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I am trying to properly understand how side effects work when traversing a list in F# using monadic style, following Scott's guide here
I have an AsyncSeq of items, and a side-effecting function that can return a Result<'a,'b> (it is saving the items to disk).
I get the general idea - split the head and tail, apply the func to the head. If it returns Ok then recurse through the tail, doing the same thing. If an Error is returned at any point then short circuit and return it.
I also get why Scott's ultimate solution uses foldBack rather than fold - it keeps the output list in the same order as the input as each processed item is prepended to the previous.
I can also follow the logic:
The result from the list's last item (processed first as we are using foldback) will be passed as the accumulator to the next item.
If it is an Error and the next item is Ok, the next item is discarded.
If the next item is an Error, it replaces any previous results and becomes the accumulator.
That means by the time you have recursed over the entire list from right to left and ended up at the start, you either have an Ok of all of the results in the correct order or the most recent Error (which would have been the first to occur if we had gone left to right).
The thing that confuses me is that surely, since we are starting at the end of the list, all side effects of processing every item will take place, even if we only get back the last Error that was created?
This seems to be confirmed here as the print output starts with [5], then [4,5], then [3,4,5] etc.
The thing that confuses me is that this isn't what I see happening when I use AsyncSeq.traverseChoiceAsync from the FSharpx lib (which I wrapped to process Result instead of Choice). I see side effects happening from left to right, stopping on the first error, which is what I want to happen.
It also looks like Scott's non-tail recursive version (which doesn't use foldBack and just recurses over the list) goes from left to right? The same goes for the AsyncSeq version. That would explain why I see it short circuit on the first error but surely if it completes Ok then the output items would be reversed, which is why we normally use foldback?
I feel I am misunderstanding or misreading something obvious! Could someone please explain it to me? :)
Edit:
rmunn has given a really great comprehensive explanation of the AsyncSeq traversal below. The TLDR was that
Scott's initial implementation and the AsyncSeq traverse both do go from left to right as I thought and so only process until they hit an error
they keep their contents in order by prepending the head to the processed tail rather than prepending each processed result to the previous (which is what the built in F# fold does).
foldback would keep things in order but would indeed execute every case (which could take forever with an async seq)
It's pretty simple: traverseChoiceAsync isn't using foldBack. Yes, with foldBack the last item would be processed first, so that by the time you get to the first item and discover that its result is Error you'd have triggered the side effects of every item. Which is, I think, precisely why whoever wrote traverseChoiceAsync in FSharpx chose not to use foldBack, because they wanted to ensure that side effects would be triggered in order, and stop at the first Error (or, in the case of the Choice version of the function, the first Choice2Of2 — but I'll pretend from this point on that that function was written to use the Result type.)
Let's look at the traverseChoieAsync function in the code you linked to, and read through it step-by-step. I'll also rewrite it to use Result instead of Choice, because the two types are basically identical in function but with different names in the DU, and it'll be a little easier to tell what's going on if the DU cases are called Ok and Error instead of Choice1Of2 and Choice2Of2. Here's the original code:
let rec traverseChoiceAsync (f:'a -> Async<Choice<'b, 'e>>) (s:AsyncSeq<'a>) : Async<Choice<AsyncSeq<'b>, 'e>> = async {
let! s = s
match s with
| Nil -> return Choice1Of2 (Nil |> async.Return)
| Cons(a,tl) ->
let! b = f a
match b with
| Choice1Of2 b ->
return! traverseChoiceAsync f tl |> Async.map (Choice.mapl (fun tl -> Cons(b, tl) |> async.Return))
| Choice2Of2 e ->
return Choice2Of2 e }
And here's the original code rewritten to use Result. Note that it's a simple rename, and none of the logic needs to be changed:
let rec traverseResultAsync (f:'a -> Async<Result<'b, 'e>>) (s:AsyncSeq<'a>) : Async<Result<AsyncSeq<'b>, 'e>> = async {
let! s = s
match s with
| Nil -> return Ok (Nil |> async.Return)
| Cons(a,tl) ->
let! b = f a
match b with
| Ok b ->
return! traverseChoiceAsync f tl |> Async.map (Result.map (fun tl -> Cons(b, tl) |> async.Return))
| Error e ->
return Error e }
Now let's step through it. The whole function is wrapped inside an async { } block, so let! inside this function means "unwrap" in an async context (essentially, "await").
let! s = s
This takes the s parameter (of type AsyncSeq<'a>) and unwraps it, binding the result to a local name s that henceforth will shadow the original parameter. When you await the result of an AsyncSeq, what you get is the first element only, while the rest is still wrapped in an async that needs to be further awaited. You can see this by looking at the result of the match expression, or by looking at the definition of the AsyncSeq type:
type AsyncSeq<'T> = Async<AsyncSeqInner<'T>>
and AsyncSeqInner<'T> =
| Nil
| Cons of 'T * AsyncSeq<'T>
So when you do let! x = s when s is of type AsyncSeq<'T>, the value of x will either be Nil (when the sequence has run to its end) or it will be Cons(head, tail) where head is of type 'T and tail is of type AsyncSeq<'T>.
So after this let! s = s line, our local name s now refers to an AsyncSeqInner type, which contains the head item of the sequence (or Nil if the sequence was empty), and the rest of the sequence is still wrapped in an AsyncSeq so it has yet to be evaluated (and, crucially, its side effects have not yet happened).
match s with
| Nil -> return Ok (Nil |> async.Return)
There's a lot happening in this line, so it'll take a bit of unpacking, but the gist is that if the input sequence s had Nil as its head, i.e. had reached its end, then that's not an error, and we return an empty sequence.
Now to unpack. The outer return is in an async keyword, so it takes the Result (whose value is Ok something) and turns it into an Async<Result<something>>. Remembering that the return type of the function is declared as Async<Result<AsyncSeq>>, the inner something is clearly an AsyncSeq type. So what's going on with that Nil |> async.Return? Well, async isn't an F# keyword, it's the name of an instance of AsyncBuilder. Inside a computation expression foo { ... }, return x is translated into foo.Return(x). So calling async.Return x is just the same as writing async { return x }, except that it avoids nesting a computation expression inside another computation expression, which would be a little nasty to try and parse mentally (and I'm not 100% sure the F# compiler allows it syntactically). So Nil |> async.Return is async.Return Nil which means it produces a value of Async<x> where x is the type of the value Nil. And as we just saw, this Nil is a value of type AsyncSeqInner, so Nil |> async.Return produces an Async<AsyncSeqInner>. And another name for Async<AsyncSeqInner> is AsyncSeq. So this whole expression produces an Async<Result<AsyncSeq>> that has the meaning of "We're done here, there are no more items in the sequence, and there was no error".
Phew. Now for the next line:
| Cons(a,tl) ->
Simple: if the next item in the AsyncSeq named s was a Cons, we deconstruct it so that the actual item is now called a, and the tail (another AsyncSeq) is called tl.
let! b = f a
This calls f on the value we just got out of s, and then unwraps the Async part of f's return value, so that b is now a Result<'b, 'e>.
match b with
| Ok b ->
More shadowed names. Inside this branch of the match, b now names a value of type 'b rather than a Result<'b, 'e>.
return! traverseResultAsync f tl |> Async.map (Result.map (fun tl -> Cons(b, tl) |> async.Return))
Hoo boy. That's too much to tackle at once. Let's write this as if the |> operators were lined up on separate lines, and then we'll go through each step one at a time. (Note that I've wrapped an extra pair of parentheses around this, just to clarify that it's the final result of this whole expression that will be passed to the return! keyword).
return! (
traverseResultAsync f tl
|> Async.map (
Result.map (
fun tl -> Cons(b, tl) |> async.Return)))
I'm going to tackle this expression from the inside out. The inner line is:
fun tl -> Cons(b, tl) |> async.Return
The async.Return thing we've already seen. This is a function that takes a tail (we don't currently know, or care, what's inside that tail, except that by the necessity of the type signature of Cons it must be an AsyncSeq) and turns it into an AsyncSeq that is b followed by the tail. I.e., this is like b :: tl in a list: it sticks b onto the front of the AsyncSeq.
One step out from that innermost expression is:
Result.map
Remember that the function map can be thought of in two ways: one is "take a function and run it against whatever is "inside" this wrapper". The other is "take a function that operates on 'T and make it into a function that operates on Wrapper<'T>". (If you don't have both of those clear in your mind yet, https://sidburn.github.io/blog/2016/03/27/understanding-map is a pretty good article to help grok that concept). So what this is doing is taking a function of type AsyncSeq -> AsyncSeq and turning it into a function of type Result<AsyncSeq> -> Result<AsyncSeq>. Alternately, you could think of it as taking a Result<tail> and calling fun tail -> ... against that tail result, then re-wrapping the result of that function in a new Result. Important: Because this is using Result.map (Choice.mapl in the original) we know that if tail is an Error value (or if the Choice was a Choice2Of2 in the original), the function will not be called. So if traverseResultAsync produces a result that starts with an Error value, it's going to produce an <Async<Result<foo>>> where the value of Result<foo> is an Error, and so the value of the tail will be discarded. Keep that in mind for later.
Okay, next step out.
Async.map
Here, we have a Result<AsyncSeq> -> Result<AsyncSeq> function produced by the inner expression, and this converts it to an Async<Result<AsyncSeq>> -> Async<Result<AsyncSeq>> function. We've just talked about this, so we don't need to go over how map works again. Just remember that the effect of this Async<Result<AsyncSeq>> -> Async<Result<AsyncSeq>> function that we've built up will be the following:
Await the outer async.
If the result is Error, return that Error.
If the result is Ok tail, produce an Ok (Cons (b, tail)).
Next line:
traverseResultAsync f tl
I probably should have started with this, because this will actually run first, and then its value will be passed into the Async<Result<AsyncSeq>> -> Async<Result<AsyncSeq>> function that we've just analysed.
So what this whole thing will do is to say "Okay, we took the first part of the AsyncSeq we were handed, and passed it to f, and f produced an Ok result with a value we're calling b. So now we need to process the rest of the sequence similarly, and then, if the rest of the sequence produces an Ok result, we'll stick b on the front of it and return an Ok sequence with contents b :: tail. BUT if the rest of the sequence produces an Error, we'll throw away the value of b and just return that Error unchanged."
return!
This just takes the result we just got (either an Error or an Ok (b :: tail), already wrapped in an Async) and returns it unchanged. But note that the call to traverseResultAsync is NOT tail-recursive, because its value had to be passed into the Async.map (...) expression first.
And now we still have one more bit of traverseResultAsync to look at. Remember when I said "Keep that in mind for later"? Well, that time has arrived.
| Error e ->
return Error e }
Here we're back in the match b with expression. If b was an Error result, then no further recursive calls are made, and the whole traverseResultAsync returns an Async<Result> where the Result value is Error. And if we were currently nested deep inside a recursion (i.e., we're in the return! traverseResultAsync ... expression), then our return value will be Error, which means the result of the "outer" call, as we've kept in mind, will also be Error, discarding any other Ok results that might have happened "before".
Conclusion
And so the effect of all of that is:
Step through the AsyncSeq, calling f on each item in turn.
The first time f returns Error, stop stepping through, throw away any previous Ok results, and return that Error as the result of the whole thing.
If f never returns Error and instead returns Ok b every time, return an Ok result that contains an AsyncSeq of all those b values, in their original order.
Why are they in their original order? Because the logic in the Ok case is:
If sequence was empty, return an empty sequence.
Split into head and tail.
Get value b from f head.
Process the tail.
Stick value b in front of the result of processing the tail.
So if we started with (conceptually) [a1; a2; a3], which actually looks like Cons (a1, Cons (a2, Cons (a3, Nil))) we'll end up with Cons (b1, Cons (b2, Cons (b3, Nil))) which translates to the conceptual sequence [b1; b2; b3].
See #rmunn's great answer above for the explanation. I just wanted to post a little helper for anyone that reads this in the future, it allows you to use the AsyncSeq traverse with Results instead of the old Choice type it was written with:
let traverseResultAsyncM (mapping : 'a -> Async<Result<'b,'c>>) source =
let mapping' =
mapping
>> Async.map (function
| Ok x -> Choice1Of2 x
| Error e -> Choice2Of2 e)
AsyncSeq.traverseChoiceAsync mapping' source
|> Async.map (function
| Choice1Of2 x -> Ok x
| Choice2Of2 e -> Error e)
Also here is a version for non-async mappings:
let traverseResultM (mapping : 'a -> Result<'b,'c>) source =
let mapping' x = async {
return
mapping x
|> function
| Ok x -> Choice1Of2 x
| Error e -> Choice2Of2 e
}
AsyncSeq.traverseChoiceAsync mapping' source
|> Async.map (function
| Choice1Of2 x -> Ok x
| Choice2Of2 e -> Error e)
I am having some trouble in how to define no operation in scheme language
like ; in c
i want to do something like this:
(cond
[(null? exp) nop ]
.....
)
if i leave it empty it wiil return #t
thanks!
Note that functional programs are different to imperative programs. You should always think in every expression/function to return something (the result or value of that expression). With conditionals, you have to be careful to maintain this "something" through all the different branches, as your expression has to yield a value in any case.
Then, you have to decide what you want to return in that case, and build the code accordingly. If you don't want to return #t, you can return just #f or the empty list:
(cond
[(null? exp) #f]
.....
)
In fact, if you think carefuly, the concept of a "no op" (that is, do nothing), in C, is almost the same than "producing some value", because you're doing nothing apart from producing the value, that doesn't induce any change at all in your program.
Scheme has no statements, only expressions. Each expression returns a value -or may never return-
So you want an expression which does not compute much. You could use nil (or #f, or any other value) for that purpose:
(cond
((null? exp) ())
....
)
If you wrote a condition with only a test - and no "then" body sub-expressions
(cond
((null? exp))
)
then the result of the cond when exp is nil is the result of the (null? exp) test which is #t
actually, when exp is null, you could just return exp itself.
You should probably change your program to eliminate useless condition. But if you just need a nil value, the standard answer is to say (if #f #f), which is a one-legged if that is always false but has no false expression, hence returns nothing.
I'm getting myself introduced to Erlang by Armstrongs "Programming Erlang". One Exercise is to write a reeimplementation of the tuple_to_list/1 BIF. My solution seems rather inelegant to me, especially because of the helper function I use. Is there a more Erlang-ish way of doing this?
tup2lis({}) -> [];
tup2lis(T) -> tup2list_help(T,1,tuple_size(T)).
tup2list_help(T,Size,Size) -> [element(Size,T)];
tup2list_help(T,Pos,Size) -> [element(Pos,T)|tup2list_help(T,Pos+1,Size)].
Thank you very much for your ideas. :)
I think your function is ok, and more if your goal is to learn the language.
As a matter of style, usually the base case when constructing lists is just the empty list [].
So I'd write
tup2list(Tuple) -> tup2list(Tuple, 1, tuple_size(Tuple)).
tup2list(Tuple, Pos, Size) when Pos =< Size ->
[element(Pos,Tuple) | tup2list(Tuple, Pos+1, Size)];
tup2list(_Tuple,_Pos,_Size) -> [].
you can write pretty much the same with list comprehension
[element(I,Tuple) || I <- lists:seq(1,tuple_size(Tuple))].
it will work as expected when the tuple has no elements, as lists:seq(1,0) gives an empty list.
Your code is good and also idiomatic way how to make this sort of stuff. You can also build this list backward which in this case will be a little bit faster because of tail call but not significant.
tup2list(T) -> tup2list(T, size(T), []).
tup2list(T, 0, Acc) -> Acc;
tup2list(T, N, Acc) -> tup2list(T, N-1, [element(N,T)|Acc]).
I am attempting exercises from Joe Armstrong book, Here is what I came up with
my_tuple_to_list(Tuple) -> [element(T, Tuple) || T <- lists:seq(1, tuple_size(Tuple))].
In Erlang R16B you can also use erlang:delete_element/2 function like this:
tuple2list({}) -> [];
tuple2list(T) when is_tuple(T) ->
[element(1, T) | tuple2list(erlang:delete_element(1, T))].
Strangely enough I'm learning this right now using the same book by Joe Armstrong the second edition and Chapter 4 is about modules and functions which covers List Comprehension, Guards, Accumulators etc. Anyways using this knowledge my solution is the code below :
-module(my_tuple_to_list).
-export([convert/1]).
convert(T) when is_tuple(T) -> [element(Pos,T) || Pos <- lists:seq(1,tuple_size(T))].
Most of the answers were already given in the same way mine just adds the Guard to ensure that a Tuple was given.
Erlang 17.0, you should build list in natural order, solutions above is incorrect from the point of efficiency. Always add elements to the head of an existing list:
%% ====================================================================
%% API functions
%% ====================================================================
my_tuple_to_list({}) ->
[];
my_tuple_to_list(Tuple) ->
tuple_to_list_iter(1, size(Tuple), Tuple, [])
.
%% ====================================================================
%% Internal functions
%% ====================================================================
tuple_to_list_iter(N, N, Tuple, List) ->
lists:reverse([element(N, Tuple)|List]);
tuple_to_list_iter(N, Tuplesize, Tuple, List) ->
L = [element(N, Tuple)|List],
tuple_to_list_iter(N + 1, Tuplesize, Tuple, L)
.
mytuple_to_list(T) when tuple_size(T) =:= 0 -> [];
mytuple_to_list(T) -> [element(1, T)|mytuple_to_list(erlang:delete_element(1, T))].
I've been looking around and can't find examples of this and all of my syntax wrestling skills are failing me. Can someone please tell me how to make this compile?? My ,s ;s or .s are just wrong I guess for defining a nested function...
I'm aware there is a function for doing string replaces already so I don't need to implement this, but I'm playing with Erlang trying to pick it up so I'm hand spinning some of the basics I need to use..
replace(Whole,Old,New) ->
OldLen = length(Old),
ReplaceInit = fun(Next, NewWhole) ->
if
lists:prefix(Old, [Next|NewWhole]) -> {_,Rest} = lists:split(OldLen-1, NewWhole), New ++ Rest;
true -> [Next|NewWhole]
end,
lists:foldr(ReplaceInit, [], Whole).
Basically I'm trying to write this haskell (also probably bad but beyond the point):
repl xs ys zs =
foldr replaceInit [] xs
where
ylen = length ys
replaceInit y newxs
| take ylen (y:newxs) == ys = zs ++ drop (ylen-1) newxs
| otherwise = y:newxs
The main problem is that in an if you are only allowed to use guards as tests. Guards are very restricted and, amongst other things, calls to general Erlang functions are not allowed. Irrespective of whether they are part of the OTP release or written by you. The best solution for your function is to use case instead of if. For example:
replace(Whole,Old,New) ->
OldLen = length(Old),
ReplaceInit = fun (Next, NewWhole) ->
case lists:prefix(Old, [Next|NewWhole]) of
true ->
{_,Rest} = lists:split(OldLen-1, NewWhole),
New ++ Rest;
false -> [Next|NewWhole]
end
end,
lists:foldr(ReplaceInit, [], Whole).
Because of this if is not used that often in Erlang. See about if and about guards in the Erlang documentation.
It seems quite a few mainstream languages support function literals these days. They are also called anonymous functions, but I don't care if they have a name. The important thing is that a function literal is an expression which yields a function which hasn't already been defined elsewhere, so for example in C, &printf doesn't count.
EDIT to add: if you have a genuine function literal expression <exp>, you should be able to pass it to a function f(<exp>) or immediately apply it to an argument, ie. <exp>(5).
I'm curious which languages let you write function literals which are recursive. Wikipedia's "anonymous recursion" article doesn't give any programming examples.
Let's use the recursive factorial function as the example.
Here are the ones I know:
JavaScript / ECMAScript can do it with callee:
function(n){if (n<2) {return 1;} else {return n * arguments.callee(n-1);}}
it's easy in languages with letrec, eg Haskell (which calls it let):
let fac x = if x<2 then 1 else fac (x-1) * x in fac
and there are equivalents in Lisp and Scheme. Note that the binding of fac is local to the expression, so the whole expression is in fact an anonymous function.
Are there any others?
Most languages support it through use of the Y combinator. Here's an example in Python (from the cookbook):
# Define Y combinator...come on Gudio, put it in functools!
Y = lambda g: (lambda f: g(lambda arg: f(f)(arg))) (lambda f: g(lambda arg: f(f)(arg)))
# Define anonymous recursive factorial function
fac = Y(lambda f: lambda n: (1 if n<2 else n*f(n-1)))
assert fac(7) == 5040
C#
Reading Wes Dyer's blog, you will see that #Jon Skeet's answer is not totally correct. I am no genius on languages but there is a difference between a recursive anonymous function and the "fib function really just invokes the delegate that the local variable fib references" to quote from the blog.
The actual C# answer would look something like this:
delegate Func<A, R> Recursive<A, R>(Recursive<A, R> r);
static Func<A, R> Y<A, R>(Func<Func<A, R>, Func<A, R>> f)
{
Recursive<A, R> rec = r => a => f(r(r))(a);
return rec(rec);
}
static void Main(string[] args)
{
Func<int,int> fib = Y<int,int>(f => n => n > 1 ? f(n - 1) + f(n - 2) : n);
Func<int, int> fact = Y<int, int>(f => n => n > 1 ? n * f(n - 1) : 1);
Console.WriteLine(fib(6)); // displays 8
Console.WriteLine(fact(6));
Console.ReadLine();
}
You can do it in Perl:
my $factorial = do {
my $fac;
$fac = sub {
my $n = shift;
if ($n < 2) { 1 } else { $n * $fac->($n-1) }
};
};
print $factorial->(4);
The do block isn't strictly necessary; I included it to emphasize that the result is a true anonymous function.
Well, apart from Common Lisp (labels) and Scheme (letrec) which you've already mentioned, JavaScript also allows you to name an anonymous function:
var foo = {"bar": function baz() {return baz() + 1;}};
which can be handier than using callee. (This is different from function in top-level; the latter would cause the name to appear in global scope too, whereas in the former case, the name appears only in the scope of the function itself.)
In Perl 6:
my $f = -> $n { if ($n <= 1) {1} else {$n * &?BLOCK($n - 1)} }
$f(42); # ==> 1405006117752879898543142606244511569936384000000000
F# has "let rec"
You've mixed up some terminology here, function literals don't have to be anonymous.
In javascript the difference depends on whether the function is written as a statement or an expression. There's some discussion about the distinction in the answers to this question.
Lets say you are passing your example to a function:
foo(function(n){if (n<2) {return 1;} else {return n * arguments.callee(n-1);}});
This could also be written:
foo(function fac(n){if (n<2) {return 1;} else {return n * fac(n-1);}});
In both cases it's a function literal. But note that in the second example the name is not added to the surrounding scope - which can be confusing. But this isn't widely used as some javascript implementations don't support this or have a buggy implementation. I've also read that it's slower.
Anonymous recursion is something different again, it's when a function recurses without having a reference to itself, the Y Combinator has already been mentioned. In most languages, it isn't necessary as better methods are available. Here's a link to a javascript implementation.
In C# you need to declare a variable to hold the delegate, and assign null to it to make sure it's definitely assigned, then you can call it from within a lambda expression which you assign to it:
Func<int, int> fac = null;
fac = n => n < 2 ? 1 : n * fac(n-1);
Console.WriteLine(fac(7));
I think I heard rumours that the C# team was considering changing the rules on definite assignment to make the separate declaration/initialization unnecessary, but I wouldn't swear to it.
One important question for each of these languages / runtime environments is whether they support tail calls. In C#, as far as I'm aware the MS compiler doesn't use the tail. IL opcode, but the JIT may optimise it anyway, in certain circumstances. Obviously this can very easily make the difference between a working program and stack overflow. (It would be nice to have more control over this and/or guarantees about when it will occur. Otherwise a program which works on one machine may fail on another in a hard-to-fathom manner.)
Edit: as FryHard pointed out, this is only pseudo-recursion. Simple enough to get the job done, but the Y-combinator is a purer approach. There's one other caveat with the code I posted above: if you change the value of fac, anything which tries to use the old value will start to fail, because the lambda expression has captured the fac variable itself. (Which it has to in order to work properly at all, of course...)
You can do this in Matlab using an anonymous function which uses the dbstack() introspection to get the function literal of itself and then evaluating it. (I admit this is cheating because dbstack should probably be considered extralinguistic, but it is available in all Matlabs.)
f = #(x) ~x || feval(str2func(getfield(dbstack, 'name')), x-1)
This is an anonymous function that counts down from x and then returns 1. It's not very useful because Matlab lacks the ?: operator and disallows if-blocks inside anonymous functions, so it's hard to construct the base case/recursive step form.
You can demonstrate that it is recursive by calling f(-1); it will count down to infinity and eventually throw a max recursion error.
>> f(-1)
??? Maximum recursion limit of 500 reached. Use set(0,'RecursionLimit',N)
to change the limit. Be aware that exceeding your available stack space can
crash MATLAB and/or your computer.
And you can invoke the anonymous function directly, without binding it to any variable, by passing it directly to feval.
>> feval(#(x) ~x || feval(str2func(getfield(dbstack, 'name')), x-1), -1)
??? Maximum recursion limit of 500 reached. Use set(0,'RecursionLimit',N)
to change the limit. Be aware that exceeding your available stack space can
crash MATLAB and/or your computer.
Error in ==> create#(x)~x||feval(str2func(getfield(dbstack,'name')),x-1)
To make something useful out of it, you can create a separate function which implements the recursive step logic, using "if" to protect the recursive case against evaluation.
function out = basecase_or_feval(cond, baseval, fcn, args, accumfcn)
%BASECASE_OR_FEVAL Return base case value, or evaluate next step
if cond
out = baseval;
else
out = feval(accumfcn, feval(fcn, args{:}));
end
Given that, here's factorial.
recursive_factorial = #(x) basecase_or_feval(x < 2,...
1,...
str2func(getfield(dbstack, 'name')),...
{x-1},...
#(z)x*z);
And you can call it without binding.
>> feval( #(x) basecase_or_feval(x < 2, 1, str2func(getfield(dbstack, 'name')), {x-1}, #(z)x*z), 5)
ans =
120
It also seems Mathematica lets you define recursive functions using #0 to denote the function itself, as:
(expression[#0]) &
e.g. a factorial:
fac = Piecewise[{{1, #1 == 0}, {#1 * #0[#1 - 1], True}}] &;
This is in keeping with the notation #i to refer to the ith parameter, and the shell-scripting convention that a script is its own 0th parameter.
I think this may not be exactly what you're looking for, but in Lisp 'labels' can be used to dynamically declare functions that can be called recursively.
(labels ((factorial (x) ;define name and params
; body of function addrec
(if (= x 1)
(return 1)
(+ (factorial (- x 1))))) ;should not close out labels
;call factorial inside labels function
(factorial 5)) ;this would return 15 from labels
Delphi includes the anonymous functions with version 2009.
Example from http://blogs.codegear.com/davidi/2008/07/23/38915/
type
// method reference
TProc = reference to procedure(x: Integer);
procedure Call(const proc: TProc);
begin
proc(42);
end;
Use:
var
proc: TProc;
begin
// anonymous method
proc := procedure(a: Integer)
begin
Writeln(a);
end;
Call(proc);
readln
end.
Because I was curious, I actually tried to come up with a way to do this in MATLAB. It can be done, but it looks a little Rube-Goldberg-esque:
>> fact = #(val,branchFcns) val*branchFcns{(val <= 1)+1}(val-1,branchFcns);
>> returnOne = #(val,branchFcns) 1;
>> branchFcns = {fact returnOne};
>> fact(4,branchFcns)
ans =
24
>> fact(5,branchFcns)
ans =
120
Anonymous functions exist in C++0x with lambda, and they may be recursive, although I'm not sure about anonymously.
auto kek = [](){kek();}
'Tseems you've got the idea of anonymous functions wrong, it's not just about runtime creation, it's also about scope. Consider this Scheme macro:
(define-syntax lambdarec
(syntax-rules ()
((lambdarec (tag . params) . body)
((lambda ()
(define (tag . params) . body)
tag)))))
Such that:
(lambdarec (f n) (if (<= n 0) 1 (* n (f (- n 1)))))
Evaluates to a true anonymous recursive factorial function that can for instance be used like:
(let ;no letrec used
((factorial (lambdarec (f n) (if (<= n 0) 1 (* n (f (- n 1)))))))
(factorial 4)) ; ===> 24
However, the true reason that makes it anonymous is that if I do:
((lambdarec (f n) (if (<= n 0) 1 (* n (f (- n 1))))) 4)
The function is afterwards cleared from memory and has no scope, thus after this:
(f 4)
Will either signal an error, or will be bound to whatever f was bound to before.
In Haskell, an ad hoc way to achieve same would be:
\n -> let fac x = if x<2 then 1 else fac (x-1) * x
in fac n
The difference again being that this function has no scope, if I don't use it, with Haskell being Lazy the effect is the same as an empty line of code, it is truly literal as it has the same effect as the C code:
3;
A literal number. And even if I use it immediately afterwards it will go away. This is what literal functions are about, not creation at runtime per se.
Clojure can do it, as fn takes an optional name specifically for this purpose (the name doesn't escape the definition scope):
> (def fac (fn self [n] (if (< n 2) 1 (* n (self (dec n))))))
#'sandbox17083/fac
> (fac 5)
120
> self
java.lang.RuntimeException: Unable to resolve symbol: self in this context
If it happens to be tail recursion, then recur is a much more efficient method:
> (def fac (fn [n] (loop [count n result 1]
(if (zero? count)
result
(recur (dec count) (* result count))))))