On Sunday 30 August 2009 20:43:17 Yaron Minsky wrote:
> Float refs are not unboxed automatically, because refs 
> Are polymorphic containers. If you create your own pseudo-ref, i.e., a
> record with a single mutable float field, then I believe you should
> get the behaviour you expect.

I believe you are talking at cross purposes. Will wants his accumulator in a 
register. You are referring to float references in data structures being 
boxed because records are boxed.

Look at the compiled forms of the inner loops of these dot products, for 
example:

let dot a b =
  let x = ref 0.0 in
  for i=0 to Array.length a - 1 do
    x := !x +. a.(i) *. b.(i)
  done;
  !x

.L101:
.L103:  movl    caml_young_ptr, %eax
        subl    $12, %eax
        movl    %eax, caml_young_ptr
        cmpl    caml_young_limit, %eax
        jb      .L104
        leal    4(%eax), %eax
        movl    $2301, -4(%eax)
        fldl    -4(%edi, %ecx, 4)
        fmull   -4(%ebx, %ecx, 4)
        faddl   (%esi)
        fstpl   (%eax)
        movl    %eax, %esi
        movl    %ecx, %eax
        addl    $2, %ecx
        cmpl    %edx, %eax
        jne     .L101

let dot2 a b =
  let x = ref 0.0 in
  for i=0 to Array.length a - 1 do
    x := !x +. a.(i) *. b.(i)
  done;
  1.0 *. !x

.L107:
        fldl    -4(%eax, %ecx, 4)
        fmull   -4(%ebx, %ecx, 4)
        faddl   0(%esp)
        fstpl   0(%esp)
        movl    %ecx, %edx
        addl    $2, %ecx
        cmpl    %esi, %edx
        jne     .L107

In the latter case, "x" is unboxed into a register.

> Come to think of it, I wonder if it would be better to implement ref
> on top of a single-cell array, since then everyone would get the float
> unboxing whenever applicable. I imagine there is some runtime overhead
> to this, though.

All-float (including one-float) records are unboxed anyway.

Boxing was discussed in the book OCaml for Scientists and the OCaml Journal 
articles about optimization and the SciMark2 benchmark.

-- 
Dr Jon Harrop, Flying Frog Consultancy Ltd.
http://www.ffconsultancy.com/?e

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