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Why are functions in Ocaml/F# not recursive by default?
OCaml uses let to define a new function, or let rec to define a function that is recursive. Why does it need both of these - couldn't we just use let for everything?
For example, to define a non-recursive successor function and recursive factorial in OCaml (actually, in the OCaml interpreter) I might write
let succ n = n + 1;;
let rec fact n =
if n = 0 then 1 else n * fact (n-1);;
Whereas in Haskell (GHCI) I can write
let succ n = n + 1
let fact n =
if n == 0 then 1 else n * fact (n-1)
Why does OCaml distinguish between let and let rec? Is it a performance issue, or something more subtle?
Well, having both available instead of only one gives the programmer tighter control on the scope. With let x = e1 in e2, the binding is only present in e2's environment, while with let rec x = e1 in e2 the binding is present in both e1 and e2's environments.
(Edit: I want to emphasize that it is not a performance issue, that makes no difference at all.)
Here are two situations where having this non-recursive binding is useful:
shadowing an existing definition with a refinement that use the old binding. Something like: let f x = (let x = sanitize x in ...), where sanitize is a function that ensures the input has some desirable property (eg. it takes the norm of a possibly-non-normalized vector, etc.). This is very useful in some cases.
metaprogramming, for example macro writing. Imagine I want to define a macro SQUARE(foo) that desugars into let x = foo in x * x, for any expression foo. I need this binding to avoid code duplication in the output (I don't want SQUARE(factorial n) to compute factorial n twice). This is only hygienic if the let binding is not recursive, otherwise I couldn't write let x = 2 in SQUARE(x) and get a correct result.
So I claim it is very important indeed to have both the recursive and the non-recursive binding available. Now, the default behaviour of the let-binding is a matter of convention. You could say that let x = ... is recursive, and one must use let nonrec x = ... to get the non-recursive binder. Picking one default or the other is a matter of which programming style you want to favor and there are good reasons to make either choice. Haskell suffers¹ from the unavailability of this non-recursive mode, and OCaml has exactly the same defect at the type level : type foo = ... is recursive, and there is no non-recursive option available -- see this blog post.
¹: when Google Code Search was available, I used it to search in Haskell code for the pattern let x' = sanitize x in .... This is the usual workaround when non-recursive binding is not available, but it's less safe because you risk writing x instead of x' by mistake later on -- in some cases you want to have both available, so picking a different name can be voluntary. A good idiom would be to use a longer variable name for the first x, such as unsanitized_x. Anyway, just looking for x' literally (no other variable name) and x1 turned a lot of results. Erlang (and all language that try to make variable shadowing difficult: Coffeescript, etc.) has even worse problems of this kind.
That said, the choice of having Haskell bindings recursive by default (rather than non-recursive) certainly makes sense, as it is consistent with lazy evaluation by default, which makes it really easy to build recursive values -- while strict-by-default languages have more restrictions on which recursive definitions make sense.
Related
I'm wondering why F# allows shadowing, particularly within the same scope.
I've always thought of value binding in purely functional programming constructs as being akin to assignments in algebra/mathematics.
So for example,
y = x + 1
will be a valid mathematical expression but
y = y + 1
won't be.
However, because of shadowing,
let y = y + 1
is a perfectly valid expression in F#.
Why does the language allow this?
The most correct answer to this is because the creator, Don Syme, felt that it was a useful feature to add to the language.
There are some good uses for it, though. One is when working with F#-style optional parameters:
type C() =
member _.M(?x) =
let x = Option.defaultValue 0 x
printfn "%d" x
F# optional parameters are options, which brings a high degree of correctness and consistency. But imagine having 3 or more optional parameters to a method. It would get annoying to have to rebind the value for each of them and use a different name for them! This is one area where shadowing comes in handy.
It's also handy when writing recursive subroutines. Consider the following naiive implementation of sum for a list:
let mySum xs =
let rec loop xs acc =
match xs with
| [] -> acc
| h :: t -> loop t (h + acc)
loop xs 0
I don't need to rebind xs for the inner loop because of shadowing. Since it's generic, xs is about as good of a name as I can come up with, so it would be annoying to have had to use a different name for the inner loop.
Shadowing isn't all good news, though. If you're not careful, types from one open declaraction can shadow types from one that was declared previously. This can be confusing. F# editor tooling can distinguish bindings from the ones that shadow them, but you don't get that with plain text. So the bottom line is: think carefully when applying shadowing in F#.
I am new to the world of fixed-point combinators and I guess they are used to recurse on anonymous lambdas, but I haven't really got to use them, or even been able to wrap my head around them completely.
I have seen the example in Javascript for a Y-combinator but haven't been able to successfully run it.
The question here is, can some one give an intuitive answer to:
What are Fixed-point combinators, (not just theoretically, but in context of some example, to reveal what exactly is the fixed-point in that context)?
What are the other kinds of fixed-point combinators, apart from the Y-combinator?
Bonus Points: If the example is not just in one language, preferably in Clojure as well.
UPDATE:
I have been able to find a simple example in Clojure, but still find it difficult to understand the Y-Combinator itself:
(defn Y [r]
((fn [f] (f f))
(fn [f]
(r (fn [x] ((f f) x))))))
Though the example is concise, I find it difficult to understand what is happening within the function. Any help provided would be useful.
Suppose you wanted to write the factorial function. Normally, you would write it as something like
function fact(n) = if n=0 then 1 else n * fact(n-1)
But that uses explicit recursion. If you wanted to use the Y-combinator instead, you could first abstract fact as something like
function factMaker(myFact) = lamba n. if n=0 then 1 else n * myFact(n-1)
This takes an argument (myFact) which it calls were the "true" fact would have called itself. I call this style of function "Y-ready", meaning it's ready to be fed to the Y-combinator.
The Y-combinator uses factMaker to build something equivalent to the "true" fact.
newFact = Y(factMaker)
Why bother? Two reasons. The first is theoretical: we don't really need recursion if we can "simulate" it using the Y-combinator.
The second is more pragmatic. Sometimes we want to wrap each function call with some extra code to do logging or profiling or memoization or a host of other things. If we try to do this to the "true" fact, the extra code will only be called for the original call to fact, not all the recursive calls. But if we want to do this for every call, including all the recursive call, we can do something like
loggingFact = LoggingY(factMaker)
where LoggingY is a modified version of the Y combinator that introduces logging. Notice that we did not need to change factMaker at all!
All this is more motivation why the Y-combinator matters than a detailed explanation from how that particular implementation of Y works (because there are many different ways to implement Y).
To answer your second question about fix-point combinators other than Y. There are countably infinitely many standard fix-point combinators, that is, combinators fix that satisfy the equation
fix f = f (fix f)
There are also contably many non-standard fix-point combinators, which satisfy the equation
fix f = f (f (fix f))
etc. Standard fix-point combinators are recursively enumerable, but non-standard are not. Please see the following web page for examples, references and discussion.
http://okmij.org/ftp/Computation/fixed-point-combinators.html#many-fixes
I was thinking about pure Object Oriented Languages like Ruby, where everything, including numbers, int, floats, and strings are themselves objects. Is this the same thing with pure functional languages? For example, in Haskell, are Numbers and Strings also functions?
I know Haskell is based on lambda calculus which represents everything, including data and operations, as functions. It would seem logical to me that a "purely functional language" would model everything as a function, as well as keep with the definition that a function most always returns the same output with the same inputs and has no state.
It's okay to think about that theoretically, but...
Just like in Ruby not everything is an object (argument lists, for instance, are not objects), not everything in Haskell is a function.
For more reference, check out this neat post: http://conal.net/blog/posts/everything-is-a-function-in-haskell
#wrhall gives a good answer. However you are somewhat correct that in the pure lambda calculus it is consistent for everything to be a function, and the language is Turing-complete (capable of expressing any pure computation that Haskell, etc. is).
That gives you some very strange things, since the only thing you can do to anything is to apply it to something else. When do you ever get to observe something? You have some value f and want to know something about it, your only choice is to apply it some value x to get f x, which is another function and the only choice is to apply it to another value y, to get f x y and so on.
Often I interpret the pure lambda calculus as talking about transformations on things that are not functions, but only capable of expressing functions itself. That is, I can make a function (with a bit of Haskelly syntax sugar for recursion & let):
purePlus = \zero succ natCase ->
let plus = \m n -> natCase m n (\m' -> plus m' n)
in plus (succ (succ zero)) (succ (succ zero))
Here I have expressed the computation 2+2 without needing to know that there are such things as non-functions. I simply took what I needed as arguments to the function I was defining, and the values of those arguments could be church encodings or they could be "real" numbers (whatever that means) -- my definition does not care.
And you could think the same thing of Haskell. There is no particular reason to think that there are things which are not functions, nor is there a particular reason to think that everything is a function. But Haskell's type system at least prevents you from applying an argument to a number (anybody thinking about fromInteger right now needs to hold their tongue! :-). In the above interpretation, it is because numbers are not necessarily modeled as functions, so you can't necessarily apply arguments to them.
In case it isn't clear by now, this whole answer has been somewhat of a technical/philosophical digression, and the easy answer to your question is "no, not everything is a function in functional languages". Functions are the things you can apply arguments to, that's all.
The "pure" in "pure functional" refers to the "freedom from side effects" kind of purity. It has little relation to the meaning of "pure" being used when people talk about a "pure object-oriented language", which simply means that the language manipulates purely (only) in objects.
The reason is that pure-as-in-only is a reasonable distinction to use to classify object-oriented languages, because there are languages like Java and C++, which clearly have values that don't have all that much in common with objects, and there are also languages like Python and Ruby, for which it can be argued that every value is an object1
Whereas for functional languages, there are no practical languages which are "pure functional" in the sense that every value the language can manipulate is a function. It's certainly possible to program in such a language. The most basic versions of the lambda calculus don't have any notion of things that are not functions, but you can still do arbitrary computation with them by coming up with ways of representing the things you want to compute on as functions.2
But while the simplicity and minimalism of the lambda calculus tends to be great for proving things about programming, actually writing substantial programs in such a "raw" programming language is awkward. The function representation of basic things like numbers also tends to be very inefficient to implement on actual physical machines.
But there is a very important distinction between languages that encourage a functional style but allow untracked side effects anywhere, and ones that actually enforce that your functions are "pure" functions (similar to mathematical functions). Object-oriented programming is very strongly wed to the use of impure computations3, so there are no practical object-oriented programming languages that are pure in this sense.
So the "pure" in "pure functional language" means something very different from the "pure" in "pure object-oriented language".4 In each case the "pure vs not pure" distinction is one that is completely uninteresting applied to the other kind of language, so there's no very strong motive to standardise the use of the term.
1 There are corner cases to pick at in all "pure object-oriented" languages that I know of, but that's not really very interesting. It's clear that the object metaphor goes much further in languages in which 1 is an instance of some class, and that class can be sub-classed, than it does in languages in which 1 is something else than an object.
2 All computation is about representation anyway. Computers don't know anything about numbers or anything else. They just have bit-patterns that we use to represent numbers, and operations on bit-patterns that happen to correspond to operations on numbers (because we designed them so that they would).
3 This isn't fundamental either. You could design a "pure" object-oriented language that was pure in this sense. I tend to write most of my OO code to be pure anyway.
4 If this seems obtuse, you might reflect that the terms "functional", "object", and "language" have vastly different meanings in other contexts also.
A very different angle on this question: all sorts of data in Haskell can be represented as functions, using a technique called Church encodings. This is a form of inversion of control: instead of passing data to functions that consume it, you hide the data inside a set of closures, and to consume it you pass in callbacks describing what to do with this data.
Any program that uses lists, for example, can be translated into a program that uses functions instead of lists:
-- | A list corresponds to a function of this type:
type ChurchList a r = (a -> r -> r) --^ how to handle a cons cell
-> r --^ how to handle the empty list
-> r --^ result of processing the list
listToCPS :: [a] -> ChurchList a r
listToCPS xs = \f z -> foldr f z xs
That function is taking a concrete list as its starting point, but that's not necessary. You can build up ChurchList functions out of just pure functions:
-- | The empty 'ChurchList'.
nil :: ChurchList a r
nil = \f z -> z
-- | Add an element at the front of a 'ChurchList'.
cons :: a -> ChurchList a r -> ChurchList a r
cons x xs = \f z -> f z (xs f z)
foldChurchList :: (a -> r -> r) -> r -> ChurchList a r -> r
foldChurchList f z xs = xs f z
mapChurchList :: (a -> b) -> ChurchList a r -> ChurchList b r
mapChurchList f = foldChurchList step nil
where step x = cons (f x)
filterChurchList :: (a -> Bool) -> ChurchList a r -> ChurchList a r
filterChurchList pred = foldChurchList step nil
where step x xs = if pred x then cons x xs else xs
That last function uses Bool, but of course we can replace Bool with functions as well:
-- | A Bool can be represented as a function that chooses between two
-- given alternatives.
type ChurchBool r = r -> r -> r
true, false :: ChurchBool r
true a _ = a
false _ b = b
filterChurchList' :: (a -> ChurchBool r) -> ChurchList a r -> ChurchList a r
filterChurchList' pred = foldChurchList step nil
where step x xs = pred x (cons x xs) xs
This sort of transformation can be done for basically any type, so in theory, you could get rid of all "value" types in Haskell, and keep only the () type, the (->) and IO type constructors, return and >>= for IO, and a suitable set of IO primitives. This would obviously be hella impractical—and it would perform worse (try writing tailChurchList :: ChurchList a r -> ChurchList a r for a taste).
Is getChar :: IO Char a function or not? Haskell Report doesn't provide us with a definition. But it states that getChar is a function (see here). (Well, at least we can say that it is a function.)
So I think the answer is YES.
I don't think there can be correct definition of "function" except "everything is a function". (What is "correct definition"? Good question...) Consider the next example:
{-# LANGUAGE NoMonomorphismRestriction #-}
import Control.Applicative
f :: Applicative f => f Int
f = pure 1
g1 :: Maybe Int
g1 = f
g2 :: Int -> Int
g2 = f
Is f a function or datatype? It depends.
I understand what the concept of currying is, and know how to use it. These are not my questions, rather I am curious as to how this is actually implemented at some lower level than, say, Haskell code.
For example, when (+) 2 4 is curried, is a pointer to the 2 maintained until the 4 is passed in? Does Gandalf bend space-time? What is this magic?
Short answer: yes a pointer is maintained to the 2 until the 4 is passed in.
Longer than necessary answer:
Conceptually, you're supposed to think about Haskell being defined in terms of the lambda calculus and term rewriting. Lets say you have the following definition:
f x y = x + y
This definition for f comes out in lambda calculus as something like the following, where I've explicitly put parentheses around the lambda bodies:
\x -> (\y -> (x + y))
If you're not familiar with the lambda calculus, this basically says "a function of an argument x that returns (a function of an argument y that returns (x + y))". In the lambda calculus, when we apply a function like this to some value, we can replace the application of the function by a copy of the body of the function with the value substituted for the function's parameter.
So then the expression f 1 2 is evaluated by the following sequence of rewrites:
(\x -> (\y -> (x + y))) 1 2
(\y -> (1 + y)) 2 # substituted 1 for x
(1 + 2) # substituted 2 for y
3
So you can see here that if we'd only supplied a single argument to f, we would have stopped at \y -> (1 + y). So we've got a whole term that is just a function for adding 1 to something, entirely separate from our original term, which may still be in use somewhere (for other references to f).
The key point is that if we implement functions like this, every function has only one argument but some return functions (and some return functions which return functions which return ...). Every time we apply a function we create a new term that "hard-codes" the first argument into the body of the function (including the bodies of any functions this one returns). This is how you get currying and closures.
Now, that's not how Haskell is directly implemented, obviously. Once upon a time, Haskell (or possibly one of its predecessors; I'm not exactly sure on the history) was implemented by Graph reduction. This is a technique for doing something equivalent to the term reduction I described above, that automatically brings along lazy evaluation and a fair amount of data sharing.
In graph reduction, everything is references to nodes in a graph. I won't go into too much detail, but when the evaluation engine reduces the application of a function to a value, it copies the sub-graph corresponding to the body of the function, with the necessary substitution of the argument value for the function's parameter (but shares references to graph nodes where they are unaffected by the substitution). So essentially, yes partially applying a function creates a new structure in memory that has a reference to the supplied argument (i.e. "a pointer to the 2), and your program can pass around references to that structure (and even share it and apply it multiple times), until more arguments are supplied and it can actually be reduced. However it's not like it's just remembering the function and accumulating arguments until it gets all of them; the evaluation engine actually does some of the work each time it's applied to a new argument. In fact the graph reduction engine can't even tell the difference between an application that returns a function and still needs more arguments, and one that has just got its last argument.
I can't tell you much more about the current implementation of Haskell. I believe it's a distant mutant descendant of graph reduction, with loads of clever short-cuts and go-faster stripes. But I might be wrong about that; maybe they've found a completely different execution strategy that isn't anything at all like graph reduction anymore. But I'm 90% sure it'll still end up passing around data structures that hold on to references to the partial arguments, and it probably still does something equivalent to factoring in the arguments partially, as it seems pretty essential to how lazy evaluation works. I'm also fairly sure it'll do lots of optimisations and short cuts, so if you straightforwardly call a function of 5 arguments like f 1 2 3 4 5 it won't go through all the hassle of copying the body of f 5 times with successively more "hard-coding".
Try it out with GHC:
ghc -C Test.hs
This will generate C code in Test.hc
I wrote the following function:
f = (+) 16777217
And GHC generated this:
R1.p[1] = (W_)Hp-4;
*R1.p = (W_)&stg_IND_STATIC_info;
Sp[-2] = (W_)&stg_upd_frame_info;
Sp[-1] = (W_)Hp-4;
R1.w = (W_)&integerzmgmp_GHCziInteger_smallInteger_closure;
Sp[-3] = 0x1000001U;
Sp=Sp-3;
JMP_((W_)&stg_ap_n_fast);
The thing to remember is that in Haskell, partially applying is not an unusual case. There's technically no "last argument" to any function. As you can see here, Haskell is jumping to stg_ap_n_fast which will expect an argument to be available in Sp.
The stg here stands for "Spineless Tagless G-Machine". There is a really good paper on it, by Simon Peyton-Jones. If you're curious about how the Haskell runtime is implemented, go read that first.
Is it impossible to know if two functions are equivalent? For example, a compiler writer wants to determine if two functions that the developer has written perform the same operation, what methods can he use to figure that one out? Or can what can we do to find out that two TMs are identical? Is there a way to normalize the machines?
Edit: If the general case is undecidable, how much information do you need to have before you can correctly say that two functions are equivalent?
Given an arbitrary function, f, we define a function f' which returns 1 on input n if f halts on input n. Now, for some number x we define a function g which, on input n, returns 1 if n = x, and otherwise calls f'(n).
If functional equivalence were decidable, then deciding whether g is identical to f' decides whether f halts on input x. That would solve the Halting problem. Related to this discussion is Rice's theorem.
Conclusion: functional equivalence is undecidable.
There is some discussion going on below about the validity of this proof. So let me elaborate on what the proof does, and give some example code in Python.
The proof creates a function f' which on input n starts to compute f(n). When this computation finishes, f' returns 1. Thus, f'(n) = 1 iff f halts on input n, and f' doesn't halt on n iff f doesn't. Python:
def create_f_prime(f):
def f_prime(n):
f(n)
return 1
return f_prime
Then we create a function g which takes n as input, and compares it to some value x. If n = x, then g(n) = g(x) = 1, else g(n) = f'(n). Python:
def create_g(f_prime, x):
def g(n):
return 1 if n == x else f_prime(n)
return g
Now the trick is, that for all n != x we have that g(n) = f'(n). Furthermore, we know that g(x) = 1. So, if g = f', then f'(x) = 1 and hence f(x) halts. Likewise, if g != f' then necessarily f'(x) != 1, which means that f(x) does not halt. So, deciding whether g = f' is equivalent to deciding whether f halts on input x. Using a slightly different notation for the above two functions, we can summarise all this as follows:
def halts(f, x):
def f_prime(n): f(n); return 1
def g(n): return 1 if n == x else f_prime(n)
return equiv(f_prime, g) # If only equiv would actually exist...
I'll also toss in an illustration of the proof in Haskell (GHC performs some loop detection, and I'm not really sure whether the use of seq is fool proof in this case, but anyway):
-- Tells whether two functions f and g are equivalent.
equiv :: (Integer -> Integer) -> (Integer -> Integer) -> Bool
equiv f g = undefined -- If only this could be implemented :)
-- Tells whether f halts on input x
halts :: (Integer -> Integer) -> Integer -> Bool
halts f x = equiv f' g
where
f' n = f n `seq` 1
g n = if n == x then 1 else f' n
Yes, it is undecidable. This is a form of the halting problem.
Note that I mean that it's undecidable for the general case. Just as you can determine halting for sufficiently simple programs, you can determine equivalency for sufficiently simple functions, and it's not inconceivable that this could be of some use for an application. But you cannot make a general method for determining equivalency of any two possible functions.
The general case is undecidable by Rice's Theorem, as others have already said (Rice's Theorem essentially says that any nontrivial property of a Turing-complete formalism is undecidable).
There are special cases where equivalence is decidable, the best-known example is probably equivalence of finite state automata. If I remember correctly equivalence of pushdown automata is already undecidable by reduction to Post's Correspondence Problem.
To prove that two given functions are equivalent you would require as input a proof of the equivalence in some formalism, which you can then check for correctness. The essential parts of this proof are the loop invariants, as these cannot be derived automatically.
In the general case it's undecidable whether two turing machines have always the same output for the identical input. Since you can't even decide whether a tm will halt on the input, I don't see how it should be possible to decide whether both halt AND output the same result...
It depends on what you mean by "function."
If the functions you are talking about are guaranteed to terminate -- for example, because they are written in a language in which all functions terminate -- and operate over finite domains, it's "easy" (although it might still take a very, very long time): two functions are equivalent if and only if they have the same value at every point in their shared domain.
This is called "extensional" equivalence to distinguish it from syntactic or "intensional" equivalence. Two functions are extensionally equivalent if they are intensionally equivalent, but the converse does not hold.
(All the other people above noting that it is undecidable in the general case are quite correct, of course, this is a fairly uncommon -- and usually uninteresting in practice -- special case.)
Note that the halting problem is decidable for linear bounded automata. Real computers are always bounded, and programs for them will always loop back to a previous configuration after sufficiently many steps. If you are using an unbounded (imaginary) computer to keep track of the configurations, you can detect that looping and take it into account.
You could check in your compiler to see if they are "exactly" identical, sure, but determining if they return identical values would be difficult and time consuming. You would have to basically call that method and perform its routine over an infinite number of possible calls and compare the value with that from the other routine.
Even if you could do the above, you would have to account for what global values change within the function, what objects are destroyed / changed in the function that do not affect the outcome.
You can really only compare the compiled code. So compile the compiled code to refactor?
Imagine the run time on trying to compile the code with "that" compiler. You could spend a LOT of time on here answering questions saying: "busy compiling..." :)
I think if you allow side effects, you can show that the problem can be morphed into the Post correspondence problem so you can't, in general, show if two functions are even capable of having the same side effects.
Is it impossible to know if two functions are equivalent?
No. It is possible to know that two functions are equivalent. If you have f(x), you know f(x) is equivalent to f(x).
If the question is "it is possible to determine if f(x) and g(x) are equivalent with f and g being any function and for all functions g and f", then the answer is no.
However, if the question is "can a compiler determine that if f(x) and g(x) are equivalent that they are equivalent?", then the answer is yes if they are equivalent in both output and side effects and order of side effects. In other words, if one is a transformation of the other that preserves behavior, then a compiler of sufficient complexity should be able to detect it. It also means that the compiler can transform a function f into a more optimal and equivalent function g given a particular definition of equivalent. It gets even more fun if f includes undefined behavior, because then g can also include undefined (but different) behavior!