I want to write a function that will take an arbitrary number of (curried) arguments and simply print them out (or perform some other unspecified action with them). Here is what I have come up with:
let print arg =
let rec print args arg =
if not (FSharpType.IsFunction(typeof<'t>)) then
printfn "%A" args
Unchecked.defaultof<'t>
else
print (box arg::args)
print []
When I try to compile this I get the error The resulting type would be infinite when unifying ''t' and ''a -> 't.
I know I could just pass the arguments as a list, but I am trying to develop an API of sorts where this would be a useful idiom to have.
Is there some clever compiler trick to make such a function possible in F# or is it a lost cause?
It seems that the two branches of the inner print want to return different types: the "then" part wants to return 't, but the "else" part wants to return 'a -> 't, where 't is necessarily the same in both branches. That is, your function tries to return either its own return type or a function from another type to its own return type. Such combined return type would, indeed, be infinite, which is in perfect accordance with what you set out to do - namely, create a function with infinite number of arguments. Though I do not know how to formally prove it, I would say this is indeed impossible.
If your goal is to simply create a list of boxed values, you could get away with defining a few infix operators.
let (<+>) a b = a # [(box b)]
let (<&>) a b = [(box a); (box b)]
let xs = 5 <&> "abc" <+> 3.0 <+> None <+> true
>> val xs : obj list = [5; "abc"; 3.0; null; true]
Alternatively, with carefully chosen operator precedence, you can apply a function (but then you'll need a terminator):
let (^>) a b = (box a)::b
let (<&>) f xs = f xs
let print xs = sprintf "%A" xs
let xs = print <&> 5 ^> "abc" ^> 3.0 ^> None ^> true ^> []
>> val xs : string = "[5; "abc"; 3.0; null; true]"
Related
I have 2 Optionals (or Maybe objects) that I would like to combine so that I get the following results:
|| first operand
second ++-------------+-------------
operand || empty | optional(x)
============||=============|=============
empty || empty | optional(x)
------------++-------------+-------------
optional(y) || optional(y) |optional(x+y)
In other words, a non-empty Optional always replaces/overwrites an empty one, and two non-empty Optionals are combined according to some + function.
Initially, I assumed that the standard monadic flatMap method would do the trick, but (at least in Java) Optional.flatMap always returns an empty optional when the original Optional was already empty (and I'm not sure if any other implementation would comply with the Monad Laws).
Then, as both operands are wrapped in the same monadic type, I figured that this might be a good job for an Applicative Functor. I tried a couple different functional libraries, but I couldn't implement the desired behavior with any of the zip/ap methods that I tried.
What I'm trying to do seems to me a fairly common operation that one might do with Optionals, and I realize that I could just write my own operator with the desired behavior. Still, I am wondering if there is a standard function/method in functional programming to achieve this common operation?
Update: I removed the java tag, as I'm curious how other languages handle this situation
In a functional language, you'd do this with pattern matching, such as (Haskell):
combine :: Maybe t -> Maybe t -> (t -> t -> t) -> Maybe t
combine (Some x) (Some y) f = Some (f x y)
combine (Some x) _ _ = (Some x)
combine _ (Some y) _ = (Some y)
combine None None _ = None
There are other ways to write it, but you are basically pattern matching on the cases. Note that this still involves "unpacking" the optionals, but because its built into the language, it is less obvious.
In Haskell you can do this by wrapping any semigroup in a Maybe. Specifically, if you want to add numbers together:
Prelude> import Data.Semigroup
Prelude Data.Semigroup> Just (Sum 1) <> Just (Sum 2)
Just (Sum {getSum = 3})
Prelude Data.Semigroup> Nothing <> Just (Sum 2)
Just (Sum {getSum = 2})
Prelude Data.Semigroup> Just (Sum 1) <> Nothing
Just (Sum {getSum = 1})
Prelude Data.Semigroup> Nothing <> Nothing
Nothing
The above linked article contains more explanations, and also some C# examples.
It's not possible to combine optional objects without "unpacking" them.
I don't know the specifics of your case. For me, creating such a logic just in order to fuse the two optionals is an overkill.
But nevertheless, there's a possible solution with streams.
I assume that you're not going to pass optional objects as arguments (because such practice is discouraged). Therefore, there are two dummy methods returning Optional<T>.
Method combine() expects a BinaryOperator<T> as an argument and creates a stream by concatenating singleton-streams produced from each of the optional objects returned by getX() and getY().
The flavor of reduce(BinaryOperator) will produce an optional result.
public static <T> Optional<T> getX(Class<T> t) {
return // something
}
public static <T> Optional<T> getY(Class<T> t) {
return // something
}
public static <T> Optional<T> combine(BinaryOperator<T> combiner,
Class<T> t) {
return Stream.concat(getX(t).stream(), getY(t).stream())
.reduce(combiner);
}
If we generalize the problem to "how to combine N optional objects" then it can be solved like this:
#SafeVarargs
public static <T> Optional<T> combine(BinaryOperator<T> combiner,
Supplier<Optional<T>>... suppliers) {
return Arrays.stream(suppliers)
.map(Supplier::get) // fetching Optional<T>
.filter(Optional::isPresent) // filtering optionals that contain results to avoid NoSuchElementException while invoking `get()`
.map(Optional::get) // "unpacking" optionals
.reduce(combiner);
}
Here's one way:
a.map(x -> b.map(y -> x + y).orElse(x)).or(() -> b)
Ideone Demo
OptionalInt x = ...
OptionalInt y = ...
OptionalInt sum = IntStream.concat(x.stream(), y.stream())
.reduce(OptionalInt.empty(),
(opt, z) -> OptionalInt.of(z + opt.orElse(0)));
Since java 9 you can turn an Optional into a Stream.
With concat you get a Stream of 0, 1 or 2 elements.
Reduce it to an empty when 0 elements,and for more add it to the previous OptionalInt, defaulting to 0.
Not very straight (.sum()) because of the need for an empty().
You can implement your function in Java by combining flatMap and map:
optA.flatMap(a -> optB.map(b -> a + b));
More general example:
public static void main(String[] args) {
test(Optional.empty(), Optional.empty());
test(Optional.of(3), Optional.empty());
test(Optional.empty(), Optional.of(4));
test(Optional.of(3), Optional.of(4));
}
static void test(Optional<Integer> optX, Optional<Integer> optY) {
final Optional<Integer> optSum = apply(Integer::sum, optX, optY);
System.out.println(optX + " + " + optY + " = " + optSum);
}
static <A, B, C> Optional<C> apply(BiFunction<A, B, C> fAB, Optional<A> optA, Optional<B> optB) {
return optA.flatMap(a -> optB.map(b -> fAB.apply(a, b)));
}
Since flatMap and map are standard functions for Optional/Maybe (and monad types generally), this approach should work in any other language (though most FP languages will have a more concise solution). E.g. in Haskell:
combine ma mb = do a <- ma ; b <- mb ; return (a + b)
In F#, i would call this logic reduce.
Reason:
The function must be of type 'a -> 'a -> 'a as it only can combine thinks of equal type.
Like other reduce operations, like on list, you always need at least one value, otherwise it fails.
With a option and two of them, you just need to cover four cases. In F# it will be written this way.
(* Signature: ('a -> 'a -> 'a) -> option<'a> -> option<'a> -> option<'a> *)
let reduce fn x y =
match x,y with
| Some x, Some y -> Some (fn x y)
| Some x, None -> Some x
| None , Some y -> Some y
| None , None -> None
printfn "%A" (reduce (+) (Some 3) (Some 7)) // Some 10
printfn "%A" (reduce (+) (None) (Some 7)) // Some 7
printfn "%A" (reduce (+) (Some 3) (None)) // Some 3
printfn "%A" (reduce (+) (None) (None)) // None
In another lets say Pseudo-like C# language, it would look like.
Option<A> Reduce(Action<A,A,A> fn, Option<A> x, Option<A> y) {
if ( x.isSome ) {
if ( y.isSome ) {
return Option.Some(fn(x.Value, y.Value));
}
else {
return x;
}
}
else {
if ( y.isSome ) {
return y;
}
else {
return Option.None;
}
}
}
If I have one or more recursive functions inside an Ocaml function how can I call them without exit from the main function taking their value as return of the main function?
I'm new in Ocaml so I'll try to explain me better...
If I have :
let function =
let rec recursive1 = ...
...
let rec recursive2 = ...
...
How can I call them inside function to tell it "Hey, do you see this recursive function? Now call it and takes its value."
Because my problem is that Ocaml as return of my functions sees Unit instead of the right return.
I will post the code below :
let change k v list_ =
let rec support k v list_ =
match list_ with
| [] -> []
| (i,value) :: tl -> if i = k
then (k,v) :: tl
else (i,value) :: support k v tl in
let inserted = support k v list_ in inserted
let () =
let k = [ (1,"ciao");(2,"Hola");(3,"Salut") ] in
change 2 "Aufwidersen" k
Change takes as input a key, a value and a (int * string )list and should return the same list of the input but changing the value linked to the key selected ( if in list ).
support, instead, makes the dirty job. It builds a new list and when k is found i = k it changes value and attach the tile, closing the function.
The return of change is unit when it should be (int * string) list. I think because inserted isn't taken as return of the function.
change does not return unit. The error in fact tells you exactly the opposite, that it returns (int * string) list but that it expects unit. And it expects unit because you're assigning it to a () pattern.
I don't know what you actually intend to do with the return value, as right now you don't seem to care about it, but you can fix the error by just assigning it to a name:
let result: (int * string) list =
let k = [ (1,"ciao");(2,"Hola");(3,"Salut") ] in
change 2 "Aufwidersen" k
Since it's not used I've added a type annotation to make sure we're getting what we expect here, as otherwise result could be anything and the compiler wouldn't complain. You don't typically need this if you're going to use result however, as you'd then get an error if the type doesn't unify with its usage.
I want to be able to repeat a process where a collection that we are iterating over is altered an n number of times. n is only known at runtime, and can be specified by the user, so we cannot hard-code it into the type.
An approach that uses intermediate data structures by collect-ing between iterations is possible, like so:
let n = 10;
let mut vec1 = vec![1, 2, 3];
{
for _index in 0..n {
let temp_vec = vec1.into_iter().flat_map(|x| vec![x, x * 2]).collect();
vec1 = temp_vec;
}
}
However, this seems wasteful, because we are creating intermediate datastructures, so I went on looking for a solution that chains iterators directly.
At first I thought one could just do something like:
let mut iter = vec![1, 2, 3].into_iter();
for index in 0..n {
iter = iter.flat_map(|x| vec![x, x * 2].into_iter());
}
However, this does not work because in Rust, all functions on iterators return their own kind of 'compound iterator' struct. (In for instance Haskell, functions on iterators return the appropriate kind of result iterator, which does not become a 'bigger and bigger compound type'.)
Rewriting this as a recursive function had similar problems because (a) I was returning 'some kind of Iterator' whose type was (near?)-impossible to write out by hand because of the recursion, and (b) this type was different in the base case from the recursive case.
I found this question about conditionally returning either one or the other iterator type, as well as using impl Iterator to indicate that we return some concrete type that implements the Iterator trait, but we do not care about its exact nature.
A similar example to the code in the linked answer has been implemented in the code below as maybe_flatmap. This works.
However, I don't want to run flat_map zero or one time, but rather N times on the incoming iterator. Therefore, I adapted the code to call itself recursively up to a depth of N.
Attempting to do that, then makes the Rust compiler complain with an error[E0720]: opaque type expands to a recursive type:
use either::Either; // 1.5.3
/// Later we want to work with any appropriate items,
/// but for simplicity's sake, just use plain integers for now.
type I = u64;
/// Works, but limited to single level.
fn maybe_flatmap<T: Iterator<Item = I>>(iter: T, flag: bool) -> impl Iterator<Item = I> {
match flag {
false => Either::Left(iter),
true => Either::Right(iter.flat_map(move |x| vec![x, x * 2].into_iter())),
}
}
/// Does not work: opaque type expands to a recursive type!
fn rec_flatmap<T: Iterator<Item = I>>(iter: T, depth: usize) -> impl Iterator<Item = I> {
match depth {
0 => Either::Left(iter),
_ => {
let iter2 = iter.flat_map(move |x| vec![x, x * 2]).into_iter();
Either::Right(rec_flatmap(iter2, depth - 1))
}
}
}
fn main() {
let xs = vec![1, 2, 3, 4];
let xs2 = xs.into_iter();
let xs3 = maybe_flatmap(xs2, true);
let xs4: Vec<_> = xs3.collect();
println!("{:?}", xs4);
let ys = vec![1, 2, 3, 4];
let ys2 = ys.into_iter();
let ys3 = rec_flatmap(ys2, 5);
let ys4: Vec<_> = ys3.collect();
println!("{:?}", ys4);
}
Rust playground
error[E0720]: opaque type expands to a recursive type
--> src/main.rs:16:65
|
16 | fn rec_flatmap<T: Iterator<Item = I>>(iter: T, depth: usize) -> impl Iterator<Item = I> {
| ^^^^^^^^^^^^^^^^^^^^^^^ expands to a recursive type
|
= note: expanded type is `either::Either<T, impl std::iter::Iterator>`
I am stuck.
Since regardless of how often you flat_map, the final answer is going to be an (iterator over) a vector of integers, it seems like there ought to be a way of writing this function using only a single concrete return type.
Is this possible? Is there a way out of this situation without resorting to runtime polymorphism?
I believe/hope that a solution without dynamic polymorphism (trait objects or the like) is possible because regardless of how often you call flat_map the end result should have (at least morally) have the same type. I hope there is a way to shoehorn the (non-matching) nested FlatMap struct in a matching single static type somehow.
Is there a way to resolve this without runtime polymorphism?
No.
To solve it using a trait object:
let mut iter: Box<dyn Iterator<Item = i32>> = Box::new(vec![1, 2, 3].into_iter());
for _ in 0..n {
iter = Box::new(iter.flat_map(|x| vec![x, x * 2].into_iter()));
}
regardless of how often you call flat_map the end result should have (at least morally) have the same type
I don't know which morality to apply to type systems, but the literal size in memory is (very likely to be) different for FlatMap<...> and FlatMap<FlatMap<...>>. They are different types.
See also:
Conditionally iterate over one of several possible iterators
Creating Diesel.rs queries with a dynamic number of .and()'s
How do I iterate over a Vec of functions returning Futures in Rust?
How can I extend the lifetime of a temporary variable inside of an iterator adaptor in Rust?
Why does Iterator::take_while take ownership of the iterator?
I was looking at the examples of writer monad to understand how it works and almost all of those looks like a list writer monad. I know a list writer monad is a type of writer monad. But what really is a writer monad in lay-mans terms.
In lay terms, the writer monad is the monad that lets you "write" items to a "log" while you produce a value. When you're done, you end up with the value you produced and the log that contains all the stuff you wrote. To put it another way, it is the monad whose side effects are "writing things to a log".
Let's make this more concrete with examples of both the list writer and the (generic) writer monads. I'll use Haskell here, since it is the original context in which Monads for Functional Programming were described.
The List Writer Monad
I assume that the "list writer" monad is one that logs an item (of some type we'll call w) into a list of items (of type [w], of course). It also produces a value of type a. (See the note at the bottom if you get errors using this code yourself.)
newtype ListWriter w a = ListWriter { runListWriter :: ([w], a) }
instance Monad (ListWriter w) where
return a = ListWriter ([], a) -- produce an a, don't log anything
ListWriter (ws, a) >>= k =
let ListWriter (xs, a') = k a -- run the action 'k' on the existing value,
in ListWriter (ws ++ xs, a') -- add anything it logs to the existing log,
-- and produce a new result value
-- Add an item to the log and produce a boring value.
-- When sequenced with >>, this will add the item to existing log.
tell :: w -> ListWriter w ()
tell w = ListWriter ([w], ())
ex1 :: ListWriter String Int
ex1 = do
tell "foo"
tell "bar"
return 0
(NB: This is equivalent to ex1 = tell "foo" >> tell "bar" >> return 0, demonstrating the use of tell with >> to add an item to the log.)
If we evaluate runListWriter ex1 in GHCi, we see that it wrote "foo" and "bar" to the log and produced the result value 0.
λ> runListWriter ex1
(["foo","bar"],0)
The (Generic) Writer Monad
Now, let's see how we turn this into the generic writer monad. The writer monad works with any sort of thing that can be combined together, not just a list. Specifically, it works with any Monoid:
class Monoid m where
mempty :: m -- an empty m
mappend :: m -> m -> m -- combine two m's into a single m
Lists are a Monoid with [] and (++) as mempty and mappend respectively. A non-list example of a Monoid is sums of integers:
λ> Sum 1 <> Sum 2 -- (<>) = mappend
Sum {getSum = 3}
The writer monad is then
newtype Writer w m = Writer { runWriter :: (w, m) }
Instead of a list of w's, we just have a single w. But when we define the Monad, we ensure that w is a Monoid so we can start with an empty log and append a new entry to the log:
instance Monoid w => Monad (Writer w) where
return a = Writer (mempty, a) -- produce an a, don't log anything
Writer (w, a) >>= k =
let Writer (x, a') = k a -- we combine the two w's rather than
in Writer (w <> x, a') -- (++)-ing two lists
Note the differences here: we use mempty instead of [] and (<>) instead of (++). This is how we generalize from lists to any Monoid.
So the writer monad is really a generalization of the list monad to arbitrary things that can be combined rather than just lists. You can use lists with the Writer to get something (almost) equivalent to ListWriter. The only difference is that you have to wrap your logged item in a list when you append it to the log:
ex2 :: Writer [String] Int
ex2 = do
tell ["foo"]
tell ["bar"]
return 0
but you get the same result:
λ> runWriter ex2
(["foo","bar"],0)
This is because instead of logging "an item that will be put in a list", you are logging "a list". (This does mean that you can log multiple items at the same time by passing a list of more than one element.)
For an example of a non-list use of Writer, consider counting the comparisons a sort function makes. Each time your function make a comparison, you can tell (Sum 1). (You can tell someone. Get it? Is this thing on?) Then, at the end, you'll get back the total count (i.e., the sum) of all of the comparisons along with the sorted list.
NOTE: If you try to use these ListWriter and Writer definitions yourself, GHC will tell you that you are missing Functor and Applicative instances. Once you have the Monad instance, you can write the others in its terms:
import Control.Monad (ap, liftM)
instance Functor (ListWriter w) where
fmap = liftM
instance Applicative (ListWriter w) where
pure = return
(<*>) = ap
And likewise for Writer. I elided them above for clarity.
i'm studying the programming language Standard ML and i am wondering how i can iterate a list with a check condition.
In other languages we have for loops like :
var input;
for(var i = 0; i < arr.length; i++) {
if(arr[i] == input) {
//you have arrived at your condition...
} else {
//other case
}
}
f.ex
i want to iterate through a list and check if the input variable matches a existing element in the list.
i = 5
xs = [1,5,2,3,6] --> the element matches after one iteration.
fun check i nil = []
| check i (x::xs) = if i=x
then //dowork
else //iterate;
I've gone through many documentations on how to implement this without success.
It would be really helpful if someone could give me some explaining regarding how i can use let val A in B end; inside or outside of if conditions for this kind of work.
how i can iterate a list with a check condition
fun check i nil = []
| check i (x::xs) = if i=x
then //dowork
else //iterate;
i want to iterate through a list and check if the input variable matches a existing element in the list.
I would call this a predicate combinator. It already exists in the standard library and is called List.exists. But you can also make it yourself:
fun exists p [] = false
| exists p (x::xs) = p x orelse exists p xs
This is a simplification of the if-then-else you're attempting, which would look like:
fun exists p [] = false
| exists p (x::xs) = if p x then true else exists p xs
If-then-else isn't really necessary when the result type is a boolean, since orelse, andalso and not are short-circuiting (will not evaluate their second operand if the result can be determined with the first).
Using this List.exists function to check if a list contains a specific element, you have to construct a p that compares the list element with some given value, e.g.:
fun check y xs = List.exists (fn x => ...) xs
This may seem a bit more complicated than simply writing check recursively from scratch,
fun check y [] = false
| check y (x::xs) = ... orelse check y xs
but a solution using higher-order functions is preferred for several reasons.
One is that a seasoned reader will quickly detect what you're doing when seeing List.exists: Ah, you're scanning a list for an element given a predicate. Whereas if your function is explicitly recursive, the reader will have to read the entire recursion scheme: OK, the function doesn't do anything funky, which I'd have known if I'd seen e.g. List.exists.