recursion - 带尾递归的慢字节码

Question

灵感来自 this SO question 的回答我使用代码来检查针对尾递归的命令式循环:

let rec nothingfunc i =
  match i with
  | 1000000000 -> 1
  | _ -> nothingfunc (i+1)

let nothingloop1 () =
  let i = ref 0 in
   while !i < 1000000000 do incr i done;
   1

let timeit f v =
  let t1 = Unix.gettimeofday() in
  let _ = f v in
  let t2 =  Unix.gettimeofday() in
    t2 -. t1

let () =
  Printf.printf "recursive function: %g s\n%!" (timeit nothingfunc 0);
  Printf.printf "while loop with ref counter buitin incr: %g s\n%!" (timeit nothingloop1 ());

对于字节码和 native 代码，结果是

str@s131-intel:~> ./bench_loop
recursive function: 20.7656 s
while loop with ref counter buitin incr: 12.0642 s
str@s131-intel:~> ./bench_loop.opt 
recursive function: 0.755594 s
while loop with ref counter buitin incr: 0.753947 s

问题是:20 到 12 秒的执行时间相差很大的原因是什么？

编辑，我的结论:

一个函数调用 apply (在字节码中)涉及堆栈大小检查、可能的堆栈扩大和信号检查。为了获得最佳性能， native 代码编译器将提供。

(旁注:在这里询问 SO 因为它对搜索引擎友好。)

Answer 1

查看 ocamlfind ocamlc -package unix test.ml -dlambda 的输出

(nothingloop1/1010 =
     (function param/1022
       (let (i/1011 =v 0)
         (seq (while (< i/1011 100000000) (assign i/1011 (1+ i/1011))) 1)))

(nothingfunc/1008
   (function i/1009
     (if (!= i/1009 100000000) (apply nothingfunc/1008 (+ i/1009 1)) 1)))

显然 assign比 apply 快.在函数调用时似乎会检查堆栈溢出和信号，但不会检查简单的分配。详情请看: https://github.com/ocaml/ocaml/blob/trunk/byterun/interp.c