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Passing a function in recursion
f#compiler

I'm trying to make an expanded version of List.map3, that takes 5 inputs. In the following code I've followed the template of List.map3. I cannot access Helpers. which seems to look like a list but act like it throws an exception. So I stubbed that out.

More seriously, I cannot access OptimizedClosures.FastFunc5<_,_,_,_,_,_>.Adapt(f) which seems to allow the function f to be passed correctly in the recursion. Commenting it out causes the code not to work.

let rec map5a (f:OptimizedClosures.FastFunc5<_,_,_,_,_,_>) l1 l2 l3 l4 l5 acc =
match l1,l2,l3,l4,l5 with
| [],[],[],[],[] -> List.rev acc
| (h1::t1),(h2::t2),(h3::t3),(h4::t4),(h5::t5) -> let x = f.Invoke(h1,h2,h3,h4,h5) in map5a f t1 t2 t3 t4 t5 (x :: acc)
| _ -> [printfn "the lists had different lengths"] // Helpers.invalid_arg2 "l5" "the lists had different lengths"

let map5 f l1 l2 l3 l4 l5 =
//let f = OptimizedClosures.FastFunc5<_,_,_,_,_,_>.Adapt(f)
map5a f l1 l2 l3 l4 l5 []

I tested it with the following code:

let list1 = [0;1;2;3;4;5]
let list2 = [0;1;2;3;4;5]
let list3 = [0;1;2;3;4;5]
let list4 = [0;1;2;3;4;5]
let list5 = [0;1;2;3;4;5]
let output = map5 (fun x1 x2 x3 x4 x5 -> x1+x2+x3+x4+x5) list1 list2 list3 list4 list5

and the compiler complains that the closure (fun ...) is not where a function is expected or has too many arguments. I think it's not expecting a function.

.

The docs on OptimizedClosures say to avoid using it, so why use it?

[link:research.microsoft.com]

This works fine:

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let map5 f l1 l2 l3 l4 l5 = 
    let rec map5 f l1 l2 l3 l4 l5 acc = 
        match l1,l2,l3,l4,l5 with
        | [],[],[],[],[] -> List.rev acc
        | (h1::t1),(h2::t2),(h3::t3),(h4::t4),(h5::t5) -> let x = f h1 h2 h3 h4 h5 in map5 f t1 t2 t3 t4 t5 (x :: acc)
        | _ -> failwith "the lists had different lengths"
    map5 f l1 l2 l3 l4 l5 []

let list1 = [0;1;2;3;4;5]
let list2 = [0;1;2;3;4;5]
let list3 = [0;1;2;3;4;5]
let list4 = [0;1;2;3;4;5]
let list5 = [0;1;2;3;4;5]
let output = map5 (fun x1 x2 x3 x4 x5 -> x1+x2+x3+x4+x5) list1 list2 list3 list4 list5
printfn "%A" output
By on 3/26/2009 9:49 AM ()

Because I thought I had to use it or there was some special trick to this. Thanks for the correction; the function really can be passed just like any argument.

I am going to start trying to write my functions in a more general way like this.

I didn't know the inner recursive function could have the same name as the outer non-recursive function without confusion.

By on 3/26/2009 10:46 AM ()

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