.
ReifiedLens is such a rank-2 function; so when you say
ReifiedLens l
you get a type lambda around the argument to ReifiedLens, and then l is applied immediately to the the lambda-bound type argument. So l is effectively just eta-expanded. (Compilers are free to eta-reduce this and just use l directly).
Then, when you say
f (ReifiedLens l) = ...
on the right-hand side, l is a variable with polymorphic type, so every use of l is immediately implicitly assigned to whatever type arguments are needed for the expression to type-check. So everything works the way you expect.
The other way to think about is that, if you say
newtype ReifiedLens s t a b = ReifiedLens { unReify :: Lens s t a b }
the two functions ReifiedLens and unReify act like explicit type abstraction and application operators; this allows the compiler to identify where you want the abstractions and applications to take place well enough that the issues with impredicative type systems don't come up.
[1] In lens terminology, this is apparently called something other than a 'lens'; my entire knowledge of lenses comes from SPJ's presentation on them so I have no way to verify that. The point remains, since the polymorphism is still necessary to make it work as both a getter and a setter.

Can all recursive structures be replaced by a non recursive solution?

For example, could you define a list in Haskell without defining a recursive structure? Or replace all lists by some function(s)?
data List a = Empty | (a, List a) -- <- recursive definition
EDIT
I gave the list as an example, but I was really asking about all data structures in general.
Maybe we only need one recursive data structure for all cases where recursion is needed? Like the Y combinator being the only recursive function needed. #TikhonJelvis 's answer made me think about that.
Now I'm pretty sure this post is better suited for cs.stackexchange.
About current selected answer
I was really looking for answers that looked more like the ones given by #DavidYoung & #TikhonJelvis, but they only give a partial answer and I appreciate them.
So, if any has an answer that uses functional concepts, please share.
That's a bit of an odd question. I think the answer is not really, but the definition of the data type does not have to be recursive directly.
Ultimately, lists are recursive data structures. You can't define them without having some sort of recursion somewhere. It's core to their essence.
However, we don't have to make the actual definition of List recursive. Instead, we can factor out recursion into a single data type Fix and then define all other recursive types with it. In a sense, Fix just captures the essence of what it means for a data structure to be recursive. (It's the type-level version of the fix function, which does the same thing for functions.)
data Fix f = Roll (f (Fix f))
The idea is that Fix f corresponds to f applied to itself repeatedly. To make it work with Haskell's algebraic data types, we have to throw in a Roll constructor at every level, but this does not change what the type represents.
Essentially, f applied to itself repeatedly like this is the essence of recursion.
Now we can define a non-recursive analog to List that takes an extra type argument f that replaces our earlier recursion:
data ListF a f = Empty | Cons a f
This is a straightforward data type that is not recursive.
If we combine the two, we get our old List type except with some extra Roll constructors at each recursive step.
type List a = Fix (ListF a)
A value of this type looks like this:
Roll (Cons 1 (Roll (Cons 2 (Roll Empty))))
It carries the same information as (Cons 1 (Cons 2 Empty)) or even just [1, 2], but a few extra constructors sprinkled through.
So if you were given Fix, you could define List without using recursion. But this isn't particularly special because, in a sense, Fix is recursion.
I'm not sure if all recursive structures can be replaced by a non-recursive version but some certainly can, including lists. One possible way to do this is with what is called a Boehm-Berarducci encoding. This a way to represent a structure as a function, specifically the fold over that structure (foldr in the case of a list):
{-# LANGUAGE RankNTypes #-}
type List a = forall x . (a -> x -> x) -> x -> x
-- ^^^^^^^^^^^^^ ^
-- Cons branch Nil branch
(From the above link with slightly different formatting)
This type is also something like a case analysis over the list. The first argument represents the cons case and the second argument represents the nil case.
In general, the branches of a sum type become different arguments to the function and fields of a product type become function types with an argument for each field. Note that in the encoding above, the nil branch is (in general) a non-function because the nil constructor takes no arguments, while the cons branch has two arguments since the cons constructor takes two arguments. The recursion parts of the definition are "replaced" with a Rank N type (called x here).
I think this question breaks down into considering three distinct feature subsets that Haskell provides:
Facilities for defining new data types.
A repertoire of built-in types.
A foreign function interface that allows interfacing with functionality external to the language.
Looking only at (1), the native type definition facilities don't really provide for defining any infinitely-large types other than by recursion.
Looking at (2), however, Haskell 2010 provides the Data.Array module, which provides array types that together with (1) can be used to build non-recursive definitions of many different structures.
And even if the language did not provide arrays, (3) means that we could bolt them to the language as an FFI extension. Haskell implementations are also allowed to provide extra functionality that can be used for this in stead of the FFI, and many libraries for GHC exploit those (e.g., vector).
So I'd say that the best answer is that Haskell only allows you to define nonrecursive collection types only to the extent that it provides you with basic built-in ones that you can use as building blocks for more complex ones.

Given a Haskell type signature, is it possible to generate the code automatically?

What it says in the title. If I write a type signature, is it possible to algorithmically generate an expression which has that type signature?
It seems plausible that it might be possible to do this. We already know that if the type is a special-case of a library function's type signature, Hoogle can find that function algorithmically. On the other hand, many simple problems relating to general expressions are actually unsolvable (e.g., it is impossible to know if two functions do the same thing), so it's hardly implausible that this is one of them.
It's probably bad form to ask several questions all at once, but I'd like to know:
Can it be done?
If so, how?
If not, are there any restricted situations where it becomes possible?
It's quite possible for two distinct expressions to have the same type signature. Can you compute all of them? Or even some of them?
Does anybody have working code which does this stuff for real?
Djinn does this for a restricted subset of Haskell types, corresponding to a first-order logic. It can't manage recursive types or types that require recursion to implement, though; so, for instance, it can't write a term of type (a -> a) -> a (the type of fix), which corresponds to the proposition "if a implies a, then a", which is clearly false; you can use it to prove anything. Indeed, this is why fix gives rise to ⊥.
If you do allow fix, then writing a program to give a term of any type is trivial; the program would simply print fix id for every type.
Djinn is mostly a toy, but it can do some fun things, like deriving the correct Monad instances for Reader and Cont given the types of return and (>>=). You can try it out by installing the djinn package, or using lambdabot, which integrates it as the #djinn command.
Oleg at okmij.org has an implementation of this. There is a short introduction here but the literate Haskell source contains the details and the description of the process. (I'm not sure how this corresponds to Djinn in power, but it is another example.)
There are cases where is no unique function:
fst', snd' :: (a, a) -> a
fst' (a,_) = a
snd' (_,b) = b
Not only this; there are cases where there are an infinite number of functions:
list0, list1, list2 :: [a] -> a
list0 l = l !! 0
list1 l = l !! 1
list2 l = l !! 2
-- etc.
-- Or
mkList0, mkList1, mkList2 :: a -> [a]
mkList0 _ = []
mkList1 a = [a]
mkList2 a = [a,a]
-- etc.
(If you only want total functions, then consider [a] as restricted to infinite lists for list0, list1 etc, i.e. data List a = Cons a (List a))
In fact, if you have recursive types, any types involving these correspond to an infinite number of functions. However, at least in the case above, there is a countable number of functions, so it is possible to create an (infinite) list containing all of them. But, I think the type [a] -> [a] corresponds to an uncountably infinite number of functions (again restrict [a] to infinite lists) so you can't even enumerate them all!
(Summary: there are types that correspond to a finite, countably infinite and uncountably infinite number of functions.)
This is impossible in general (and for languages like Haskell that does not even has the strong normalization property), and only possible in some (very) special cases (and for more restricted languages), such as when a codomain type has the only one constructor (for example, a function f :: forall a. a -> () can be determined uniquely). In order to reduce a set of possible definitions for a given signature to a singleton set with just one definition need to give more restrictions (in the form of additional properties, for example, it is still difficult to imagine how this can be helpful without giving an example of use).
From the (n-)categorical point of view types corresponds to objects, terms corresponds to arrows (constructors also corresponds to arrows), and function definitions corresponds to 2-arrows. The question is analogous to the question of whether one can construct a 2-category with the required properties by specifying only a set of objects. It's impossible since you need either an explicit construction for arrows and 2-arrows (i.e., writing terms and definitions), or deductive system which allows to deduce the necessary structure using a certain set of properties (that still need to be defined explicitly).
There is also an interesting question: given an ADT (i.e., subcategory of Hask) is it possible to automatically derive instances for Typeable, Data (yes, using SYB), Traversable, Foldable, Functor, Pointed, Applicative, Monad, etc (?). In this case, we have the necessary signatures as well as additional properties (for example, the monad laws, although these properties can not be expressed in Haskell, but they can be expressed in a language with dependent types). There is some interesting constructions:
http://ulissesaraujo.wordpress.com/2007/12/19/catamorphisms-in-haskell
which shows what can be done for the list ADT.
The question is actually rather deep and I'm not sure of the answer, if you're asking about the full glory of Haskell types including type families, GADT's, etc.
What you're asking is whether a program can automatically prove that an arbitrary type is inhabited (contains a value) by exhibiting such a value. A principle called the Curry-Howard Correspondence says that types can be interpreted as mathematical propositions, and the type is inhabited if the proposition is constructively provable. So you're asking if there is a program that can prove a certain class of propositions to be theorems. In a language like Agda, the type system is powerful enough to express arbitrary mathematical propositions, and proving arbitrary ones is undecidable by Gödel's incompleteness theorem. On the other hand, if you drop down to (say) pure Hindley-Milner, you get a much weaker and (I think) decidable system. With Haskell 98, I'm not sure, because type classes are supposed to be able to be equivalent to GADT's.
With GADT's, I don't know if it's decidable or not, though maybe some more knowledgeable folks here would know right away. For example it might be possible to encode the halting problem for a given Turing machine as a GADT, so there is a value of that type iff the machine halts. In that case, inhabitability is clearly undecidable. But, maybe such an encoding isn't quite possible, even with type families. I'm not currently fluent enough in this subject for it to be obvious to me either way, though as I said, maybe someone else here knows the answer.
(Update:) Oh a much simpler interpretation of your question occurs to me: you may be asking if every Haskell type is inhabited. The answer is obviously not. Consider the polymorphic type
a -> b
There is no function with that signature (not counting something like unsafeCoerce, which makes the type system inconsistent).

How do you do generic programming in Haskell?

Coming from C++, I find generic programming indispensable. I wonder how people approach that in Haskell?
Say how do write generic swap function in Haskell?
Is there an equivalent concept of partial specialization in Haskell?
In C++, I can partially specialize the generic swap function with a special one for a generic map/hash_map container that has a special swap method for O(1) container swap. How do you do that in Haskell or what's the canonical example of generic programming in Haskell?
This is closely related to your other question about Haskell and quicksort. I think you probably need to read at least the introduction of a book about Haskell. It sounds as if you haven't yet grasped the key point about it which is that it bans you from modifying the values of existing variables.
Swap (as understood and used in C++) is, by its very nature, all about modifying existing values. It's so we can use a name to refer to a container, and replace that container with completely different contents, and specialize that operation to be fast (and exception-free) for specific containers, allowing us to implement a modify-and-publish approach (crucial for writing exception-safe code or attempting to write lock-free code).
You can write a generic swap in Haskell, but it would probably take a pair of values and return a new pair containing the same values with their positions reversed, or something like that. Not really the same thing, and not having the same uses. It wouldn't make any sense to try and specialise it for a map by digging inside that map and swapping its individual member variables, because you're just not allowed to do things like that in Haskell (you can do the specialization, but not the modifying of variables).
Suppose we wanted to "measure" a list in Haskell:
measure :: [a] -> Integer
That's a type declaration. It means that the function measure takes a list of anything (a is a generic type parameter because it starts with a lowercase letter) and returns an Integer. So this works for a list of any element type - it's what would be called a function template in C++, or a polymorphic function in Haskell (not the same as a polymorphic class in C++).
We can now define that by providing specializations for each interesting case:
measure [] = 0
i.e. measure the empty list and you get zero.
Here's a very general definition that covers all other cases:
measure (h:r) = 1 + measure r
The bit in parentheses on the LHS is a pattern. It means: take a list, break off the head and call it h, call the remaining part r. Those names are then parameters we can use. This will match any list with at least one item on it.
If you've tried template metaprogramming in C++ this will all be old hat to you, because it involves exactly the same style - recursion to do loops, specialization to make the recursion terminate. Except that in Haskell it works at runtime (specialization of the function for particular values or patterns of values).
As Earwicker sais, the example is not as meaningful in Haskell. If you absolutely want to have it anyway, here is something similar (swapping the two parts of a pair), c&p from an interactive session:
GHCi, version 6.8.2: http://www.haskell.org/ghc/ :? for help
Loading package base ... linking ... done.
Prelude> let swap (a,b) = (b,a)
Prelude> swap("hello", "world")
("world","hello")
Prelude> swap(1,2)
(2,1)
Prelude> swap("hello",2)
(2,"hello")
In Haskell, functions are as generic (polymorphic) as possible - the compiler will infer the "Most general type". For example, TheMarko's example swap is polymorphic by default in the absence of a type signature:
*Main> let swap (a,b) = (b,a)
*Main> :t swap
swap :: (t, t1) -> (t1, t)
As for partial specialization, ghc has a non-98 extension:
file:///C:/ghc/ghc-6.10.1/doc/users_guide/pragmas.html#specialize-pragma
Also, note that there's a mismatch in terminology. What's called generic in c++, Java, and C# is called polymorphic in Haskell. "Generic" in Haskell usually means polytypic:
http://haskell.readscheme.org/generic.html
But, aboe i use the c++ meaning of generic.
In Haskell you would create type classes. Type classes are not like classes in OO languages. Take the Numeric type class It says that anything that is an instance of the class can perform certain operations(+ - * /) so Integer is a member of Numeric and provides implementations of the functions necessary to be considered Numeric and can be used anywhere a Numeric is expected.
Say you want to be able to foo Ints and Strings. Then you would declare Int and String to be
instances of the type class Foo. Now anywhere you see the type (Foo a) you can now use Int or String.
The reason why you can't add ints and floats directly is because add has the type (Numeric a) a -> a -> a a is a type variable and just like regular variables it can only be bound once so as soon as you bind it to Int every a in the list must be Int.
After reading enough in a Haskell book to really understand Earwicker's answer I'd suggest you also read about type classes. I'm not sure what “partial specialization” means, but it sounds like they could come close.

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