[erlang-questions] anyway to make recursion inside anonymous funs?
Robert Virding
<>
Thu Mar 29 23:53:33 CEST 2007
Well, that is all well and good but if we really want to do it properly
(:-)) we should of course use the Y-combinator:
Y = fun (F) ->
G = fun (X) -> F(fun(Y) -> (X(X))(Y) end) end,
G(G)
end.
Yes I know it is ugly with the G but I couldn't be bothered to write it
out twice.
Then you get:
R = fun (F) ->
fun (X) ->
receive {P,V} -> P!(X+V), F(X) end
end
end.
and the call:
(Y(R))(X).
Much nicer!
Unfortunately you need a separate Y-combinator for each arity your
recursing function has. Unless of course you always put them in a tuple
and so always have one argument.
Robert
Håkan Huss wrote:
> On 3/29/07, ok <> wrote:
>> On 29 Mar 2007, at 3:45 am, June Kim wrote:
>>
>>> "fun" expressions can be used to declare a function.
>>> Can [a fun] make a recursive call inside?
>> Let's take
>> factorial 0 = 1
>> factorial n = n * factorial (n-1)
>> as an example.
>>
>>
>> G = fun (_, 0) -> ; (G, N) -> N * G(G, N-1) end,
>> F = fun (N) -> G(G, N) end
>>
>> We *do* have to give G a name (because F uses G twice);
>> we *don't* have to give F a name.
>>
> Well, obviously we don't *have to* give G a name, or lambda-calculus
> would have severe problems. Not doing so, however, requires you to
> write out the fun previously known as G twice:
> F = fun (N) ->
> fun (_, 0) -> 1;
> (G, N) -> N * G(G, N-1)
> end
> (fun (_, 0) -> 1;
> (G, N) -> N * G(G, N-1)
> end,
> N)
> end.
>
>>> Can I somehow declare the following with callself, which is an
>>> imaginary functionality that calls the anonymous function itself?
>>>
>>> fun(X) -> receive {P,V} -> P!(X+V), callself(X+V) end end.
>> This example becomes
>> ( G = fun (G, X) -> receive {P,V} -> P!(X+V), G(G, X+V) end end,
>> fun (X) -> G(G, X) end
>> )
>>
> Without G:
> fun (X) ->
> fun (F) ->
> receive {P,V} -> P!(X+V), F(F) end
> end (fun (F) ->
> receive {P,V} -> P!(X+V), F(F) end
> end)
> end.
>
> /Håkan
>
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