How does currying work? - haskell

I'm very new to Haskell and FP in general. I've read many of the writings that describe what currying is, but I haven't found an explanation to how it actually works.
Here is a function: (+) :: a -> (a -> a)
If I do (+) 4 7, the function takes 4 and returns a function that takes 7 and returns 11. But what happens to 4 ? What does that first function do with 4? What does (a -> a) do with 7?
Things get more confusing when I think about a more complicated function:
max' :: Int -> (Int -> Int)
max' m n | m > n = m
| otherwise = n
what does (Int -> Int) compare its parameter to? It only takes one parameter, but it needs two to do m > n.

Understanding higher-order functions
Haskell, as a functional language, supports higher-order functions (HOFs). In mathematics HOFs are called functionals, but you don't need any mathematics to understand them. In usual imperative programming, like in Java, functions can accept values, like integers and strings, do something with them, and return back a value of some other type.
But what if functions themselves were no different from values, and you could accept a function as an argument or return it from another function? f a b c = a + b - c is a boring function, it sums a and b and then substracts c. But the function could be more interesting, if we could generalize it, what if we'd want sometimes to sum a and b, but sometimes multiply? Or divide by c instead of subtracting?
Remember, (+) is just a function of 2 numbers that returns a number, there's nothing special about it, so any function of 2 numbers that returns a number could be in place of it. Writing g a b c = a * b - c, h a b c = a + b / c and so on just doesn't cut it for us, we need a general solution, we are programmers after all! Here how it is done in Haskell:
let f g h a b c = a `g` b `h` c in f (*) (/) 2 3 4 -- returns 1.5
And you can return functions too. Below we create a function that accepts a function and an argument and returns another function, which accepts a parameter and returns a result.
let g f n = (\m -> m `f` n); f = g (+) 2 in f 10 -- returns 12
A (\m -> m `f` n) construct is an anonymous function of 1 argument m that applies f to that m and n. Basically, when we call g (+) 2 we create a function of one argument, that just adds 2 to whatever it receives. So let f = g (+) 2 in f 10 equals 12 and let f = g (*) 5 in f 5 equals 25.
(See also my explanation of HOFs using Scheme as an example.)
Understanding currying
Currying is a technique that transforms a function of several arguments to a function of 1 argument that returns a function of 1 argument that returns a function of 1 argument... until it returns a value. It's easier than it sounds, for example we have a function of 2 arguments, like (+).
Now imagine that you could give only 1 argument to it, and it would return a function? You could use this function later to add this 1st argument, now encased in this new function, to something else. E.g.:
f n = (\m -> n - m)
g = f 10
g 8 -- would return 2
g 4 -- would return 6
Guess what, Haskell curries all functions by default. Technically speaking, there are no functions of multiple arguments in Haskell, only functions of one argument, some of which may return new functions of one argument.
It's evident from the types. Write :t (++) in interpreter, where (++) is a function that concatenates 2 strings together, it will return (++) :: [a] -> [a] -> [a]. The type is not [a],[a] -> [a], but [a] -> [a] -> [a], meaning that (++) accepts one list and returns a function of type [a] -> [a]. This new function can accept yet another list, and it will finally return a new list of type [a].
That's why function application syntax in Haskell has no parentheses and commas, compare Haskell's f a b c with Python's or Java's f(a, b, c). It's not some weird aesthetic decision, in Haskell function application goes from left to right, so f a b c is actually (((f a) b) c), which makes complete sense, once you know that f is curried by default.
In types, however, the association is from right to left, so [a] -> [a] -> [a] is equivalent to [a] -> ([a] -> [a]). They are the same thing in Haskell, Haskell treats them exactly the same. Which makes sense, because when you apply only one argument, you get back a function of type [a] -> [a].
On the other hand, check the type of map: (a -> b) -> [a] -> [b], it receives a function as its first argument, and that's why it has parentheses.
To really hammer down the concept of currying, try to find the types of the following expressions in the interpreter:
(+)
(+) 2
(+) 2 3
map
map (\x -> head x)
map (\x -> head x) ["conscience", "do", "cost"]
map head
map head ["conscience", "do", "cost"]
Partial application and sections
Now that you understand HOFs and currying, Haskell gives you some syntax to make code shorter. When you call a function with 1 or multiple arguments to get back a function that still accepts arguments, it's called partial application.
You understand already that instead of creating anonymous functions you can just partially apply a function, so instead of writing (\x -> replicate 3 x) you can just write (replicate 3). But what if you want to have a divide (/) operator instead of replicate? For infix functions Haskell allows you to partially apply it using either of arguments.
This is called sections: (2/) is equivalent to (\x -> 2 / x) and (/2) is equivalent to (\x -> x / 2). With backticks you can take a section of any binary function: (2`elem`) is equivalent to (\xs -> 2 `elem` xs).
But remember, any function is curried by default in Haskell and therefore always accepts one argument, so sections can be actually used with any function: let (+^) be some weird function that sums 4 arguments, then let (+^) a b c d = a + b + c in (2+^) 3 4 5 returns 14.
Compositions
Other handy tools to write concise and flexible code are composition and application operator. Composition operator (.) chains functions together. Application operator ($) just applies function on the left side to the argument on the right side, so f $ x is equivalent to f x. However ($) has the lowest precedence of all operators, so we can use it to get rid of parentheses: f (g x y) is equivalent to f $ g x y.
It is also helpful when we need to apply multiple functions to the same argument: map ($2) [(2+), (10-), (20/)] would yield [4,8,10]. (f . g . h) (x + y + z), f (g (h (x + y + z))), f $ g $ h $ x + y + z and f . g . h $ x + y + z are equivalent, but (.) and ($) are different things, so read Haskell: difference between . (dot) and $ (dollar sign) and parts from Learn You a Haskell to understand the difference.

You can think of it like that the function stores the argument and returns a new function that just demands the other argument(s). The new function already knows the first argument, as it is stored together with the function. This is handled internally by the compiler. If you want to know how this works exactly, you may be interested in this page although it may be a bit complicated if you are new to Haskell.
If a function call is fully saturated (so all arguments are passed at the same time), most compilers use an ordinary calling scheme, like in C.

Does this help?
max' = \m -> \n -> if (m > n)
then m
else n
Written as lambdas. max' is a value of a lambda that itself returns a lambda given some m, which returns the value.
Hence max' 4 is
max' 4 = \n -> if (4 > n)
then 4
else n

Something that may help is to think about how you could implement curry as a higher order function if Haskell didn't have built in support for it. Here is a Haskell implementation that works for a function on two arguments.
curry :: (a -> b -> c) -> a -> (b -> c)
curry f a = \b -> f a b
Now you can pass curry a function on two arguments and the first argument and it will return a function on one argument (this is an example of a closure.)
In ghci:
Prelude> let curry f a = \b -> f a b
Prelude> let g = curry (+) 5
Prelude> g 10
15
Prelude> g 15
20
Prelude>
Fortunately we don't have to do this in Haskell (you do in Lisp if you want currying) because support is built into the language.

If you come from C-like languages, their syntax might help you to understand it. For example in PHP the add function could be implemented as such:
function add($a) {
return function($b) use($a) {
return $a + $b;
};
}

Haskell is based on Lambda calculus. Internally what happens is that everything gets converted into a function. So your compiler evaluates (+) as follows
(+) :: Num a => a -> a -> a
(+) x y = \x -> (\y -> x + y)
That is, (+) :: a -> a -> a is essentially the same as (+) :: a -> (a -> a). Hope this helps.

Related

Is a section the result of currying?

In Programming in Haskell by Hutton
In general, if # is an operator, then expressions of the form (#), (x #), and (# y) for arguments x and
y are called sections, whose meaning as functions can be
formalised using lambda expressions as follows:
(#) = \x -> (\y -> x # y)
(x #) = \y -> x # y
(# y) = \x -> x # y
What are the difference and relation between "section" and "currying"?
Is a section the result of applying the currying operation to a multi-argument function?
Thanks.
A section is just special syntax for applying an infix operator to a single argument. (# y) is the more useful of the two, as (x #) is equivalent to (#) x (which is just applying the infix operator as a function to a single argument in the usual fashion).
curry f x y = f (x,y). uncurry g (x,y) = g x y.
(+ 3) 4 = (+) 4 3 = 4 + 3. (4 +) 3 = (+) 4 3 = 4 + 3.
A section is a result of partial application of a curried function: (+ 3) = flip (+) 3, (4 +) = (+) 4.
A curried function (like g or (+)) expects its arguments one at a time. An uncurried function (like f) expects its arguments in a tuple.
To partially apply an uncurried function we have first to turn it into a curried function, with curry. To partially apply a curried function we don't need to do anything, just apply it to an argument.
curry :: ((a, b) -> c ) -> ( a -> (b -> c))
uncurry :: (a -> (b -> c)) -> ((a, b) -> c )
x :: a
g :: a -> (b -> c)
--------------------
g x :: b -> c
x :: a
f :: (a, b) -> c
---------------------------
curry f :: a -> (b -> c)
curry f x :: b -> c
Left sections and right sections are syntactical devices for partially applying an infix operator to a single argument (see also chepner's answer). For the sake of accuracy, we should note that currying is not the same thing as partial application:
Currying is converting a function that takes N arguments into a function that takes a single argument and returns a function that takes N-1 arguments.
Partial application is making a function that takes N-1 arguments out of a function that takes N arguments by supplying one of the arguments.
In Haskell, it happens that everything is curried; all functions take just one argument (even uncurried functions in Haskell take a tuple, which is, strictly speaking, a single argument -- you might want to play with the curry and uncurry functions to see how that works). Still, we very often think informally of functions that return functions as functions of multiple arguments. From that vantage point, a nice consequence of currying by default is that partial application of a function to its first argument becomes trivial: while, for instance, elem takes a value and a container and tests if the value is an element of the contaier, elem "apple" takes a container (of strings) and tests if "apple" is an element of it.
As for operators, when we write, for instance...
5 / 2
... we are applying the operator / to the arguments 5 and 2. The operator can also be used in prefix form, rather than infix:
(/) 5 2
In prefix form, the operator can be partially applied in the usual way:
(/) 5
That, however, arguably looks a little awkward -- after all, 5 here is the numerator, and not the denominator. I'd say left section syntax is easier on the eye in this case:
(5 /)
Furthermore, partial application to the second argument is not quite as straightforward to write, requiring a lambda, or flip. In the case of operators, a right section can help with that:
(/ 2)
Note that sections also work with functions made into operators through backtick syntax, so this...
(`elem` ["apple", "grape", "orange"])
... takes a string and tests whether it can be found in ["apple", "grape", "orange"].

Understanding `ap` in a point-free function in Haskell

I am able to understand the basics of point-free functions in Haskell:
addOne x = 1 + x
As we see x on both sides of the equation, we simplify it:
addOne = (+ 1)
Incredibly it turns out that functions where the same argument is used twice in different parts can be written point-free!
Let me take as a basic example the average function written as:
average xs = realToFrac (sum xs) / genericLength xs
It may seem impossible to simplify xs, but http://pointfree.io/ comes out with:
average = ap ((/) . realToFrac . sum) genericLength
That works.
As far as I understand this states that average is the same as calling ap on two functions, the composition of (/) . realToFrac . sum and genericLength
Unfortunately the ap function makes no sense whatsoever to me, the docs http://hackage.haskell.org/package/base-4.8.1.0/docs/Control-Monad.html#v:ap state:
ap :: Monad m => m (a -> b) -> m a -> m b
In many situations, the liftM operations can be replaced by uses of ap,
which promotes function application.
return f `ap` x1 `ap` ... `ap` xn
is equivalent to
liftMn f x1 x2 ... xn
But writing:
let average = liftM2 ((/) . realToFrac . sum) genericLength
does not work, (gives a very long type error message, ask and I'll include it), so I do not understand what the docs are trying to say.
How does the expression ap ((/) . realToFrac . sum) genericLength work? Could you explain ap in simpler terms than the docs?
Any lambda term can be rewritten to an equivalent term that uses just a set of suitable combinators and no lambda abstractions. This process is called abstraciton elimination. During the process you want to remove lambda abstractions from inside out. So at one step you have λx.M where M is already free of lambda abstractions, and you want to get rid of x.
If M is x, you replace λx.x with id (id is usually denoted by I in combinatory logic).
If M doesn't contain x, you replace the term with const M (const is usually denoted by K in combinatory logic).
If M is PQ, that is the term is λx.PQ, you want to "push" x inside both parts of the function application so that you can recursively process both parts. This is accomplished by using the S combinator defined as λfgx.(fx)(gx), that is, it takes two functions and passes x to both of them, and applies the results together. You can easily verify that that λx.PQ is equivalent to S(λx.P)(λx.Q), and we can recursively process both subterms.
As described in the other answers, the S combinator is available in Haskell as ap (or <*>) specialized to the reader monad.
The appearance of the reader monad isn't accidental: When solving the task of replacing λx.M with an equivalent function is basically lifting M :: a to the reader monad r -> a (actually the reader Applicative part is enough), where r is the type of x. If we revise the process above:
The only case that is actually connected with the reader monad is when M is x. Then we "lift" x to id, to get rid of the variable. The other cases below are just mechanical applications of lifting an expression to an applicative functor:
The other case λx.M where M doesn't contain x, it's just lifting M to the reader applicative, which is pure M. Indeed, for (->) r, pure is equivalent to const.
In the last case, <*> :: f (a -> b) -> f a -> f b is function application lifted to a monad/applicative. And this is exactly what we do: We lift both parts P and Q to the reader applicative and then use <*> to bind them together.
The process can be further improved by adding more combinators, which allows the resulting term to be shorter. Most often, combinators B and C are used, which in Haskell correspond to functions (.) and flip. And again, (.) is just fmap/<$> for the reader applicative. (I'm not aware of such a built-in function for expressing flip, but it'd be viewed as a specialization of f (a -> b) -> a -> f b for the reader applicative.)
Some time ago I wrote a short article about this: The Monad Reader Issue 17, The Reader Monad and Abstraction Elimination.
When the monad m is (->) a, as in your case, you can define ap as follows:
ap f g = \x -> f x (g x)
We can see that this indeed "works" in your pointfree example.
average = ap ((/) . realToFrac . sum) genericLength
average = \x -> ((/) . realToFrac . sum) x (genericLength x)
average = \x -> (/) (realToFrac (sum x)) (genericLength x)
average = \x -> realToFrac (sum x) / genericLength x
We can also derive ap from the general law
ap f g = do ff <- f ; gg <- g ; return (ff gg)
that is, desugaring the do-notation
ap f g = f >>= \ff -> g >>= \gg -> return (ff gg)
If we substitute the definitions of the monad methods
m >>= f = \x -> f (m x) x
return x = \_ -> x
we get the previous definition of ap back (for our specific monad (->) a). Indeed:
app f g
= f >>= \ff -> g >>= \gg -> return (ff gg)
= f >>= \ff -> g >>= \gg -> \_ -> ff gg
= f >>= \ff -> g >>= \gg _ -> ff gg
= f >>= \ff -> \x -> (\gg _ -> ff gg) (g x) x
= f >>= \ff -> \x -> (\_ -> ff (g x)) x
= f >>= \ff -> \x -> ff (g x)
= f >>= \ff x -> ff (g x)
= \y -> (\ff x -> ff (g x)) (f y) y
= \y -> (\x -> f y (g x)) y
= \y -> f y (g y)
The Simple Bit: fixing liftM2
The problem in the original example is that ap works a bit differently from the liftM functions. ap takes a function wrapped up in a monad, and applies it to an argument wrapped up in a monad. But the liftMn functions take a "normal" function (one which is not wrapped up in a monad) and apply it to argument(s) that are wrapped up in monads.
I'll explain more about what that means below, but the upshot is that if you want to use liftM2, then you have to pull (/) out and make it a separate argument at the beginning. (So in this case (/) is the "normal" function.)
let average = liftM2 ((/) . realToFrac . sum) genericLength -- does not work
let average = liftM2 (/) (realToFrac . sum) genericLength -- works
As posted in the original question, calling liftM2 should involve three agruments: liftM2 f x1 x2. Here the f is (/), x1 is (realToFrac . sum) and x2 is genericLength.
The version posted in the question (the one which doesn't work) was trying to call liftM2 with only two arguments.
The explanation
I'll build this up in a few stages. I'll start with some specific values, and build up to a function that can take any set of values. Jump to the last section for the TL:DR
In this example, lets assume the list of numbers is [1,2,3,4]. The sum of these numbers is 10, and the length of the list is 4. The average is 10/4 or 2.5.
To shoe-horn this into the right form for ap, we're going to break this into a function, an input, and a result.
ourFunction = (10/) -- "divide 10 by"
ourInput = 4
ourResult = 2.5
Three kinds of Function Application
ap and listM both involve monads. At this point in the explanation, you can think of a monad as something that a value can be "wrapped up in". I'll give a better definition below.
Normal function application applies a normal function to a normal input. liftM applies a normal function to an input wrapped in a monad, and ap applies a function wrapped in a monad to an input wrapped in a monad.
(10/) 4 -- returns 2.5
liftM (10/) monad(4) -- returns monad(2.5)
ap monad(10/) monad(4) -- returns monad(2.5)
(Note that this is pseudocode. monad(4) is not actually valid Haskell).
(Note that liftM is a different function from liftM2, which was used earlier. liftM takes a function and only one argument, which is a better fit for the pattern i'm describing.)
In the average function defined above, the monads were functions, but "functions-as-monads" can be hard to talk about, so I'll start with simpler examples.
So what's a monad?
A better description of a monad is "something which contains a value, or produces a value, or which you can somehow extract a value from, but which also has something more complicated going on."
That's a really vague description, but it kind of has to be, because the "something more complicated" can be a lot of different things.
Monads can be confusing, but the point of them is that when you use monad operations (like ap and liftM) they will take care of the "something more complicated" for you, so you can just concentrate on the values.
That's probably still not very clear, so let's do some examples:
The Maybe monad
ap (Just (10/)) (Just 4) -- result is (Just 2.5)
One of the simplest monads is 'Maybe'. The value is whatever is contained inside a Just. So if we call ap and give it (Just ourFunction) and (Just ourInput) then we get back (Just ourResult).
The "something more complicated" is the fact that there might not be a value there at all, and you have to allow for the Nothing case.
As mentioned, the point of using a function like ap is that it takes care of these extra complications for us. With the Maybe monad, ap handles this by returning Nothing if either the Maybe-function or the Maybe-input were Nothing.
ap (Just (10/)) Nothing -- result is Nothing
ap Nothing (Just 4) -- result is Nothing
The List Monad
ap [(10/)] [4] -- result is [2.5]
With the list Monad, the value is whatever is inside the list. So ap [ourfunction] [ourInput] returns [ourResult].
The "something more complicated" is that there may be more than one thing inside the list (or exactly one thing, or nothing at all).
With lists, that means ap takes a list of zero or more functions, and a list of zero or more inputs. It handles that by returning a list of zero or more results: one result for every possible combination of function and input.
ap [(10/), (100/)] [5,4,2] -- result is [2.0, 2.5, 5.0, 20.0, 25.0, 50.0]
Functions as Monads
A function like genericLength is considered a Monad because it has a value (the function's output), and it has a "something more complicated" (the fact that you have to supply an input before you can get the value).
This is where it gets a little confusing, because we're dealing with multiple functions, multiple inputs, and multiple results. It is all well defined, it's just hard to describe, so we have to be careful with our terminology.
Lets start with the list [1,2,3,4], and call that our "original input". That's the list we're trying to find the average of. It's the xs argument in the original average function.
If we give our original input ([1,2,3,4]) to genericLength then we get a value of '4'.
Our other function is ((/) . realToFrac . sum). It takes our list [1,2,3,4] and finds the sum (10), turns that into a fractional value, and then feeds it as the first argument to (/). The result is an incomplete division function that is waiting for another argument. ie it takes [1,2,3,4] as an input, and produces (10/) as its output.
This all fits with the way ap is defined for functions. With functions, ap takes two things. The first is a function that reads the original input and produces a new function. The second is a function that reads the original input and produces a new input. The final result is a function that takes the original input, and returns the same thing you would get if you applied the new function to the new input.
You might have to read that a few times to make sense of it. Alternatively, here it is in pseudocode:
average =
ap
(functionThatTakes [1,2,3,4] and returns "(10/)" )
(functionThatTakes [1,2,3,4] and returns " 4 " )
-- which means:
average =
(functionThatTakes [1,2,3,4] and returns "2.5" )
If you compare this to the simpler examples above, you'll see that it still has our function (10/), our input 4 and our result 2.5. And each of them is once again wrapped up in the "something more complicated". In this case, the "something more complicated" is the "function that takes [1,2,3,4] and returns...".
Of course, since they're functions, they don't have to take [1,2,3,4] as their input. If they took a different list of integers (eg [1,2,3,4,5]) then we would get different results (e.g. new function: (15/), new input 5 and new value 3).
Other examples
minPlusMax = ap ((+) . minimum) maximum
-- a function that adds the minimum element of a list, to the maximum element
upperAndLower = ap ((,) . toUpper) toLower
-- a function that takes a Char and returns a tuple, with the upper case and lower case versions of a character
These could all also be defined using liftM2.
average = liftM2 (/) sum genericLength
minPlusMax = liftM2 (+) minimum maximum
upperAndLower = liftM2 (,) toUpper toLower

Haskell High order functions. passing two functions as arguments.

I have recently been teaching myself Haskell, and one of my exercises was to implement a function that takes two functions as arguments and passes number 3 to the first function and the result to the second function. I have more experience with racket. The function that i created for racket is
(define (nestfun3 function function2)
(function2 (function 3)))
I'm trying to replicate this function in racket.
i know that high order functions with only one function as argument can be like
twice function = function . function
How can I pass two functions? I tried
twice function = function . function
three function = function . twice
the direct translation would be:
apply3 :: Num a => (a -> b) -> (b -> c) -> c
apply3 f g = g (f 3)
remember: you don't want to just compose functions - you want to first apply 3 and then apply the result to the other function
sorry for the mixing ... this should work now:
λ> let f = (+1)
λ> let g = (*2)
λ> apply3 f g
8
λ> apply3 g f
7
in case you wondered - you don't have to actually start with the signature (although it's good practice to add it) - if you do this in ghci:
λ> let apply3 f g = g (f 3)
λ> :t apply3
apply3 :: Num a => (a -> t1) -> (t1 -> t) -> t
it will give it do you (or you can use ghc-mod or whatever) - I actually just renamed the type-parameters (those t are ugly)
This way it should look very similar to what you did in Scheme
of course you might want to make sure not to swap functions while renaming :|

Composing a chain of 2-argument functions

So I have a list of a functions of two arguments of the type [a -> a -> a]
I want to write a function which will take the list and compose them into a chain of functions which takes length+1 arguments composed on the left. For example if I have [f,g,h] all of types [a -> a -> a] I need to write a function which gives:
chain [f,g,h] = \a b c d -> f ( g ( h a b ) c ) d
Also if it helps, the functions are commutative in their arguments ( i.e. f x y = f y x for all x y ).
I can do this inside of a list comprehension given that I know the the number of functions in question, it would be almost exactly like the definition. It's the stretch from a fixed number of functions to a dynamic number that has me stumped.
This is what I have so far:
f xs = f' xs
where
f' [] = id
f' (x:xs) = \z -> x (f' xs) z
I think the logic is along the right path, it just doesn't type-check.
Thanks in advance!
The comment from n.m. is correct--this can't be done in any conventional way, because the result's type depends on the length of the input list. You need a much fancier type system to make that work. You could compromise in Haskell by using a list that encodes its length in the type, but that's painful and awkward.
Instead, since your arguments are all of the same type, you'd be much better served by creating a function that takes a list of values instead of multiple arguments. So the type you want is something like this: chain :: [a -> a -> a] -> [a] -> a
There are several ways to write such a function. Conceptually you want to start from the front of the argument list and the end of the function list, then apply the first function to the first argument to get something of type a -> a. From there, apply that function to the next argument, then apply the next function to the result, removing one element from each list and giving you a new function of type a -> a.
You'll need to handle the case where the list lengths don't match up correctly, as well. There's no way around that, other than the aforementioned type-encoded-lengths and the hassle associate with such.
I wonder, whether your "have a list of a functions" requirement is a real requirement or a workaround? I was faced with the same problem, but in my case set of functions was small and known at compile time. To be more precise, my task was to zip 4 lists with xor. And all I wanted is a compact notation to compose 3 binary functions. What I used is a small helper:
-- Binary Function Chain
bfc :: (c -> d) -> (a -> b -> c) -> a -> b -> d
bfc f g = \a b -> f (g a b)
For example:
ghci> ((+) `bfc` (*)) 5 3 2 -- (5 * 3) + 2
17
ghci> ((+) `bfc` (*) `bfc` (-)) 5 3 2 1 -- ((5 - 3) * 2) + 1
5
ghci> zipWith3 ((+) `bfc` (+)) [1,2] [3,4] [5,6]
[9,12]
ghci> getZipList $ (xor `bfc` xor `bfc` xor) <$> ZipList [1,2] <*> ZipList [3,4] <*> ZipList [5,6] <*> ZipList [7,8]
[0,8]
That doesn't answers the original question as it is, but hope still can be helpful since it covers pretty much what question subject line is about.

The composition of functions in a list of functions!

I need to define a function 'Compose' which takes a list 'L' which is a list of functions. When I specify a parameter that will suit all the functions in the list, the last function evaluates itself using this param. The result is then passed to the second last function and so on until we get to the first item (function) in the list and we get the final result.
E.g.
Compose ( ( fn N -> N + 1 ) ^ ( fn N -> 2 * N ) ^ # ) 3 .
give the answer 7.
I have to write this in a functional programming language called SAL (simple applicative language) devised by a lecturer in my college (hence funny syntax above ( ^ seperates list items and # marks end of list)).
If any solutions could be written in pseudo-code bearing in mind I can't use loops, variables etc. that would be much appreciated. Apparently the solution is a one-line answer. I imagine it involves recursion (99% of our task functions do!).
Also I don't understand Haskell (guess I'll have to learn!) so psuedo code or even plain English would be great. –
Thanks a bunch.
If the solution is a one-line answer, it could be something involving a fold:
compose :: [a -> a] -> a -> a
compose fs v = foldl (flip (.)) id fs $ v
http://haskell.org/haskellwiki/Compose
You can also implement it as a right fold, which works the way you want:
compose = foldr (.) id
*Main> let compose = foldr (.) id
*Main> compose [\x -> x+1, \x -> 2 * x, id] 3
7
in haskell:
compose :: a -> [a -> a] -> a
compose a (x:xs) = x (compose a xs)
compose a [] = a
Dan kind of gives it away, but here's a hint on how to do it yourself. You can recurse over numbers:
0! = 1
n! = (n-1)! * n
You can also recurse over structure. A list, for example, has a recursive structure, broken down into two cases: an empty list, and an item followed by the rest of the list. In no particular language:
List := Item x List
| Nil
Item marks the head of the list, x is the value stored in the head, and List is the tail. In this grammar, your list would be written:
Item ( fn N -> N + 1 ) Item ( fn N -> 2 * N ) Nil
The rule for a list in the syntax your professor invented could be written recursively as:
List := x ^ List
| #
A function on a list must recurse over this structure, which means it handles each of the two cases:
sum l:List = Nil -> 0
| Item x xs:List = x + sum xs
The recursion, specifically, is the term sum l:List = x + sum xs. Writing this function using the professor's syntax left as an exercise.
In your problem, your metafunction takes a list of functions and must return a function. Consider each case, the empty list and an item (the head) followed by a list (the tail). In the latter case, you can recursively use your function to get a function from the tail, then combine that somehow with the head to return a function. That's the gist, at any rate.
The same using monoids, point-free
import Data.Monoid
compose :: [a -> a] -> a -> a
compose = appEndo . mconcat . map Endo
Or somewhat more generally:
import Data.Monoid
compose :: (Functor t, Foldable t) => t (a -> a) -> a -> a
compose = appEndo . foldl1 (<>) . fmap Endo
Here's what I used:
compose :: [a -> a] -> a -> a
compose list startingvalue = foldl (\x f -> f x) startingvalue list

Resources