Francesco Abbate <segfault@email.it> writes:

> Dear Ocamlers,
>
> ok, I confess that I'm a little bit paranoid and I often
> look to the assembler code generated by Ocaml to get an
> idea of real efficience of the compiler.
>
> While, generally speaking, the ASM code generated by ocaml
> is pretty good, I wonder why the following function
> is not decently assembled by ocaml :
> -----------------------------------------
> let rec conv n =
>   let r = n mod 10
>   and n' = n / 10 in
>   if n' = 0 then r
>   else r + 8 * (conv n')
> -----------------------------------------
> nor the C version is decently assembled by GCC
> -----------------------------------------
> int
> conv (int n)
> {
>   int m = n / 10, r = n % 10;
>   if (m > 0)
>     return r + 8 * conv (m);
>   return m;
> }
> -----------------------------------------

> So my answer is why nor Ocaml nor GCC does generate efficient
> assembler code ?
>
> I will attempt to give a tentative answer
> - for some reason the compiler does not understands the (n mod 10)
>   and (n /10) both can be avaluated with a simgle "idiv"
>   instruction

This require some analysis that isn't needed in general

> - for some reason the compilers does not conceive to have a loop
>   which push something on the stack at each cycle.

Ocaml (and I believe GCC) only optimize code witch is tail recursive,
that is the result of the function is the result of the recursive
case. 

You should transform your code into a tail rec function by hand :
let conv n =
   let rec helper n mult accu =
      if n = 0 then accu
      else
         let r = n mod 10
         and n' = n / 10 in
         helper n' (mult * 8) (accu + r * mult) in
   helper n 1 0

As you can see, the result of the recursive function helper is the
result of the recursive call.

-- 
Rémi Vanicat

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