]> 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 ------------------- To unsubscribe, mail caml-list-request@inria.fr Archives: http://caml.inria.fr Bug reports: http://caml.inria.fr/bin/caml-bugs FAQ: http://caml.inria.fr/FAQ/ Beginner's list: http://groups.yahoo.com/group/ocaml_beginners